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Different mathematical solving methods can affect how information is memorized

by University of Geneva

The way we memorize information—a mathematical problem statement, for example—reveals the way we process it. A team from the University of Geneva (UNIGE), in collaboration with CY Cergy Paris University (CYU) and Bourgogne University (uB), has shown how different solving methods can alter the way information is memorized and even create false memories.

By identifying learners' unconscious deductions, this study opens up new perspectives for mathematics teaching. These results are published in the Journal of Experimental Psychology: Learning, Memory, and Cognition .

Remembering information goes through several stages: perception, encoding—the way it is processed to become an easily accessible memory trace—and retrieval (or reactivation). At each stage, errors can occur, sometimes leading to the formation of false memories .

Scientists from the UNIGE, CYU and Bourgogne University set out to determine whether solving arithmetic problems could generate such memories and whether they could be influenced by the nature of the problems.

Unconscious deductions create false memories

When solving a mathematical problem , it is possible to call upon either the ordinal property of numbers, i.e., the fact that they are ordered, or their cardinal property, i.e., the fact that they designate specific quantities. This can lead to different solving strategies and, when memorized, to different encoding.

In concrete terms, the representation of a problem involving the calculation of durations or differences in heights (ordinal problem) can sometimes allow unconscious deductions to be made, leading to a more direct solution. This is in contrast to the representation of a problem involving the calculation of weights or prices (cardinal problem), which can lead to additional steps in the reasoning, such as the intermediate calculation of subsets.

The scientists therefore hypothesized that, as a result of spontaneous deductions, participants would unconsciously modify their memories of ordinal problem statements, but not those of cardinal problems.

To test this, a total of 67 adults were asked to solve arithmetic problems of both types, and then to recall the wording in order to test their memories. The scientists found that in the majority of cases (83%), the statements were correctly recalled for cardinal problems.

In contrast, the results were different when the participants had to remember the wording of ordinal problems, such as: "Sophie's journey takes 8 hours. Her journey takes place during the day. When she arrives, the clock reads 11. Fred leaves at the same time as Sophie. Fred's journey is 2 hours shorter than Sophie's. What time does the clock show when Fred arrives?"

In more than half the cases, information deduced by the participants when solving these problems was added unintentionally to the statement. In the case of the problem mentioned above, for example, they could be convinced—wrongly—that they had read: "Fred arrived 2 hours before Sophie" (an inference made because Fred and Sophie left at the same time, but Fred's journey took 2 hours less, which is factually true but constitutes an alteration to what the statement indicated).

"We have shown that when solving specific problems, participants have the illusion of having read sentences that were never actually presented in the statements, but were linked to unconscious deductions made when reading the statements. They become confused in their minds with the sentences they actually read," explains Hippolyte Gros, former post-doctoral fellow at UNIGE's Faculty of Psychology and Educational Sciences, lecturer at CYU, and first author of the study.

Invoking memories to understand reasoning

In addition, the experiments showed that the participants with the false memories were only those who had discovered the shortest strategy, thus revealing their unconscious reasoning that had enabled them to find this resolution shortcut. On the other hand, the others, who had operated in more stages, were unable to "enrich" their memory because they had not carried out the corresponding reasoning.

"This work can have applications for learning mathematics. By asking students to recall statements, we can identify their mental representations and therefore the reasoning they used when solving the problem, based on the presence or absence of false memories in their restitution," explains Emmanuel Sander, full professor at the UNIGE's Faculty of Psychology and Educational Sciences, who directed this research.

It is difficult to access mental constructs directly. Doing so indirectly, by analyzing memorization processes, could lead to a better understanding of the difficulties encountered by students in solving problems, and provide avenues for intervention in the classroom.

Provided by University of Geneva

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A general method for solving differential equations of motion using physics-informed neural networks.

1. Introduction

2. physics-informed neural networks, 2.1. fully connected neural network, 2.2. differential equations, 2.3. training process of neural network, 3. numerical studies, 3.1. two-degree-of-freedom system, 3.1.1. training sample number, 3.1.2. number of hidden layers and neurons, 3.2. four-layer framework structure, 3.2.1. activation function, 3.2.2. the weight coefficients of the loss function, 3.3. cantilever beam, 4. conclusions, author contributions, institutional review board statement, informed consent statement, data availability statement, acknowledgments, conflicts of interest.

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Zhang, W.; Ni, P.; Zhao, M.; Du, X. A General Method for Solving Differential Equations of Motion Using Physics-Informed Neural Networks. Appl. Sci. 2024 , 14 , 7694. https://doi.org/10.3390/app14177694

Zhang W, Ni P, Zhao M, Du X. A General Method for Solving Differential Equations of Motion Using Physics-Informed Neural Networks. Applied Sciences . 2024; 14(17):7694. https://doi.org/10.3390/app14177694

Zhang, Wenhao, Pinghe Ni, Mi Zhao, and Xiuli Du. 2024. "A General Method for Solving Differential Equations of Motion Using Physics-Informed Neural Networks" Applied Sciences 14, no. 17: 7694. https://doi.org/10.3390/app14177694

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NIM Agent Blueprints give developers a head start when creating applications with AI agents , Nvidia said.

Blueprints include sample applications built with Nvidia NeMo , Nvidia NIM, and partner microservices.

NIM Agent Blueprints follow a GenAI market trend where many enterprises seek to move from ideation to implementation.

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Thus, Nvidia NIM attempts to solve some of the problems enterprises face implementing GenAI technology , including the speed at which enterprises integrate GenAI workflows, said Chirag Dekate, an analyst with Gartner.

"It's about creating pluggable, modular components that enterprises can build off," Dekate added.

While Nvidia could have chosen to give developers AI tools to build their applications, with NIMs such as NIM Agent Blueprints, the vendor provides enterprises with pretrained models that allow them to move quicker in building generative AI products and services, Futurum Group analyst Olivier Blanchard said.

While this is cost-effective, it might not be what every enterprise needs, Blanchard said.

"For enterprises that are super unique, NIM might not be the model," he said. "They might start somewhere else first and then use them to build additional layers of capabilities or features on top of their other product."

What Nvidia is providing is not unique to GenAI technology, said Rob Enderle, an analyst with Enderle Group.

"Part of what's required to be successful in the market, you have to have a complete solution, not just the technology," Enderle said. "Initially, generative AI was a technology. What Nvidia is doing is turning it into a complete solution."

Nvidia is not the only vendor endeavoring to make generative AI more than just a technology.

AMD, a competitor to Nvidia, is one of the leading companies in the movement, Enderle said.

The vendor last week purchased ZT Systems along with several other companies, including Mipsology , an AI software startup in France, to help with AI inference capabilities.

Google, AWS and Microsoft also are building tools to help enterprises easily deploy GenAI systems.

Nvidia software services

NIM Agent Blueprints allows Nvidia to beef up its software services, making services such as NIMs available regardless of the cloud provider.

It's about creating pluggable, modular components that enterprises can build off. Chirag Dekate Analyst, Gartner

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Russia is turning to crypto as sanctions snarl payments, report says

• Russia will start cryptocurrency exchange trials on September 1, Bloomberg reported.
• The nation is look for new ways to transact payments amid sanctions.
• The trials mark a turnaround in Russia's anti-crypto attitude.

Russia, home to anti-crypto legislation, plans on testing digital assets as a way to break away from Western sanctions, Bloomberg reported.

Moscow will initiate cryptocurrency exchange trials as soon as September 1, anonymous sources told the outlet. The intention is to ease payment difficulties for Russian exporters, caused by restrictions related to Moscow's war in Ukraine.

Sources said the National Payment Card System will be used to swap between rubles and cryptocurrencies in these trials. The network was chosen due to its preexisting functions, such as an interbank settlement and clearing feature, and its full regulation by the central bank.

If successful, this experiment could eventually provide Russia with an alternative payment method for cross-border transactions. Payment issues have made it difficult to buy foreign supplies and have left exporters struggling to get reimbursed.

This became problematic in June, when the new US sanctions started targeting Russia-friendly foreign lenders. Given that global institutions rely on access to the dollar, the threat of secondary sanctions forced many to cut off ties with Moscow.

For instance, most Chinese banks have since halted yuan payments to Russia despite the yuan's prominent role in Russia's current-day economy. Meanwhile, Moscow can't rely on Western currencies such as the dollar, after the country lost access to the SWIFT international banking system .

Moscow's possible embrace of digital currencies is a significant retreat from its usual attitude: Russia has spent years arguing against crypto and has proposed a blanket ban in 2022. Even if crypto is adopted in international payments, its use inside the country remains restricted .

The legal turnaround came on August 8th , when President Vladimir Putin legalized the creation of a cross-border crypto framework. Additionally, he signed into law a bill allowing from crypto mining in the country.

Eventually, Russia may permit the Moscow Exchange and St. Petersburg Currency Exchange to creat crypto platforms next year, sources told Bloomberg. As to traditional crypto exchanges, Finance Minister Anton Siluanov acknowledged that this looks less likely:

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By Tim Fernholz

Photographs by Ian C. Bates

Tim Fernholz reported from Twain Harte, Calif.

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They are called skidders: 10-foot-tall vehicles, on four massive wheels, with a bulldozer-like blade on the front and a tree-size grapple dangling from the back. They are the worker bees, hauling downed logs from the forest to landing sites where they are delimbed and loaded onto trucks bound for the sawmill. Usually, a single driver operates them for a 12-hour shift, grabbing logs from behind and then driving forward.

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Chapter 2: Linear Equations

2.7 Variation Word Problems

Direct variation problems.

There are many mathematical relations that occur in life. For instance, a flat commission salaried salesperson earns a percentage of their sales, where the more they sell equates to the wage they earn. An example of this would be an employee whose wage is 5% of the sales they make. This is a direct or a linear variation, which, in an equation, would look like:

[latex]\text{Wage }(x)=5\%\text{ Commission }(k)\text{ of Sales Completed }(y)[/latex]

[latex]x=ky[/latex]

A historical example of direct variation can be found in the changing measurement of pi, which has been symbolized using the Greek letter π since the mid 18th century. Variations of historical π calculations are Babylonian [latex]\left(\dfrac{25}{8}\right),[/latex] Egyptian [latex]\left(\dfrac{16}{9}\right)^2,[/latex] and Indian [latex]\left(\dfrac{339}{108}\text{ and }10^{\frac{1}{2}}\right).[/latex] In the 5th century, Chinese mathematician Zu Chongzhi calculated the value of π to seven decimal places (3.1415926), representing the most accurate value of π for over 1000 years.

Pi is found by taking any circle and dividing the circumference of the circle by the diameter, which will always give the same value: 3.14159265358979323846264338327950288419716… (42 decimal places). Using an infinite-series exact equation has allowed computers to calculate π to 10 13 decimals.

[latex]\begin{array}{c} \text{Circumference }(c)=\pi \text{ times the diameter }(d) \\ \\ \text{or} \\ \\ c=\pi d \end{array}[/latex]

All direct variation relationships are verbalized in written problems as a direct variation or as directly proportional and take the form of straight line relationships. Examples of direct variation or directly proportional equations are:

• [latex]x[/latex] varies directly as [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex]
• [latex]x[/latex] varies directly proportional to [latex]y[/latex]
• [latex]x[/latex] is proportional to [latex]y[/latex]
• [latex]x[/latex] varies directly as the square of [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex] squared
• [latex]x[/latex] is proportional to the square of [latex]y[/latex]
• [latex]x[/latex] varies directly as the cube of [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex] cubed
• [latex]x[/latex] is proportional to the cube of [latex]y[/latex]
• [latex]x[/latex] varies directly as the square root of [latex]y[/latex]
• [latex]x[/latex] varies as the root of [latex]y[/latex]
• [latex]x[/latex] is proportional to the square root of [latex]y[/latex]

Example 2.7.1

Find the variation equation described as follows:

The surface area of a square surface [latex](A)[/latex] is directly proportional to the square of either side [latex](x).[/latex]

[latex]\begin{array}{c} \text{Area }(A) =\text{ constant }(k)\text{ times side}^2\text{ } (x^2) \\ \\ \text{or} \\ \\ A=kx^2 \end{array}[/latex]

Example 2.7.2

When looking at two buildings at the same time, the length of the buildings’ shadows [latex](s)[/latex] varies directly as their height [latex](h).[/latex] If a 5-story building has a 20 m long shadow, how many stories high would a building that has a 32 m long shadow be?

The equation that describes this variation is:

[latex]h=kx[/latex]

Breaking the data up into the first and second parts gives:

[latex]\begin{array}{ll} \begin{array}{rrl} \\ &&\textbf{1st Data} \\ s&=&20\text{ m} \\ h&=&5\text{ stories} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ h&=&kx \\ 5\text{ stories}&=&k\text{ (20 m)} \\ k&=&5\text{ stories/20 m}\\ k&=&0.25\text{ story/m} \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ s&=&\text{32 m} \\ h&=&\text{find 2nd} \\ k&=&0.25\text{ story/m} \\ \\ &&\text{Find }h\text{:} \\ h&=&kx \\ h&=&(0.25\text{ story/m})(32\text{ m}) \\ h&=&8\text{ stories} \end{array} \end{array}[/latex]

Inverse Variation Problems

Inverse variation problems are reciprocal relationships. In these types of problems, the product of two or more variables is equal to a constant. An example of this comes from the relationship of the pressure [latex](P)[/latex] and the volume [latex](V)[/latex] of a gas, called Boyle’s Law (1662). This law is written as:

[latex]\begin{array}{c} \text{Pressure }(P)\text{ times Volume }(V)=\text{ constant} \\ \\ \text{ or } \\ \\ PV=k \end{array}[/latex]

Written as an inverse variation problem, it can be said that the pressure of an ideal gas varies as the inverse of the volume or varies inversely as the volume. Expressed this way, the equation can be written as:

[latex]P=\dfrac{k}{V}[/latex]

Another example is the historically famous inverse square laws. Examples of this are the force of gravity [latex](F_{\text{g}}),[/latex] electrostatic force [latex](F_{\text{el}}),[/latex] and the intensity of light [latex](I).[/latex] In all of these measures of force and light intensity, as you move away from the source, the intensity or strength decreases as the square of the distance.

In equation form, these look like:

[latex]F_{\text{g}}=\dfrac{k}{d^2}\hspace{0.25in} F_{\text{el}}=\dfrac{k}{d^2}\hspace{0.25in} I=\dfrac{k}{d^2}[/latex]

These equations would be verbalized as:

• The force of gravity [latex](F_{\text{g}})[/latex] varies inversely as the square of the distance.
• Electrostatic force [latex](F_{\text{el}})[/latex] varies inversely as the square of the distance.
• The intensity of a light source [latex](I)[/latex] varies inversely as the square of the distance.

All inverse variation relationship are verbalized in written problems as inverse variations or as inversely proportional. Examples of inverse variation or inversely proportional equations are:

• [latex]x[/latex] varies inversely as [latex]y[/latex]
• [latex]x[/latex] varies as the inverse of [latex]y[/latex]
• [latex]x[/latex] varies inversely proportional to [latex]y[/latex]
• [latex]x[/latex] is inversely proportional to [latex]y[/latex]
• [latex]x[/latex] varies inversely as the square of [latex]y[/latex]
• [latex]x[/latex] varies inversely as [latex]y[/latex] squared
• [latex]x[/latex] is inversely proportional to the square of [latex]y[/latex]
• [latex]x[/latex] varies inversely as the cube of [latex]y[/latex]
• [latex]x[/latex] varies inversely as [latex]y[/latex] cubed
• [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex]
• [latex]x[/latex] varies inversely as the square root of [latex]y[/latex]
• [latex]x[/latex] varies as the inverse root of [latex]y[/latex]
• [latex]x[/latex] is inversely proportional to the square root of [latex]y[/latex]

Example 2.7.3

The force experienced by a magnetic field [latex](F_{\text{b}})[/latex] is inversely proportional to the square of the distance from the source [latex](d_{\text{s}}).[/latex]

[latex]F_{\text{b}} = \dfrac{k}{{d_{\text{s}}}^2}[/latex]

Example 2.7.4

The time [latex](t)[/latex] it takes to travel from North Vancouver to Hope varies inversely as the speed [latex](v)[/latex] at which one travels. If it takes 1.5 hours to travel this distance at an average speed of 120 km/h, find the constant [latex]k[/latex] and the amount of time it would take to drive back if you were only able to travel at 60 km/h due to an engine problem.

[latex]t=\dfrac{k}{v}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} &&\textbf{1st Data} \\ v&=&120\text{ km/h} \\ t&=&1.5\text{ h} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ k&=&tv \\ k&=&(1.5\text{ h})(120\text{ km/h}) \\ k&=&180\text{ km} \end{array} & \hspace{0.5in} \begin{array}{rrl} \\ \\ \\ &&\textbf{2nd Data} \\ v&=&60\text{ km/h} \\ t&=&\text{find 2nd} \\ k&=&180\text{ km} \\ \\ &&\text{Find }t\text{:} \\ t&=&\dfrac{k}{v} \\ \\ t&=&\dfrac{180\text{ km}}{60\text{ km/h}} \\ \\ t&=&3\text{ h} \end{array} \end{array}[/latex]

Joint or Combined Variation Problems

In real life, variation problems are not restricted to single variables. Instead, functions are generally a combination of multiple factors. For instance, the physics equation quantifying the gravitational force of attraction between two bodies is:

[latex]F_{\text{g}}=\dfrac{Gm_1m_2}{d^2}[/latex]

• [latex]F_{\text{g}}[/latex] stands for the gravitational force of attraction
• [latex]G[/latex] is Newton’s constant, which would be represented by [latex]k[/latex] in a standard variation problem
• [latex]m_1[/latex] and [latex]m_2[/latex] are the masses of the two bodies
• [latex]d^2[/latex] is the distance between the centres of both bodies

To write this out as a variation problem, first state that the force of gravitational attraction [latex](F_{\text{g}})[/latex] between two bodies is directly proportional to the product of the two masses [latex](m_1, m_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating the two masses. From this information, the necessary equation can be derived. All joint variation relationships are verbalized in written problems as a combination of direct and inverse variation relationships, and care must be taken to correctly identify which variables are related in what relationship.

Example 2.7.5

The force of electrical attraction [latex](F_{\text{el}})[/latex] between two statically charged bodies is directly proportional to the product of the charges on each of the two objects [latex](q_1, q_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating these two charged bodies.

[latex]F_{\text{el}}=\dfrac{kq_1q_2}{d^2}[/latex]

Solving these combined or joint variation problems is the same as solving simpler variation problems.

First, decide what equation the variation represents. Second, break up the data into the first data given—which is used to find [latex]k[/latex]—and then the second data, which is used to solve the problem given. Consider the following joint variation problem.

Example 2.7.6

[latex]y[/latex] varies jointly with [latex]m[/latex] and [latex]n[/latex] and inversely with the square of [latex]d[/latex]. If [latex]y = 12[/latex] when [latex]m = 3[/latex], [latex]n = 8[/latex], and [latex]d = 2,[/latex] find the constant [latex]k[/latex], then use [latex]k[/latex] to find [latex]y[/latex] when [latex]m=-3[/latex], [latex]n = 18[/latex], and [latex]d = 3[/latex].

[latex]y=\dfrac{kmn}{d^2}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} \\ \\ \\ && \textbf{1st Data} \\ y&=&12 \\ m&=&3 \\ n&=&8 \\ d&=&2 \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ 12&=&\dfrac{k(3)(8)}{(2)^2} \\ \\ k&=&\dfrac{12(2)^2}{(3)(8)} \\ \\ k&=& 2 \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ y&=&\text{find 2nd} \\ m&=&-3 \\ n&=&18 \\ d&=&3 \\ k&=&2 \\ \\ &&\text{Find }y\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ y&=&\dfrac{(2)(-3)(18)}{(3)^2} \\ \\ y&=&12 \end{array} \end{array}[/latex]

For questions 1 to 12, write the formula defining the variation, including the constant of variation [latex](k).[/latex]

• [latex]x[/latex] is jointly proportional to [latex]y[/latex] and [latex]z[/latex]
• [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex]
• [latex]x[/latex] is jointly proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
• [latex]x[/latex] is inversely proportional to [latex]y[/latex] to the sixth power
• [latex]x[/latex] is jointly proportional with the cube of [latex]y[/latex] and inversely to the square root of [latex]z[/latex]
• [latex]x[/latex] is inversely proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
• [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex] and is inversely proportional to the cube of [latex]p[/latex]
• [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex] and square of [latex]z[/latex]

For questions 13 to 22, find the formula defining the variation and the constant of variation [latex](k).[/latex]

• If [latex]A[/latex] varies directly as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=15[/latex] and [latex]B=5.[/latex]
• If [latex]P[/latex] is jointly proportional to [latex]Q[/latex] and [latex]R,[/latex] find [latex]k[/latex] when [latex]P=12, Q=8[/latex] and [latex]R=3.[/latex]
• If [latex]A[/latex] varies inversely as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=7[/latex] and [latex]B=4.[/latex]
• If [latex]A[/latex] varies directly as the square of [latex]B,[/latex] find [latex]k[/latex] when [latex]A=6[/latex] and [latex]B=3.[/latex]
• If [latex]C[/latex] varies jointly as [latex]A[/latex] and [latex]B,[/latex] find [latex]k[/latex] when [latex]C=24, A=3,[/latex] and [latex]B=2.[/latex]
• If [latex]Y[/latex] is inversely proportional to the cube of [latex]X,[/latex] find [latex]k[/latex] when [latex]Y=54[/latex] and [latex]X=3.[/latex]
• If [latex]X[/latex] is directly proportional to [latex]Y,[/latex] find [latex]k[/latex] when [latex]X=12[/latex] and [latex]Y=8.[/latex]
• If [latex]A[/latex] is jointly proportional with the square of [latex]B[/latex] and the square root of [latex]C,[/latex] find [latex]k[/latex] when [latex]A=25, B=5[/latex] and [latex]C=9.[/latex]
• If [latex]y[/latex] varies jointly with [latex]m[/latex] and the square of [latex]n[/latex] and inversely with [latex]d,[/latex] find [latex]k[/latex] when [latex]y=10, m=4, n=5,[/latex] and [latex]d=6.[/latex]
• If [latex]P[/latex] varies directly as [latex]T[/latex] and inversely as [latex]V,[/latex] find [latex]k[/latex] when [latex]P=10, T=250,[/latex] and [latex]V=400.[/latex]

For questions 23 to 37, solve each variation word problem.

• The electrical current [latex]I[/latex] (in amperes, A) varies directly as the voltage [latex](V)[/latex] in a simple circuit. If the current is 5 A when the source voltage is 15 V, what is the current when the source voltage is 25 V?
• The current [latex]I[/latex] in an electrical conductor varies inversely as the resistance [latex]R[/latex] (in ohms, Ω) of the conductor. If the current is 12 A when the resistance is 240 Ω, what is the current when the resistance is 540 Ω?
• Hooke’s law states that the distance [latex](d_s)[/latex] that a spring is stretched supporting a suspended object varies directly as the mass of the object [latex](m).[/latex] If the distance stretched is 18 cm when the suspended mass is 3 kg, what is the distance when the suspended mass is 5 kg?
• The volume [latex](V)[/latex] of an ideal gas at a constant temperature varies inversely as the pressure [latex](P)[/latex] exerted on it. If the volume of a gas is 200 cm 3 under a pressure of 32 kg/cm 2 , what will be its volume under a pressure of 40 kg/cm 2 ?
• The number of aluminum cans [latex](c)[/latex] used each year varies directly as the number of people [latex](p)[/latex] using the cans. If 250 people use 60,000 cans in one year, how many cans are used each year in a city that has a population of 1,000,000?
• The time [latex](t)[/latex] required to do a masonry job varies inversely as the number of bricklayers [latex](b).[/latex] If it takes 5 hours for 7 bricklayers to build a park wall, how much time should it take 10 bricklayers to complete the same job?
• The wavelength of a radio signal (λ) varies inversely as its frequency [latex](f).[/latex] A wave with a frequency of 1200 kilohertz has a length of 250 metres. What is the wavelength of a radio signal having a frequency of 60 kilohertz?
• The number of kilograms of water [latex](w)[/latex] in a human body is proportional to the mass of the body [latex](m).[/latex] If a 96 kg person contains 64 kg of water, how many kilograms of water are in a 60 kg person?
• The time [latex](t)[/latex] required to drive a fixed distance [latex](d)[/latex] varies inversely as the speed [latex](v).[/latex] If it takes 5 hours at a speed of 80 km/h to drive a fixed distance, what speed is required to do the same trip in 4.2 hours?
• The volume [latex](V)[/latex] of a cone varies jointly as its height [latex](h)[/latex] and the square of its radius [latex](r).[/latex] If a cone with a height of 8 centimetres and a radius of 2 centimetres has a volume of 33.5 cm 3 , what is the volume of a cone with a height of 6 centimetres and a radius of 4 centimetres?
• The centripetal force [latex](F_{\text{c}})[/latex] acting on an object varies as the square of the speed [latex](v)[/latex] and inversely to the radius [latex](r)[/latex] of its path. If the centripetal force is 100 N when the object is travelling at 10 m/s in a path or radius of 0.5 m, what is the centripetal force when the object’s speed increases to 25 m/s and the path is now 1.0 m?
• The maximum load [latex](L_{\text{max}})[/latex] that a cylindrical column with a circular cross section can hold varies directly as the fourth power of the diameter [latex](d)[/latex] and inversely as the square of the height [latex](h).[/latex] If an 8.0 m column that is 2.0 m in diameter will support 64 tonnes, how many tonnes can be supported by a column 12.0 m high and 3.0 m in diameter?
• The volume [latex](V)[/latex] of gas varies directly as the temperature [latex](T)[/latex] and inversely as the pressure [latex](P).[/latex] If the volume is 225 cc when the temperature is 300 K and the pressure is 100 N/cm 2 , what is the volume when the temperature drops to 270 K and the pressure is 150 N/cm 2 ?
• The electrical resistance [latex](R)[/latex] of a wire varies directly as its length [latex](l)[/latex] and inversely as the square of its diameter [latex](d).[/latex] A wire with a length of 5.0 m and a diameter of 0.25 cm has a resistance of 20 Ω. Find the electrical resistance in a 10.0 m long wire having twice the diameter.
• The volume of wood in a tree [latex](V)[/latex] varies directly as the height [latex](h)[/latex] and the diameter [latex](d).[/latex] If the volume of a tree is 377 m 3 when the height is 30 m and the diameter is 2.0 m, what is the height of a tree having a volume of 225 m 3 and a diameter of 1.75 m?

Answer Key 2.7

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Modeling using Variation

• 1.1 A General Note: Direct Variation
• 1.2 How To: Given a description of a direct variation problem, solve for an unknown.
• 1.3 Example 1: Solving a Direct Variation Problem
• 1.4 Solution
• 1.5 Analysis of the Solution
• 2.1 A General Note: Inverse Variation
• 2.2 Example 2: Writing a Formula for an Inversely Proportional Relationship
• 2.3 Solution
• 2.4 How To: Given a description of an indirect variation problem, solve for an unknown.
• 2.5 Example 3: Solving an Inverse Variation Problem
• 2.6 Solution
• 2.7 Analysis of the Solution
• 3.1 A General Note: Joint Variation
• 3.2 Example 4: Solving Problems Involving Joint Variation
• 3.3 Solution
• 4 Key Equations
• 5.1 Glossary
• 5.2 Resources
• 5.3 Licensing

Direct Variation

A general note: direct variation.

If x and y are related by an equation of the form

How To: Given a description of a direct variation problem, solve for an unknown.

• Identify the input, x , and the output, y .
• Determine the constant of variation. You may need to divide y by the specified power of x to determine the constant of variation.
• Use the constant of variation to write an equation for the relationship.
• Substitute known values into the equation to find the unknown.

Example 1: Solving a Direct Variation Problem

The quantity y varies directly with the cube of x . If y = 25 when x = 2, find y when x is 6.

Now use the constant to write an equation that represents this relationship.

Substitute x = 6 and solve for y .

Analysis of the Solution

The graph of this equation is a simple cubic, as shown below.

Inverse variation

A general note: inverse variation.

Example 2: Writing a Formula for an Inversely Proportional Relationship

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.

Recall that multiplying speed by time gives distance. If we let t represent the drive time in hours, and v represent the velocity (speed or rate) at which the tourist drives, then vt = distance. Because the distance is fixed at 100 miles, vt = 100. Solving this relationship for the time gives us our function.

We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.

How To: Given a description of an indirect variation problem, solve for an unknown.

• Determine the constant of variation. You may need to multiply y by the specified power of x to determine the constant of variation.

Example 3: Solving an Inverse Variation Problem

A quantity y varies inversely with the cube of x . If y = 25 when x = 2, find y when x is 6.

Now we use the constant to write an equation that represents this relationship.

The graph of this equation is a rational function.

Joint variation

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation . For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c , cost, varies jointly with the number of students, n , and the distance, d .

A General Note: Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables.

Example 4: Solving Problems Involving Joint Variation

A quantity x varies directly with the square of y and inversely with the cube root of z . If x = 6 when y = 2 and z = 8, find x when y = 1 and z = 27.

Begin by writing an equation to show the relationship between the variables.

Substitute x = 6, y = 2, and z = 8 to find the value of the constant k .

Now we can substitute the value of the constant into the equation for the relationship.

To find x when y = 1 and z = 27, we will substitute values for y and z into our equation.

Key Equations

Key Concepts

• A relationship where one quantity is a constant multiplied by another quantity is called direct variation.
• Two variables that are directly proportional to one another will have a constant ratio.
• A relationship where one quantity is a constant divided by another quantity is called inverse variation.
• Two variables that are inversely proportional to one another will have a constant multiple.
• In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation.
• Modeling Using Variation , Lumen Learning
• Variation Word Problems , Open Text Intermediate Algebra
• Intro to Direct & Inverse Variation , Khan Academy
• Direct Inverse and Joint Variation Word Problems , The Organic Chemistry Tutor

Content obtained and/or adapted from:

• Modeling Using Variation, Lumen Learning Precalculus 1 under a CC BY license

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• Direct Variation

When two variables change in proportion it is called as direct variation. In direct variation one variable is constant times of other. If one variable increases other will increase, if one decrease other will also decease. This means that the variables change in a same ratio which is called as constant of variation.

Direct variation is the simplest type of variation and in practical life we can find many situations which can be co-related with direct variation.

If two variables A and B are so related that when A increases ( or decreases ) in a given ratio, B also increases ( or, decreases ) in the same ratio, then  A is said to vary directly as B  ( or, A is said to vary as B ).

This is symbolically written as, A ∝ B (read as, ‘A varies as B’ ). 

Suppose a train moving at a uniform speed travels d km. in t minutes. Now, consider the following table:

Like in a math examination if for one problem solving we can score 10 numbers, so five problems solving we can get 50 numbers. This can be explained with a direct variation equation. If T denotes total numbers scored, N denotes numbers of problem solved and K denotes numbers can be scored for solving a problem, then the direct variation equation for this situation will be T = KN. 

As the numbers for a problem can be scored is fixed, it is a constant = K = \(\frac{T}{N}\) = 10

For solving 5 problems total numbers scored T = KN = 10 x 5 = 50.

From the above example we can understand that the ratio of two variables is a constant K and T, N are the variables which changes in proportion with value of constant.

Direct variation can be by a linear equation Y = KX where K is a constant. When the value of constant is higher, the change of variable Y is significantly for small change of X. But when the value of K is very small, Y changes very less with change of X. For this case K is equivalent to the ratio of change of two variables. So \(\frac{σY}{σX}\) = K when K is very small.

Now we will solve some problems on direct variation:

1.  If P varies directly as Q and the value of P is 60 and Q is 40, what is the equation that describes this direct variation of P and Q?

As P varies directly with Q, ratio of P and Q is constant for any value of P and Q.

So constant K = \(\frac{P}{Q}\) = \(\frac{60}{40}\) = \(\frac{3}{2}\)

So the equation that describes the direct variation of P and Q is P = \(\frac{3}{2}\)Q.

2.  If a car runs at a constant speed and takes 3 hrs to run a distance of 180 km, what time it will take to run 100 km?

If T is the time taken to cover the distance and S is the distance and V is the speed of the car, the direct variation equation is S= VT where V is constant.

For the case given in the problem,

180 = V × 3 or V = \(\frac{180}{3}\) = 60

So speed of the car is 60kmph and it is constant.

For 100 km distance 

S = VT or 100 = 60 × T

T = \(\frac{100}{60}\) = \(\frac{5}{3}\) hrs = 1 hr 40 mins.

So it will take 1 hr 40 mins time.

3.  In X is in direct variation with square of Y and when X is 4, Y is 3. What is the value of X when Y is 6?

From the given problem direct variation equation can be expressed as

For the given case

4 = K × 3 2

or, K = \(\frac{4}{9}\)

So when Y is 6,

X = \(\frac{4}{9}\) × 6 = \(\frac{8}{3}\)

So the value of X is \(\frac{8}{3}\).

●   Variation

• What is Variation?
• Inverse Variation
• Joint Variation
• Theorem of Joint Variation
• Worked out Examples on Variation
• Problems on Variation

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Variation Equations

Equations Word Problems More Prob's

Variation problems aren't hard once you get the hang of the lingo. The only real difficulty is learning the somewhat specialized vocabulary and the techniques for this classification of problems.

But there is always just the one term in the formula; this term will be multiplied by some number and, if you don't yet know what that number is, the number is usually denoted by " k ". This number k is called "the constant of variation".

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Direct Variation

So variation equations may have complicated expressions, but they'll only ever have the one term. Their equations will never have two or more terms added together. For instance, y = 3 x is a variation equation, but y = 3 x + 2 is not.

What are the keywords for variation equations?

The keywords for variation equations are:

What do these variation-equation keywords mean?

• directly: things are working together; you are multiplying everything.
• inversely: things working against each other; you are dividing something somewhere.
• jointly: many things are working together; you are multiplying lots of stuff.

An example of a variation equation would be the formula for the area of the circle:

A = π r 2

In the language of variation, this equation means:

the area A varies directly with the square of the radius r

...and the constant of variation is k = π . This formula is an example of "direct" variation."Direct variation" means that, in the one term of the formula, the variable is "on top".

(By the way, if you see a formula with a " ∝ " character in it instead of an "equals" sign, that character is pronounced as "is directly proportional to", and indicates that they've given you a direct-variation equation, but without telling you what the variation constant is.)

On the other hand, "inverse variation" means that the variable is underneath, in the bottom of a fraction. Suppose, for instance, that you inherit a money market account containing $100,000 , and you wonder how much money your rich uncle initially invested eight years ago. Depending on the average interest rate " r ", the formula you would use would be:

...where P is the principal that your uncle invested. (This formula is a variant of the compound-interest formula , by the way.) In the language of variation, this formula reads as:

...with the constant of variation being k = 100,000 .

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The other case of variation is "jointly". "Joint variation" means "directly, but with two or more variables". An example would be the formula for the area of a triangle with base " b " and height " h ":

the area A varies jointly with b and h

• " F varies as x " means F =  kx
• " F varies jointly as x and y " means F =  kxy
• " F varies as x + y " means F =  k ( x  +  y )

Be careful with those middle two statements above. Almost always, when you translate word problems from English into math, "and" means "plus" or "added to". But in joint variation, "and" just means "both of these are together on the same side of the fraction" (usually on top), and you multiply. If you are supposed to add two variables, they'll use the format in that third bulleted example above, or they'll say "varies as the sum of x  and y .

(If they have you working with something that varies "directly" with one thing and "inversely" with another thing, this may be referred to as a "combined" variation.)

Translating variation problems isn't so bad, once you get the hang of it. But then they want you to move on to setting up and solving word problems. These generally fall into two categories: the ones where they want you to find the value of " k ", and the ones where they want you to find some other value, but only after you've found " k " first.

What are some examples of solving variation equations?

If y varies directly as x 2 and y  = 8 when x  = 2 , find y when x  = 1 ..

Since this is direct variation, the formula is " y = kx 2 ".The reason they've given me the data point ( x ,  y ) = (2, 8) is that I have to be able to find the value of " k ". So I'll plug in the information they've given me, and solve for k :

y = k x 2

8 = k  (2 2 )

Now that I have k , I can rewrite the formula completely:

With this, I can answer the question they actually asked: "Find y when x  = 1 ."

y = 2 (1) 2

y = 2 (1)

Then the answer is:

If y varies jointly as x and z , and y  = 5 when x  = 3 and z  = 4 , then find y when x  = 2 and z  = 3 .

Translating the formula from English to math, I get:

Plugging in the data point they gave me, and solving for the value of k , I get:

5 = k (3)(4)

Now that I have the value of k , I can plug in the new values, and solve for the new value of y :

y = ( 5 / 12 ) xz

y = ( 5 / 12 )(2)(3)

y = ( 5 / 12 )(6)

Given that y varies inversely as the square of the difference of w and x , and that y  = 6 when w  = 3 and x  = 1 , find the equation for y .

This one is somewhat complex. I'll work in small pieces.

They've told me that y varies inversely with the specified expression, so I know that the form of the expression will have a constant of variation on top of a rational expression (that is, a polynomial fraction), and a variable expression on the bottom of that expression. But what is the variable expression?

"The difference of w and x " means:

w − x

The square of this difference is:

( w − x ) 2

This expression will go on the bottom of the fraction, and the variation constant k will go on top, so my equation so far is:

y = k /( w − x ) 2

In order to find the complete equation for y , I'll need to find the value of the constant. To do this, I'll plug in the values they gave me:

6 = k /(3 − 1) 2

6 = k /(2) 2

Now I have all the parts of my equation, and my answer is:

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Direct Variation

In these lessons, we will learn about direct variation and how to solve applications that involve direct variation.

Related Pages: Proportions Proportion Word Problems Inverse Variation Joint And Combined Variation More Algebra Lessons

The following diagrams show Direct Variation and Indirect Variation. Scroll down the page for examples and solutions.

What Is Direct Variation?

There are many situations in our daily lives that involve direct variation .

We say that y varies directly with x. Let us represent the constant by k, i.e.

If y varies directly as x, this relation is written as y ∝ x and read as y varies as x. The sign “ ∝ ” is read “varies as” and is called the sign of variation .

Example: If y varies directly as x and given y = 9 when x = 5, find: a) the equation connecting x and y b) the value of y when x = 15 c) the value of x when y = 6

Solution: a) y ∝ x i.e. y = kx where k is a constant Substitute x = 5 and y = 9 into the equation:

Example: The cost of a taxi fare (C) varies directly as the distance (D) travelled. When the distance is 60 km, the cost is $35. Find the cost when the distance is 95 km.

How To Define Direct Variation And Solve Direct Variation Word Problems?

Some examples of direct variation problems in real life:

• The number of hours you work and the amount of your paycheck.
• The amount of weight on a spring and the distance the spring will stretch.
• The speed of a car and the distance traveled in a certain amount of time.

The following statements are equivalent:

• y varies directly as x
• y is directly proportional to x
• y = kx for some constant k

What Is The Direct Variation Formula?

A direct variation is a linear equation that can be written in the form y = kx , where k is a nonzero constant. The number k is called the constant of proportionality or constant of variation.

Graphically, we have a line that passes through the origin with the slope of k.

• y varies directly with x. y = 54 when x = 9. Determine the direct variation equation and then determine y when x = 3.5
• Hooke’s Law states that the displacement, d, that a spring is stretched by a hanging object varies directly as the mass of the object. If the distance is 10 cm when the mass is kg, what is the distance when the mass is 5 kg?
• y varies directly with x. y = 32 when x = 4. Determine the direct variation equation and then determine y when x = 6

How To Determine A Direct Variation Equation From The Given Information And Then Determine y With A Given Value Of x?

Example: y varies directly with x. Given that y = 6 when x = 30, determine the direct variation equation and then determine y when x = 8.

Real Life Example Of A Direct Variation Problem

Example: The total cost of filling up your car with gas varies directly with the number of gallons of gasoline that you are purchasing. If a gallon of gas costs $2.25, how many gallons could you purchase for $18?

Other Forms Of Direct Variation

The area A of a circle of radius r is given by the equation A = pr 2 , where p is a constant

In this situation, A is not directly proportional to r but A is directly proportional to r 2 . We say that ‘A varies directly as the square of r ’ or A ∝ r 2 .

Example: Given that y varies directly as the cube of x and that y = 21 when x = 3, calculate the value of y when x = 8.

Solution: y ∝ x 3 that is y = kx 3 where k is a constant

The following video gives some practical examples of direct variation and indirect/inverse variation.

How To Solve Word Problems Involving Direct And Inverse Variation Squared?

• On planet X, an object falls 18 feet in 2 seconds. Knowing that the distance it falls varies directly with the square of the time of the fall, how long does it take an object to fall 29 feet? Round your answer to three decimal places.
• The Intensity, I, of light received from a source varies inversely as the square of the distance, d, from the source. If the light intensity is 4-foot candles at 11 feet, find the light intensity at 13 feet.

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Polynomial and Rational Functions

Modeling using variation, learning objectives.

In this section, you will:

• Solve direct variation problems.
• Solve inverse variation problems.
• Solve problems involving joint variation.

A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.

Solving Direct Variation Problems

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula[latex]\,e=0.16s\,[/latex]tells us her earnings,[latex]\,e,\,[/latex]come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See (Figure) .

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation . Each variable in this type of relationship varies directly with the other.

(Figure) represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula[latex]\,y=k{x}^{n}\,[/latex]is used for direct variation. The value[latex]\,k\,[/latex]is a nonzero constant greater than zero and is called the constant of variation . In this case,[latex]\,k=0.16\,[/latex]and[latex]\,n=1.\,[/latex]We saw functions like this one when we discussed power functions.

• Direct Variation

If[latex]\,x\,\text{and}\,y\,[/latex]are related by an equation of the form

then we say that the relationship is direct variation and [latex]\,y\,[/latex] varies directly with, or is proportional to, the[latex]\,n\text{th}\,[/latex]power of[latex]\,x.\,[/latex]In direct variation relationships, there is a nonzero constant ratio[latex]\,k=\frac{y}{{x}^{n}},\,[/latex]where[latex]\,k\,[/latex]is called the constant of variation, which help defines the relationship between the variables.

Given a description of a direct variation problem, solve for an unknown.

• Identify the input,[latex]\,x,[/latex]and the output,[latex]\,y.\,[/latex]
• Determine the constant of variation. You may need to divide[latex]\,y\,[/latex]by the specified power of[latex]\,x\,[/latex]to determine the constant of variation.
• Use the constant of variation to write an equation for the relationship.
• Substitute known values into the equation to find the unknown.

Solving a Direct Variation Problem

The quantity[latex]\,y\,[/latex]varies directly with the cube of[latex]\,x.\,[/latex]If[latex]\,y=25\,[/latex]when[latex]\,x=2,\,[/latex]find[latex]\,y\,[/latex]when[latex]\,x\,[/latex]is 6.

The general formula for direct variation with a cube is[latex]\,y=k{x}^{3}.\,[/latex]The constant can be found by dividing[latex]\,y\,[/latex]by the cube of[latex]\,x.\,[/latex]

Now use the constant to write an equation that represents this relationship.

Substitute[latex]\,x=6\,[/latex]and solve for[latex]\,y.[/latex]

The graph of this equation is a simple cubic, as shown in (Figure) .

Do the graphs of all direct variation equations look like (Figure) ?

No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through[latex]\,\left(0,0\right).[/latex]

The quantity[latex]\,y\,[/latex]varies directly with the square of[latex]\,x.\,[/latex]If[latex]\,y=24\,[/latex]when[latex]\,x=3,\,[/latex]find[latex]\,y\,[/latex]when[latex]\,x\,[/latex]is 4.

[latex]\frac{128}{3}[/latex]

Solving Inverse Variation Problems

Water temperature in an ocean varies inversely to the water’s depth. The formula[latex]\,T=\frac{14,000}{d}\,[/latex]gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.

If we create (Figure) , we observe that, as the depth increases, the water temperature decreases.

We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations .

For our example, (Figure) depicts the inverse variation . We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula[latex]\,y=\frac{k}{x}\,[/latex]for inverse variation in this case uses[latex]\,k=14,000.\,[/latex]

• Inverse Variation

If[latex]\,x\,[/latex]and[latex]\,y\,[/latex]are related by an equation of the form

where[latex]\,k\,[/latex]is a nonzero constant, then we say that[latex]\,y[/latex] varies inversely with the[latex]\,n\text{th}\,[/latex]power of[latex]\,x.\,[/latex]In inversely proportional relationships, or inverse variations, there is a constant multiple[latex]\,k={x}^{n}y.\,[/latex]

Writing a Formula for an Inversely Proportional Relationship

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.

Recall that multiplying speed by time gives distance. If we let[latex]\,t\,[/latex]represent the drive time in hours, and[latex]\,v\,[/latex]represent the velocity (speed or rate) at which the tourist drives, then[latex]\,vt=\text{distance}\text{.}\,[/latex]Because the distance is fixed at 100 miles,[latex]\,vt=100\,[/latex] so[latex]t=100/v.\,[/latex]Because time is a function of velocity, we can write[latex]\,t\left(v\right).[/latex]

We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction. We say that time varies inversely with velocity.

Given a description of an indirect variation problem, solve for an unknown.

• Identify the input,[latex]\,x,\,[/latex]and the output,[latex]\,y.[/latex]
• Determine the constant of variation. You may need to multiply[latex]\,y\,[/latex]by the specified power of[latex]\,x\,[/latex]to determine the constant of variation.

Solving an Inverse Variation Problem

A quantity[latex]\,y\,[/latex]varies inversely with the cube of[latex]\,x.\,[/latex]If[latex]\,y=25\,[/latex]when[latex]\,x=2,\,[/latex]find[latex]\,y\,[/latex]when[latex]\,x\,[/latex]is 6.

The general formula for inverse variation with a cube is[latex]\,y=\frac{k}{{x}^{3}}.\,[/latex]The constant can be found by multiplying[latex]\,y\,[/latex]by the cube of[latex]\,x.\,[/latex]

Now we use the constant to write an equation that represents this relationship.

The graph of this equation is a rational function, as shown in (Figure) .

A quantity[latex]\,y\,[/latex]varies inversely with the square of[latex]\,x.\,[/latex]If[latex]\,y=8\,[/latex]when[latex]\,x=3,\,[/latex]find[latex]\,y\,[/latex]when[latex]\,x\,[/latex]is 4.

[latex]\,\frac{9}{2}\,[/latex]

Solving Problems Involving Joint Variation

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable[latex]\,c,[/latex]cost, varies jointly with the number of students,[latex]\,n,[/latex]and the distance,[latex]\,d.\,[/latex]

Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables.

For instance, if[latex]\,x\,[/latex]varies directly with both[latex]\,y\,[/latex]and[latex]\,z,\,[/latex] we have[latex]\,x=kyz.\,[/latex]If[latex]\,x\,[/latex]varies directly with[latex]\,y\,[/latex]and inversely with[latex]z,[/latex]we have[latex]\,x=\frac{ky}{z}.\,[/latex]Notice that we only use one constant in a joint variation equation.

A quantity[latex]\,x\,[/latex]varies directly with the square of[latex]\,y\,[/latex]and inversely with the cube root of[latex]\,z.\,[/latex]If[latex]\,x=6\,[/latex]when[latex]\,y=2\,[/latex] and[latex]\,z=8,\,[/latex]find[latex]\,x\,[/latex]when[latex]\,y=1\,[/latex] and [latex]\,z=27.\,[/latex]

Substitute[latex]\,x=6,\,[/latex][latex]y=2,\,[/latex]and[latex]\,z=8\,[/latex]to find the value of the constant[latex]\,k.\,[/latex]

Now we can substitute the value of the constant into the equation for the relationship.

To find[latex]\,x\,[/latex]when[latex]\,y=1\,[/latex]and[latex]\,z=27,\,[/latex]we will substitute values for[latex]\,y\,[/latex]and[latex]\,z\,[/latex]into our equation.

A quantity[latex]\,x\,[/latex]varies directly with the square of[latex]\,y\,[/latex]and inversely with[latex]\,z.\,[/latex]If[latex]\,x=40\,[/latex]when[latex]\,y=4\,[/latex]and[latex]\,z=2,\,[/latex]find[latex]\,x\,[/latex]when[latex]\,y=10\,[/latex] and[latex]\,z=25.[/latex]

[latex]\,x=20\,[/latex]

Access these online resources for additional instruction and practice with direct and inverse variation.

• Direct and Inverse Variation

Visit this website for additional practice questions from Learningpod.

Key Equations

Key Concepts

• A relationship where one quantity is a constant multiplied by another quantity is called direct variation. See (Figure) .
• Two variables that are directly proportional to one another will have a constant ratio.
• A relationship where one quantity is a constant divided by another quantity is called inverse variation. See (Figure) .
• Two variables that are inversely proportional to one another will have a constant multiple. See (Figure) .
• In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation. See (Figure) .

Section Exercises

What is true of the appearance of graphs that reflect a direct variation between two variables?

The graph will have the appearance of a power function.

If two variables vary inversely, what will an equation representing their relationship look like?

Is there a limit to the number of variables that can vary jointly? Explain.

No. Multiple variables may jointly vary.

For the following exercises, write an equation describing the relationship of the given variables.

[latex]\,y\,[/latex]varies directly as[latex]\,x\,[/latex]and when[latex]\,x=6, y=12.[/latex]

[latex]\,y\,[/latex]varies directly as the square of[latex]\,x\,[/latex]and when[latex]\,x=4,\, y=80.[/latex]

[latex]y=5{x}^{2}[/latex]

[latex]\,y\,[/latex]varies directly as the square root of[latex]\,x\,[/latex]and when[latex]\,x=36, \,y=24.[/latex]

[latex]\,y\,[/latex]varies directly as the cube of[latex]\,x\,[/latex]and when[latex]\,x=36,\,y=24.[/latex]

[latex]y=10{x}^{3}[/latex]

[latex]\,y\,[/latex]varies directly as the cube root of[latex]\,x\,[/latex]and when[latex]\,x=27,\, y=15.[/latex]

[latex]\,y\,[/latex]varies directly as the fourth power of[latex]\,x\,[/latex]and when[latex]\,x=1,\, y=6.[/latex]

[latex]y=6{x}^{4}[/latex]

[latex]\,y\,[/latex]varies inversely as[latex]\,x\,[/latex]and when[latex]\,x=4,\, y=2.[/latex]

[latex]\,y\,[/latex]varies inversely as the square of[latex]\,x\,[/latex]and when[latex]\,x=3,\, y=2.[/latex]

[latex]y=\frac{18}{{x}^{2}}[/latex]

[latex]\,y\,[/latex]varies inversely as the cube of[latex]\,x\,[/latex]and when[latex]\,x=2,\, y=5.[/latex]

[latex]\,y\,[/latex]varies inversely as the fourth power of[latex]\,x\,[/latex]and when[latex]\,x=3,\, y=1.\,[/latex]

[latex]y=\frac{81}{{x}^{4}}[/latex]

[latex]\,y\,[/latex]varies inversely as the square root of[latex]\,x\,[/latex]and when[latex]\,x=25, \,y=3.[/latex]

[latex]\,y\,[/latex]varies inversely as the cube root of[latex]\,x\,[/latex]and when[latex]\,x=64, \,y=5.[/latex]

[latex]y=\frac{20}{\sqrt[3]{x}}[/latex]

[latex]\,y\,[/latex]varies jointly with[latex]\,x\,[/latex]and[latex]\,z\,[/latex]and when[latex]\,x=2\,[/latex]and[latex]\,z=3,\, y=36.[/latex]

[latex]\,y\,[/latex]varies jointly as[latex]\,x, z,\,[/latex]and[latex]\,w\,[/latex]and when[latex]\,x=1,\, z=2,\, w=5,\,[/latex]then[latex]\,y=100.[/latex]

[latex]y=10xzw[/latex]

[latex]\,y\,[/latex]varies jointly as the square of[latex]\,x\,[/latex]and the square of[latex]\,z\,[/latex]and when[latex]\,x=3\,[/latex]and[latex]\,z=4,\,[/latex]then[latex]\,y=72.[/latex]

[latex]\,y\,[/latex]varies jointly as[latex]\,x\,[/latex]and the square root of[latex]\,z\,[/latex]and when[latex]\,x=2\,[/latex]and[latex]\,z=25,\,[/latex]then[latex]\,y=100.[/latex]

[latex]y=10x\sqrt{z}[/latex]

[latex]\,y[/latex]varies jointly as the square of[latex]\,x[/latex]the cube of[latex]\,z[/latex]and the square root of[latex]\,W.\,[/latex]When[latex]\,x=1,z=2,\,[/latex]and[latex]\,w=36,\,[/latex]then[latex]\,y=48.[/latex]

[latex]\,y\,[/latex] varies jointly as[latex]\,x \text{and} z\,[/latex]and inversely as[latex]\,w.\,[/latex]When[latex]\,x=3, \,z=5,\,[/latex]and[latex]\,w=6,\,[/latex]then[latex]\,y=10.[/latex]

[latex]\,y\,[/latex]varies jointly as the square of[latex]\,x\,[/latex]and the square root of[latex]\,z\,[/latex]and inversely as the cube of[latex]\,w\text{.\hspace{0.17em}}[/latex]When[latex]\,x=3,z=4,\,[/latex]and[latex]\,w=3,\,[/latex]then[latex]\,y=6.[/latex]

[latex]\,y\,[/latex]varies jointly as[latex]\,x\,[/latex]and[latex]\,z\,[/latex]and inversely as the square root of[latex]\,w\,[/latex]and the square of[latex]\,t\text{\hspace{0.17em}.}[/latex]When[latex]\,x=3,z=1,w=25,\,[/latex]and[latex]\,t=2,\,[/latex]then[latex]\,y=6.[/latex]

For the following exercises, use the given information to find the unknown value.

[latex]\,y\,[/latex]varies directly as[latex]\,x.\,[/latex]When[latex]\,x=3,\,[/latex]then[latex]\,y=12.\,[/latex]Find[latex]\,y\,[/latex]wneh[latex]\,x=20.[/latex]

[latex]\,y\,[/latex]varies directly as the square of[latex]\,x.\,[/latex]When[latex]\,x=2,\,[/latex]then[latex]\,y=16.\,[/latex]Find[latex]\,y\,[/latex]when[latex]x=8.[/latex]

[latex]y=256[/latex]

[latex]\,y\,[/latex]varies directly as the cube of[latex]\,x.\,[/latex]When[latex]\,x=3,\,[/latex]then[latex]\,y=5. \,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=4.[/latex]

[latex]\,y\,[/latex]varies directly as the square root of[latex]\,x.\,[/latex]When[latex]\,x=16,\,[/latex]then[latex]\,y=4.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=36.[/latex]

[latex]\,y\,[/latex]varies directly as the cube root of[latex]\,x.\,[/latex]When[latex]\,x=125,\,[/latex]then[latex]\,y=15.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=1,000.[/latex]

[latex]\,y\,[/latex]varies inversely with[latex]\,x.\,[/latex]When[latex]\,x=3,\,[/latex]then[latex]\,y=2.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=1.[/latex]

[latex]y=6[/latex]

[latex]\,y\,[/latex]varies inversely with the square of[latex]\,x.\,[/latex]When[latex]\,x=4,\,[/latex]then[latex]\,y=3.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=2.[/latex]

[latex]\,y\,[/latex]varies inversely with the cube of[latex]\,x.\,[/latex]When[latex]\,x=3,\,[/latex]then[latex]\,y=1.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=1.[/latex]

[latex]y=27[/latex]

[latex]\,y\,[/latex]varies inversely with the square root of[latex]\,x.\,[/latex]When[latex]\,x=64,\,[/latex]then[latex]\,y=12.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=36.[/latex]

[latex]\,y\,[/latex]varies inversely with the cube root of[latex]\,x.\,[/latex]When[latex]\,x=27,\,[/latex]then[latex]\,y=5.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=125.[/latex]

[latex]y=3[/latex]

[latex]\,y\,[/latex]varies jointly as[latex]\,x \text{and} z.[/latex]When[latex]\,x=4\,[/latex]and[latex]\,z=2,\,[/latex]then[latex]\,y=16.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=3\,[/latex]and[latex]\,z=3.[/latex]

[latex]\,y\,[/latex]varies jointly as[latex]\,x, z, \text{and} w.[/latex]When[latex]\,x=2,\,[/latex][latex]z=1,\,[/latex]and[latex]\,w=12,\,[/latex]then[latex]\,y=72.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=1,\,[/latex][latex]z=2,\,[/latex] and[latex]\,w=3.[/latex]

[latex]y=18[/latex]

[latex]\,y\,[/latex]varies jointly as[latex]\,x\,[/latex]and the square of[latex]\,\mathrm{z.}\,[/latex]When[latex]\,x=2\,[/latex]and[latex]\,z=4,\,[/latex]then[latex]\,y=144.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=4\,[/latex]and[latex]\,z=5.[/latex]

[latex]\,y\,[/latex]varies jointly as the square of[latex]\,x\,[/latex]and the square root of[latex]\,z.\,[/latex]When[latex]\,x=2\,[/latex]and[latex]\,z=9,\,[/latex]then[latex]\,y=24.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=3\,[/latex]and[latex]\,z=25.[/latex]

[latex]y=90[/latex]

[latex]\,y\,[/latex]varies jointly as[latex]\,x\,[/latex]and[latex]\,z\,[/latex]and inversely as[latex]\,w.\,[/latex]When[latex]\,x=5,\,[/latex][latex]z=2, \,[/latex]and[latex] \,w=20,\,[/latex]then[latex]y=4.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=3\,[/latex]and[latex]\,z=8, \,[/latex]and[latex]\, w=48.[/latex]

[latex]\,y\,[/latex]varies jointly as the square of[latex]\,x\,[/latex]and the cube of[latex]\,z\,[/latex]and inversely as the square root of[latex]\,w\.[/latex]When[latex]\,x=2,\,[/latex][latex]z=2,\,[/latex]and[latex]\,w=64,\,[/latex]then[latex]\,y=12.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=1,\,[/latex][latex]z=3,\,[/latex]and[latex] \,w=4.[/latex]

[latex]y=\frac{81}{2}[/latex]

[latex]\,y\,[/latex]varies jointly as the square of[latex]\,x\,[/latex]and of[latex]\,z\,[/latex]and inversely as the square root of[latex]\,w\,[/latex]and of[latex]\,t\.[/latex]When[latex]\,x=2,\,[/latex][latex]z=3,\,[/latex][latex]w=16,\,[/latex]and[latex]\,t=3,\,[/latex]then[latex]\,y=1.\,[/latex]Find[latex]\,y\,[/latex]when[latex]\,x=3,\,[/latex][latex]z=2,\,[/latex][latex] w=36,\,[/latex]and[latex]\,t=5.[/latex]

For the following exercises, use a calculator to graph the equation implied by the given variation.

[latex]\,y\,[/latex]varies directly with the square of[latex]\,x\,[/latex]and when[latex]\,x=2, y=3.[/latex]

[latex]y=\frac{3}{4}{x}^{2}[/latex]

[latex]\,y\,[/latex]varies directly as the cube of[latex]\,x\,[/latex]and when[latex]\,x=2, y=4.[/latex]

[latex]\,y\,[/latex]varies directly as the square root of[latex]\,x\,[/latex]and when[latex]\,x=36, y=2.[/latex]

[latex]y=\frac{1}{3}\sqrt{x}[/latex]

[latex]\,y\,[/latex]varies inversely with[latex]\,x\,[/latex]and when[latex]\,x=6, y=2.[/latex]

[latex]\,y\,[/latex]varies inversely as the square of[latex]\,x\,[/latex]and when[latex]\,x=1, y=4.[/latex]

[latex]y=\frac{4}{{x}^{2}}[/latex]

For the following exercises, use Kepler’s Law, which states that the square of the time,[latex]\,T,\,[/latex]required for a planet to orbit the Sun varies directly with the cube of the mean distance,[latex]\,a,\,[/latex]that the planet is from the Sun.

Using Earth’s time of 1 year and mean distance of 93 million miles, find the equation relating [latex]\,T\,[/latex] and [latex]\,a.\,[/latex]

Use the result from the previous exercise to determine the time required for Mars to orbit the Sun if its mean distance is 142 million miles.

Using Earth’s distance of 150 million kilometers, find the equation relating[latex]\,T\,[/latex]and[latex]\,a.[/latex]

Use the result from the previous exercise to determine the time required for Venus to orbit the Sun if its mean distance is 108 million kilometers.

Using Earth’s distance of 1 astronomical unit (A.U.), determine the time for Saturn to orbit the Sun if its mean distance is 9.54 A.U.

Real-World Applications

For the following exercises, use the given information to answer the questions.

The distance[latex]\,s\,[/latex]that an object falls varies directly with the square of the time,[latex]\,t,\,[/latex]of the fall. If an object falls 16 feet in one se c ond, how long for it to fall 144 feet?

The velocity[latex]\,v\,[/latex]of a falling object varies directly to the time,[latex]\,t,\,[/latex]of the fall. If after 2 seconds, the velocity of the object is 64 feet per second, what is the velocity after 5 seconds?

The rate of vibration of a string under constant tension varies inversely with the length of the string. If a string is 24 inches long and vibrates 128 times per second, what is the length of a string that vibrates 64 times per second?

The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If the volume of a gas is 1200 cubic centimeters when the pressure is 200 millimeters of mercury, what is the volume when the pressure is 300 millimeters of mercury?

The weight of an object above the surface of Earth varies inversely with the square of the distance from the center of Earth. If a body weighs 50 pounds when it is 3960 miles from Earth’s center, what would it weigh it were 3970 miles from Earth’s center?

49.75 pounds

The intensity of light measured in foot-candles varies inversely with the square of the distance from the light source. Suppose the intensity of a light bulb is 0.08 foot-candles at a distance of 3 meters. Find the intensity level at 8 meters.

The current in a circuit varies inversely with its resistance measured in ohms. When the current in a circuit is 40 amperes, the resistance is 10 ohms. Find the current if the resistance is 12 ohms.

33.33 amperes

The force exerted by the wind on a plane surface varies jointly with the square of the velocity of the wind and with the area of the plane surface. If the area of the surface is 40 square feet surface and the wind velocity is 20 miles per hour, the resulting force is 15 pounds. Find the force on a surface of 65 square feet with a velocity of 30 miles per hour.

The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute (rpm) and the cube of the diameter. If the shaft of a certain material 3 inches in diameter can transmit 45 hp at 100 rpm, what must the diameter be in order to transmit 60 hp at 150 rpm?

2.88 inches

The kinetic energy[latex]\,K\,[/latex]of a moving object varies jointly with its mass[latex]\,m\,[/latex]and the square of its velocity[latex]\,v.\,[/latex]If an object weighing 40 kilograms with a velocity of 15 meters per second has a kinetic energy of 1000 joules, find the kinetic energy if the velocity is increased to 20 meters per second.

Chapter Review Exercises

Quadratic functions.

For the following exercises, write the quadratic function in standard form. Then give the vertex and axes intercepts. Finally, graph the function.

[latex]f\left(x\right)={x}^{2}-4x-5[/latex]

[latex]f\left(x\right)={\left(x-2\right)}^{2}-9\,\text{vertex} \left(2,–9\right), \text{intercepts} \left(5,0\right); \left(–1,0\right); \left(0,–5\right)[/latex]

[latex]f\left(x\right)=-2{x}^{2}-4x[/latex]

For the following exercises, find the equation of the quadratic function using the given information.

The vertex is[latex]\left(–2,3\right)[/latex]and a point on the graph is[latex]\,\left(3,6\right).[/latex]

[latex]f\left(x\right)=\frac{3}{25}{\left(x+2\right)}^{2}+3[/latex]

The vertex is[latex]\,\left(–3,6.5\right)\,[/latex]and a point on the graph is[latex]\,\left(2,6\right).[/latex]

For the following exercises, complete the task.

A rectangular plot of land is to be enclosed by fencing. One side is along a river and so needs no fence. If the total fencing available is 600 meters, find the dimensions of the plot to have maximum area.

300 meters by 150 meters, the longer side parallel to river.

An object projected from the ground at a 45 degree angle with initial velocity of 120 feet per second has height,[latex]\,h,\,[/latex]in terms of horizontal distance traveled,[latex]\,x,\,[/latex]given by[latex]\,h\left(x\right)=\frac{-32}{{\left(120\right)}^{2}}{x}^{2}+x.\,[/latex]Find the maximum height the object attains.

Power Functions and Polynomial Functions

For the following exercises, determine if the function is a polynomial function and, if so, give the degree and leading coefficient.

[latex]f\left(x\right)=4{x}^{5}-3{x}^{3}+2x-1[/latex]

Yes, degree = 5, leading coefficient = 4

[latex]f\left(x\right)={5}^{x+1}-{x}^{2}[/latex]

[latex]f\left(x\right)={x}^{2}\left(3-6x+{x}^{2}\right)[/latex]

For the following exercises, determine end behavior of the polynomial function.

[latex]f\left(x\right)=2{x}^{4}+3{x}^{3}-5{x}^{2}+7[/latex]

[latex]f\left(x\right)=4{x}^{3}-6{x}^{2}+2[/latex]

[latex]\text{As}\,x\to -\infty ,\,f\left(x\right)\to -\infty ,\,\text{as}\,x\to \infty ,\,f\left(x\right)\to \infty [/latex]

[latex]f\left(x\right)=2{x}^{2}\left(1+3x-{x}^{2}\right)[/latex]

Graphs of Polynomial Functions

For the following exercises, find all zeros of the polynomial function, noting multiplicities.

[latex]f\left(x\right)={\left(x+3\right)}^{2}\left(2x-1\right){\left(x+1\right)}^{3}[/latex]

–3 with multiplicity 2,[latex]-\frac{1}{2}[/latex]with multiplicity 1, –1 with multiplicity 3

[latex]f\left(x\right)={x}^{5}+4{x}^{4}+4{x}^{3}[/latex]

[latex]f\left(x\right)={x}^{3}-4{x}^{2}+x-4[/latex]

4 with multiplicity 1

For the following exercises, based on the given graph, determine the zeros of the function and note multiplicity.

[latex]\frac{1}{2}\,[/latex]with multiplicity 1, 3 with multiplicity 3

Use the Intermediate Value Theorem to show that at least one zero lies between 2 and 3 for the function[latex]\,f\left(x\right)={x}^{3}-5x+1[/latex]

Dividing Polynomials

For the following exercises, use long division to find the quotient and remainder.

[latex]\frac{{x}^{3}-2{x}^{2}+4x+4}{x-2}[/latex]

[latex]\,{x}^{2}+4\,[/latex] with remainder 12

[latex]\frac{3{x}^{4}-4{x}^{2}+4x+8}{x+1}[/latex]

For the following exercises, use synthetic division to find the quotient. If the divisor is a factor, then write the factored form.

[latex]\frac{{x}^{3}-2{x}^{2}+5x-1}{x+3}[/latex]

[latex]{x}^{2}-5x+20-\frac{61}{x+3}[/latex]

[latex]\frac{{x}^{3}+4x+10}{x-3}[/latex]

[latex]\frac{2{x}^{3}+6{x}^{2}-11x-12}{x+4}[/latex]

[latex]2{x}^{2}-2x-3[/latex], so factored form is [latex]\left(x+4\right)\left(2{x}^{2}-2x-3\right)[/latex]

[latex]\frac{3{x}^{4}+3{x}^{3}+2x+2}{x+1}[/latex]

Zeros of Polynomial Functions

For the following exercises, use the Rational Zero Theorem to help you solve the polynomial equation.

[latex]2{x}^{3}-3{x}^{2}-18x-8=0[/latex]

[latex]\left\{-2, 4, -\frac{1}{2}\right\}[/latex]

[latex]3{x}^{3}+11{x}^{2}+8x-4=0[/latex]

[latex]2{x}^{4}-17{x}^{3}+46{x}^{2}-43x+12=0[/latex]

[latex]4{x}^{4}+8{x}^{3}+19{x}^{2}+32x+12=0[/latex]

For the following exercises, use Descartes’ Rule of Signs to find the possible number of positive and negative solutions.

[latex]{x}^{3}-3{x}^{2}-2x+4=0[/latex]

0 or 2 positive, 1 negative

[latex]2{x}^{4}-{x}^{3}+4{x}^{2}-5x+1=0[/latex]

Rational Functions

For the following exercises, find the intercepts and the vertical and horizontal asymptotes, and then use them to sketch a graph of the function.

[latex]f\left(x\right)=\frac{x+2}{x-5}[/latex]

Intercepts [latex]\left(–2,0\right)\text{and}\left(0,-\frac{2}{5}\right)[/latex], Asymptotes[latex]\,x=5\,[/latex] and[latex]\,y=1.[/latex]

[latex]f\left(x\right)=\frac{{x}^{2}+1}{{x}^{2}-4}[/latex]

[latex]f\left(x\right)=\frac{3{x}^{2}-27}{{x}^{2}+x-2}[/latex]

Intercepts (3, 0), (-3, 0), and[latex]\,\left(0,\frac{27}{2}\right)\,[/latex], Asymptotes[latex]\,x=1, x=–2, y=3.[/latex]

For the following exercises, find the slant asymptote.

[latex]y=\text{ }x-2[/latex]

[latex]f\left(x\right)=\frac{2{x}^{3}-{x}^{2}+4}{{x}^{2}+1}[/latex]

Inverses and Radical Functions

For the following exercises, find the inverse of the function with the domain given.

[latex]f\left(x\right)={\left(x-2\right)}^{2},\,x\ge 2[/latex]

[latex]{f}^{-1}\left(x\right)=\sqrt{x}+2[/latex]

[latex]f\left(x\right)={\left(x+4\right)}^{2}-3,\,x\ge -4[/latex]

[latex]f\left(x\right)={x}^{2}+6x-2,\,x\ge -3[/latex]

[latex]{f}^{-1}\left(x\right)=\sqrt{x+11}-3[/latex]

[latex]f\left(x\right)=2{x}^{3}-3[/latex]

[latex]f\left(x\right)=\sqrt{4x+5}-3[/latex]

[latex]{f}^{-1}\left(x\right)=\frac{{\left(x+3\right)}^{2}-5}{4},\,x\ge -3[/latex]

[latex]f\left(x\right)=\frac{x-3}{2x+1}[/latex]

For the following exercises, find the unknown value.

[latex]\,y\,[/latex] varies directly as the square of[latex]\,x.\,[/latex] If when[latex]\,x=3,\text{ }y=36,\,[/latex]find[latex]\,y\,[/latex]if[latex]\,x=4.[/latex]

[latex]y=64[/latex]

[latex]\,y\,[/latex] varies inversely as the square root of[latex]\,x\,[/latex] If when[latex]\,x=25,\text{ }y=2,\,[/latex]find[latex]\,y\,[/latex]if[latex]\,x=4.[/latex]

[latex]\,y\,[/latex]varies jointly as the cube of[latex]\,x\,[/latex]and as[latex]\,z.\,[/latex]If when[latex]\,x=1\,[/latex]and[latex]\,z=2,\,[/latex][latex]y=6,\,[/latex]find[latex]\,y\,[/latex]if[latex]\,x=2\,[/latex]and[latex]\,z=3.[/latex]

[latex]y\text{ }=\text{ }72[/latex]

[latex]\,y\,[/latex]varies jointly as[latex]\,x\,[/latex]and the square of[latex]\,z\,[/latex]and inversely as the cube of[latex]\,w.\,[/latex]If when[latex]\,x=3,\,[/latex][latex]z=4,\,[/latex]and[latex]\,w=2,\,[/latex][latex]y=48,\,[/latex]find[latex]\,y\,[/latex]if[latex]\,x=4,\,[/latex][latex]z=5,\,[/latex]and[latex]\,w=3.[/latex]

For the following exercises, solve the application problem.

The weight of an object above the surface of the earth varies inversely with the distance from the center of the earth. If a person weighs 150 pounds when he is on the surface of the earth (3,960 miles from center), find the weight of the person if he is 20 miles above the surface.

The volume[latex]\,V\,[/latex]of an ideal gas varies directly with the temperature[latex]\,T\,[/latex]and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K.

Chapter Test

Give the degree and leading coefficient of the following polynomial function.

[latex]f\left(x\right)={x}^{3}\left(3-6{x}^{2}-2{x}^{2}\right)[/latex]

Determine the end behavior of the polynomial function.

[latex]f\left(x\right)=8{x}^{3}-3{x}^{2}+2x-4[/latex]

[latex]As\,x\to -\infty ,\,f\left(x\right)\to -\infty ,\,as\,x\to \infty ,\,f\left(x\right)\to \infty [/latex]

[latex]f\left(x\right)=-2{x}^{2}\left(4-3x-5{x}^{2}\right)[/latex]

Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the function.

[latex]f\left(x\right)={x}^{2}+2x-8[/latex]

[latex]f\left(x\right)={\left(x+1\right)}^{2}-9[/latex], vertex[latex]\,\left(-1,-9\right)[/latex], intercepts[latex]\,\left(2,0\right);\left(-4,0\right);\,\left(0,-8\right)[/latex]

Given information about the graph of a quadratic function, find its equation.

Vertex[latex]\,\left(2,0\right)\,[/latex]and point on graph[latex]\,\left(4,12\right).[/latex]

Solve the following application problem.

A rectangular field is to be enclosed by fencing. In addition to the enclosing fence, another fence is to divide the field into two parts, running parallel to two sides. If 1,200 feet of fencing is available, find the maximum area that can be enclosed.

60,000 square feet

Find all zeros of the following polynomial functions, noting multiplicities.

[latex]f\left(x\right)={\left(x-3\right)}^{3}\left(3x-1\right){\left(x-1\right)}^{2}[/latex]

[latex]f\left(x\right)=2{x}^{6}-12{x}^{5}+18{x}^{4}[/latex]

0 with multiplicity 4, 3 with multiplicity 2

Based on the graph, determine the zeros of the function and multiplicities.

Use long division to find the quotient.

[latex]\frac{2{x}^{3}+3x-4}{x+2}[/latex]

[latex]2{x}^{2}-4x+11-\frac{26}{x+2}[/latex]

Use synthetic division to find the quotient. If the divisor is a factor, write the factored form.

[latex]\frac{{x}^{4}+3{x}^{2}-4}{x-2}[/latex]

[latex]\frac{2{x}^{3}+5{x}^{2}-7x-12}{x+3}[/latex]

[latex]2{x}^{2}-x-4[/latex]. So factored form is[latex]\,\left(x+3\right)\left(2{x}^{2}-x-4\right)[/latex]

Use the Rational Zero Theorem to help you find the zeros of the polynomial functions.

[latex]f\left(x\right)=2{x}^{3}+5{x}^{2}-6x-9[/latex]

[latex]f\left(x\right)=4{x}^{4}+8{x}^{3}+21{x}^{2}+17x+4[/latex]

[latex]-\frac{1}{2}\,[/latex] (has multiplicity 2),[latex]\,\frac{-1±i\sqrt{15}}{2}\,[/latex]

[latex]f\left(x\right)={x}^{5}+6{x}^{4}+13{x}^{3}+14{x}^{2}+12x+8[/latex]

[latex]\,-2\,[/latex](has multiplicity 3),[latex]\,±i[/latex]

Given the following information about a polynomial function, find the function.

It has a double zero at[latex]\,x=3\,[/latex]and zeros at[latex]\,x=1\,[/latex]and[latex]\,x=-2\,[/latex]. Its y -intercept is[latex]\,\left(0,12\right).\,[/latex]

It has a zero of multiplicity 3 at[latex]\,x=\frac{1}{2}\,[/latex]and another zero at[latex]\,x=-3\,[/latex]. It contains the point[latex]\,\left(1,8\right).[/latex]

[latex]f\left(x\right)=2{\left(2x-1\right)}^{3}\left(x+3\right)[/latex]

Use Descartes’ Rule of Signs to determine the possible number of positive and negative solutions.

For the following rational functions, find the intercepts and horizontal and vertical asymptotes, and sketch a graph.

[latex]f\left(x\right)=\frac{x+4}{{x}^{2}-2x-3}[/latex]

Intercepts[latex]\,\left(-4,0\right),\,\left(0,-\frac{4}{3}\right)\,[/latex], Asymptotes[latex]\,x=3,\text{ }x\text{ }=-1,\text{ }y=0[/latex].

[latex]f\left(x\right)=\frac{{x}^{2}+2x-3}{{x}^{2}-4}[/latex]

Find the slant asymptote of the rational function.

[latex]f\left(x\right)=\frac{{x}^{2}+3x-3}{x-1}[/latex]

Find the inverse of the function.

[latex]f\left(x\right)=\sqrt{x-2}+4[/latex]

[latex]f\left(x\right)=3{x}^{3}-4[/latex]

[latex]{f}^{-1}\left(x\right)=\sqrt[3]{\frac{x+4}{3}}[/latex]

[latex]f\left(x\right)=\frac{2x+3}{3x-1}[/latex]

Find the unknown value.

[latex]\,y\,[/latex]varies inversely as the square of[latex]\,x\,[/latex]and when[latex]\,x=3,\,[/latex][latex]y=2.\,[/latex]Find[latex]\,y\,[/latex]if[latex]\,x=1.[/latex]

[latex]\,y\,[/latex]varies jointly with[latex]\,x\,[/latex]and the cube root of[latex]\,z.\,[/latex]If when[latex]\,x=2\,[/latex]and[latex]\,z=27,\,[/latex][latex]y=12,\,[/latex]find[latex]\,y\,[/latex]if[latex]\,x=5\,[/latex]and[latex]\,z=8.[/latex]

The distance a body falls varies directly as the square of the time it falls. If an object falls 64 feet in 2 seconds, how long will it take to fall 256 feet?

• Algebra and Trigonometry. Authored by : Jay Abramson, et. al. Provided by : OpenStax CNX. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

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3.9 Modeling Using Variation

Learning objectives.

In this section, you will:

• Solve direct variation problems.
• Solve inverse variation problems.
• Solve problems involving joint variation.

A pre-owned car dealer has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. As she considers the offer, she takes into account the typical price of the dealer's cars, the overall market, and how many she can reasonably expect to sell. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.

Solving Direct Variation Problems

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula e = 0.16 s e = 0.16 s tells us her earnings, e , e , come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 1 .

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation . Each variable in this type of relationship varies directly with the other.

Figure 1 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula y = k x n y = k x n is used for direct variation. The value k k is a nonzero constant greater than zero and is called the constant of variation . In this case, k = 0.16 k = 0.16 and n = 1. n = 1.

• Direct Variation

If x and y x and y are related by an equation of the form

then we say that the relationship is direct variation and y y varies directly with the n th n th power of x . x . In direct variation relationships, there is a nonzero constant ratio k = y x n , k = y x n , where k k is called the constant of variation , which help defines the relationship between the variables.

Given a description of a direct variation problem, solve for an unknown.

• Identify the input, x , x , and the output, y . y .
• Determine the constant of variation. You may need to divide y y by the specified power of x x to determine the constant of variation.
• Use the constant of variation to write an equation for the relationship.
• Substitute known values into the equation to find the unknown.

Solving a Direct Variation Problem

The quantity y y varies directly with the cube of x . x . If y = 25 y = 25 when x = 2 , x = 2 , find y y when x x is 6.

The general formula for direct variation with a cube is y = k x 3 . y = k x 3 . The constant can be found by dividing y y by the cube of x . x .

Now use the constant to write an equation that represents this relationship.

Substitute x = 6 x = 6 and solve for y . y .

The graph of this equation is a simple cubic, as shown in Figure 2 .

Do the graphs of all direct variation equations look like Example 1 ?

No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through ( 0,0 ) . ( 0,0 ) .

The quantity y y varies directly with the square of x . x . If y = 24 y = 24 when x = 3 , x = 3 , find y y when x x is 4.

Solving Inverse Variation Problems

Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula T = 14,000 d T = 14,000 d gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.

If we create Table 2 , we observe that, as the depth increases, the water temperature decreases.

We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations .

For our example, Figure 3 depicts the inverse variation . We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula y = k x y = k x for inverse variation in this case uses k = 14,000. k = 14,000.

• Inverse Variation

If x x and y y are related by an equation of the form

where k k is a nonzero constant, then we say that y y varies inversely with the n th n th power of x . x . In inversely proportional relationships, or inverse variations , there is a constant multiple k = x n y . k = x n y .

Writing a Formula for an Inversely Proportional Relationship

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.

Recall that multiplying speed by time gives distance. If we let t t represent the drive time in hours, and v v represent the velocity (speed or rate) at which the tourist drives, then v t = distance . v t = distance . Because the distance is fixed at 100 miles, v t = 100. v t = 100. Solving this relationship for the time gives us our function.

We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.

Given a description of an indirect variation problem, solve for an unknown.

• Determine the constant of variation. You may need to multiply y y by the specified power of x x to determine the constant of variation.

Solving an Inverse Variation Problem

A quantity y y varies inversely with the cube of x . x . If y = 25 y = 25 when x = 2 , x = 2 , find y y when x x is 6.

The general formula for inverse variation with a cube is y = k x 3 . y = k x 3 . The constant can be found by multiplying y y by the cube of x . x .

Now we use the constant to write an equation that represents this relationship.

The graph of this equation is a rational function, as shown in Figure 4 .

A quantity y y varies inversely with the square of x . x . If y = 8 y = 8 when x = 3 , x = 3 , find y y when x x is 4.

Solving Problems Involving Joint Variation

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation . For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c , c , cost, varies jointly with the number of students, n , n , and the distance, d . d .

Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables.

For instance, if x x varies directly with both y y and z , z , we have x = k y z . x = k y z . If x x varies directly with y y and inversely with z , z , we have x = k y z . x = k y z . Notice that we only use one constant in a joint variation equation.

A quantity x x varies directly with the square of y y and inversely with the cube root of z . z . If x = 6 x = 6 when y = 2 y = 2 and z = 8 , z = 8 , find x x when y = 1 y = 1 and z = 27. z = 27.

Begin by writing an equation to show the relationship between the variables.

Substitute x = 6 , x = 6 , y = 2 , y = 2 , and z = 8 z = 8 to find the value of the constant k . k .

Now we can substitute the value of the constant into the equation for the relationship.

To find x x when y = 1 y = 1 and z = 27 , z = 27 , we will substitute values for y y and z z into our equation.

x x varies directly with the square of y y and inversely with z . z . If x = 40 x = 40 when y = 4 y = 4 and z = 2 , z = 2 , find x x when y = 10 y = 10 and z = 25. z = 25.

Access these online resources for additional instruction and practice with direct and inverse variation.

• Direct and Inverse Variation

3.9 Section Exercises

What is true of the appearance of graphs that reflect a direct variation between two variables?

If two variables vary inversely, what will an equation representing their relationship look like?

Is there a limit to the number of variables that can jointly vary? Explain.

For the following exercises, write an equation describing the relationship of the given variables.

y y varies directly as x x and when x = 6 , y = 12. x = 6 , y = 12.

y y varies directly as the square of x x and when x = 4 , y = 80 .  x = 4 , y = 80 . 

y y varies directly as the square root of x x and when x = 36 , y = 24. x = 36 , y = 24.

y y varies directly as the cube of x x and when x = 36 , y = 24. x = 36 , y = 24.

y y varies directly as the cube root of x x and when x = 27 , y = 15. x = 27 , y = 15.

y y varies directly as the fourth power of x x and when x = 1 , y = 6. x = 1 , y = 6.

y y varies inversely as x x and when x = 4 , y = 2. x = 4 , y = 2.

y y varies inversely as the square of x x and when x = 3 , y = 2. x = 3 , y = 2.

y y varies inversely as the cube of x x and when x = 2 , y = 5. x = 2 , y = 5.

y y varies inversely as the fourth power of x x and when x = 3 , y = 1. x = 3 , y = 1.

y y varies inversely as the square root of x x and when x = 25 , y = 3. x = 25 , y = 3.

y y varies inversely as the cube root of x x and when x = 64 , y = 5. x = 64 , y = 5.

y y varies jointly with x x and z z and when x = 2 x = 2 and z = 3 z = 3 , y = 36. y = 36.

y y varies jointly as x x , z z , and w w and when x = 1 x = 1 , z = 2 z = 2 , w = 5 w = 5 , then y = 100. y = 100.

y y varies jointly as the square of x x and the square of z z and when x = 3 x = 3 and z = 4 z = 4 , then y = 72. y = 72.

y y varies jointly as x x and the square root of z z and when x = 2 x = 2 and z = 25 z = 25 , then y = 100. y = 100.

y y varies jointly as the square of x x the cube of z z and the square root of w . w . When x = 1 x = 1 , z = 2 z = 2 , and  w = 36 ,  then  y = 48. w = 36 ,  then  y = 48.

y y varies jointly as x x and z z and inversely as w . w . When x = 3 x = 3 , z = 5 z = 5 , and w = 6 w = 6 , then y = 10. y = 10.

y y varies jointly as the square of x x and the square root of z z and inversely as the cube of w .  w .  When x = 3 , z = 4 ,  and  w = 3 ,  then  y = 6. x = 3 , z = 4 ,  and  w = 3 ,  then  y = 6.

y y varies jointly as x x and z z and inversely as the square root of w w and the square of t . t . When x = 3 x = 3 , z = 1 , w = 25 z = 1 , w = 25 , and  t = 2 t = 2 , then  y = 6. y = 6.

For the following exercises, use the given information to find the unknown value.

y y varies directly as x x . When x = 3 x = 3 , then  y = 12 y = 12 . Find  y y when  x = 20. x = 20.

y y varies directly as the square of x x . When x = 2 x = 2 , then y = 16 y = 16 . Find  y y when  x = 8 x = 8 .

y y varies directly as the cube of x x . When x = 3 x = 3 , then  y = 5 y = 5 . Find  y y  when  x = 4 x = 4 .

y y varies directly as the square root of x . x . When x = 16 x = 16 , then  y = 4 y = 4 . Find  y  when  x = 36 y  when  x = 36 .

y y varies directly as the cube root of x . x . When x = 125 x = 125 , then  y = 15 y = 15 . Find  y y  when  x = 1 x = 1 , 000. 000.

y y varies inversely with x . x . When x = 3 x = 3 , then  y = 2 y = 2 . Find  y y when  x = 1 x = 1 .

y y varies inversely with the square of x x . When x = 4 x = 4 , then  y = 3 y = 3 . Find y y  when  x = 2 x = 2 .

y y varies inversely with the cube of x . x . When x = 3 x = 3 , then  y = 1 y = 1 . Find  y y when  x = 1 x = 1 .

y y varies inversely with the square root of x . x . When x = 64 , x = 64 , then y = 12. y = 12. Find y y when x = 36. x = 36.

y y varies inversely with the cube root of x . x . When x = 27 , x = 27 , then y = 5. y = 5. Find y y when x = 125. x = 125.

y y varies jointly as x and z . x and z . When x = 4 x = 4 and z = 2 , z = 2 , then y = 16. y = 16. Find y y when x = 3 x = 3 and z = 3. z = 3.

y y varies jointly as x , z , and w . x , z , and w . When x = 2 , x = 2 , z = 1 , z = 1 , and w = 12 , w = 12 , then y = 72. y = 72. Find y y when x = 1 , x = 1 , z = 2 , z = 2 , and w = 3. w = 3.

y y varies jointly as x x and the square of z. z. When x = 2 x = 2 and z = 4 , z = 4 , then y = 144. y = 144. Find y y when x = 4 x = 4 and z = 5. z = 5.

y y varies jointly as the square of x x and the square root of z . z . When x = 2 x = 2 and z = 9 , z = 9 , then y = 24. y = 24. Find y y when x = 3 x = 3 and z = 25. z = 25.

y y varies jointly as x x and z z and inversely as w . w . When x = 5 , x = 5 , z = 2 , z = 2 , and w = 20 , w = 20 , then y = 4. y = 4. Find y y when x = 3 x = 3 and z = 8 , z = 8 , and w = 48. w = 48.

y y varies jointly as the square of x x and the cube of z z and inversely as the square root of w .  w .  When x = 2 , x = 2 , z = 2 , z = 2 , and w = 64 , w = 64 , then y = 12. y = 12. Find y y when x = 1 , x = 1 , z = 3 , z = 3 , and w = 4. w = 4.

y y varies jointly as the square of x x and of z z and inversely as the square root of w w and of t . t . When x = 2 , x = 2 , z = 3 , z = 3 , w = 16 , w = 16 , and t = 3 , t = 3 , then y = 1. y = 1. Find y y when x = 3 , x = 3 , z = 2 , z = 2 , w = 36 , w = 36 , and t = 5. t = 5.

For the following exercises, use a calculator to graph the equation implied by the given variation.

y y varies directly with the square of x x and when x = 2 , y = 3. x = 2 , y = 3.

y y varies directly as the cube of x x and when x = 2 , y = 4. x = 2 , y = 4.

y y varies directly as the square root of x x and when x = 36 , y = 2. x = 36 , y = 2.

y y varies inversely with x x and when x = 6 , y = 2. x = 6 , y = 2.

y y varies inversely as the square of x x and when x = 1 , y = 4. x = 1 , y = 4.

For the following exercises, use Kepler’s Law, which states that the square of the time, T , T , required for a planet to orbit the Sun varies directly with the cube of the mean distance, a , a , that the planet is from the Sun.

Using the Earth’s time of 1 year and mean distance of 93 million miles, find the equation relating T T and a . a .

Use the result from the previous exercise to determine the time required for Mars to orbit the Sun if its mean distance is 142 million miles.

Using Earth’s distance of 150 million kilometers, find the equation relating T T and a. a.

Use the result from the previous exercise to determine the time required for Venus to orbit the Sun if its mean distance is 108 million kilometers.

Using Earth’s distance of 1 astronomical unit (A.U.), determine the time for Saturn to orbit the Sun if its mean distance is 9.54 A.U.

Real-World Applications

For the following exercises, use the given information to answer the questions.

The distance s s that an object falls varies directly with the square of the time, t , t , of the fall. If an object falls 16 feet in one se c ond, how long for it to fall 144 feet?

The velocity v v of a falling object varies directly to the time, t , t , of the fall. If after 2 seconds, the velocity of the object is 64 feet per second, what is the velocity after 5 seconds?

The rate of vibration of a string under constant tension varies inversely with the length of the string. If a string is 24 inches long and vibrates 128 times per second, what is the length of a string that vibrates 64 times per second?

The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If the volume of a gas is 1200 cubic centimeters when the pressure is 200 millimeters of mercury, what is the volume when the pressure is 300 millimeters of mercury?

The weight of an object above the surface of the Earth varies inversely with the square of the distance from the center of the Earth. If a body weighs 50 pounds when it is 3960 miles from Earth’s center, what would it weigh it were 3970 miles from Earth’s center?

The intensity of light measured in foot-candles varies inversely with the square of the distance from the light source. Suppose the intensity of a light bulb is 0.08 foot-candles at a distance of 3 meters. Find the intensity level at 8 meters.

The current in a circuit varies inversely with its resistance measured in ohms. When the current in a circuit is 40 amperes, the resistance is 10 ohms. Find the current if the resistance is 12 ohms.

The force exerted by the wind on a plane surface varies jointly with the square of the velocity of the wind and with the area of the plane surface. If the area of the surface is 40 square feet surface and the wind velocity is 20 miles per hour, the resulting force is 15 pounds. Find the force on a surface of 65 square feet with a velocity of 30 miles per hour.

The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute (rpm)) and the cube of the diameter. If the shaft of a certain material 3 inches in diameter can transmit 45 hp at 100 rpm, what must the diameter be in order to transmit 60 hp at 150 rpm?

The kinetic energy K K of a moving object varies jointly with its mass m m and the square of its velocity v . v . If an object weighing 40 kilograms with a velocity of 15 meters per second has a kinetic energy of 1000 joules, find the kinetic energy if the velocity is increased to 20 meters per second.

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• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Precalculus 2e
• Publication date: Dec 21, 2021
• Location: Houston, Texas
• Book URL: https://openstax.org/books/precalculus-2e/pages/1-introduction-to-functions
• Section URL: https://openstax.org/books/precalculus-2e/pages/3-9-modeling-using-variation

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