solving direct and inverse variation problems

Module 3: Algebra, Geometry, and Modeling

3.6 direct and inverse variation, learning objectives.

Solve direct variation problems.
Solve inverse variation problems.

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.

Direct Variation

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, [latex]s[/latex]. If we create a table, we observe that as the sales price increases, the earnings increase as well.

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.

The graph below 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}[/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].

Graph of y=(0.16)x where the horizontal axis is labeled,

A General Note: Direct Variation

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

[latex]y=k{x}[/latex]

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

How to Solve a Direct Variation Problem

First : Find the constant of variation , by writing the model (the equation) for direct variation: [latex]y=k{x}[/latex]

Plug the given [latex]y[/latex] and [latex]x[/latex] into the model.

Solve for [latex]k[/latex], which is the constant of variation .

Second : Write the equation of variation by plugging the constant of variation into the equation [latex]y=k{x}[/latex].

Third : Substitute the remaining [latex]x[/latex] value into the equation of variation.

Simplify to find [latex]y[/latex].

Example: Solving a Direct Variation Problem

The quantity [latex]y[/latex] varies directly with [latex]x[/latex]. If [latex]y=15[/latex] when [latex]x=6[/latex], find [latex]y[/latex] when [latex]x[/latex] is 10.

First : Find the constant of variation , by writing the model (the equation) for direct variation.

Plug [latex]y=15[/latex] and [latex]x=6[/latex] into the model.

[latex]15=k{(6)}[/latex]

Solve for [latex]k[/latex] by dividing both sides of the equation by 6.

[latex]\dfrac{15}{6}[/latex] = [latex]\dfrac{k(6)}{6}[/latex]

[latex]2.5=k{}[/latex]

So, the constant of variation is [latex]k=2.5[/latex].

Second : Write the equation of variation by plugging the constant of variation (in this case, [latex]k=2.5[/latex] ) into the equation [latex]y=k{x}[/latex].

[latex]y=2.5{x}[/latex]

Third : Substitute the remaining value (in this case, [latex]x=10[/latex]) into the equation of variation.

[latex]y=2.5{(10)}[/latex]

[latex]y=25[/latex]

Therefore, if [latex]y[/latex] varies directly with [latex]x[/latex], and if [latex]y=15[/latex] when [latex]x=6[/latex], then [latex]y=25[/latex] when [latex]x[/latex] is 10.

Watch this video to see a quick lesson in direct variation. You will see more worked examples.

1) The quantity [latex]y[/latex] varies directly with [latex]x[/latex]. If [latex]y=16.8[/latex] when [latex]x=2[/latex], find [latex]y[/latex] when [latex]x[/latex] is 3.

1a) Find the constant of variation.

1b) Find the equation of variation.

1c) Find y when [latex]x=3[/latex].

2) The quantity [latex]y[/latex] varies directly with [latex]x[/latex]. If [latex]y=20.7[/latex] when [latex]x=18[/latex], find [latex]y[/latex] when [latex]x[/latex] is 11.

2a) Find the constant of variation.

Show Answer

2b) Find the equation of variation.

2c) Find y when [latex]x=11[/latex].

3) The cost to fill your car’s gas tank varies directly with the number of gallons you put in your tank. Let [latex]x[/latex] be the number of gallons you put in the tank, and let [latex]y[/latex] be the cost, in dollars, of the gasoline. Suppose that the car in front of you pumped 12 gallons of gas. You were able to see that the pump said $36.48. You are planning on pumping 15 gallons of gas into your car.

3a) Find the constant of variation for this scenario.

3b) What does that constant of variation represent in this scenario?

3c) Find the equation of variation for this scenario.

3d) Find the amount you will pay for pumping 15 gallons of gas into your car.

Inverse Variation

Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula [latex]T=\dfrac{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 a table 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, the graph 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=\dfrac{k}{x}[/latex] for inverse variation in this case uses [latex]k=14,000[/latex].

Graph of y=(14000)/x where the horizontal axis is labeled,

A General Note: Inverse Variation

[latex]y=\dfrac{k}{{x}}[/latex]

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

How to Solve an Inverse Variation Problem

First : Find the constant of variation , by writing the model (the equation) for inverse variation: [latex]y=\dfrac{k}{{x}}[/latex]

Second : Write the equation of variation by plugging the constant of variation into the equation [latex]y=\dfrac{k}{{x}}[/latex].

The following video presents a short lesson on inverse variation and includes more worked examples.

Example: Solving an Inverse Variation Problem

A quantity [latex]y[/latex] varies inversely with [latex]x[/latex]. If [latex]y=4.65[/latex] when [latex]x=2[/latex], find [latex]y[/latex] when [latex]x[/latex] is 5.

First : Find the constant of variation , by writing the model (the equation) for inverse variation.

Plug [latex]y=4.65[/latex] and [latex]x=2[/latex] into the model.

[latex]4.65=\dfrac{k}{2}[/latex]

Solve for [latex]k[/latex] by multiplying both sides of the equation by 2.

[latex]4.65×2=\dfrac{k}{2}[/latex]x2

[latex]9.3=k{}[/latex]

So, the constant of variation is [latex]k=9.3[/latex].

Second : Write the equation of variation by plugging the constant of variation (in this case, [latex]k=9.3[/latex] ) into the equation [latex]y=\dfrac{k}{{x}}[/latex]

[latex]y=\dfrac{9.3}{{x}}[/latex]

Third : Substitute the remaining value (in this case, [latex]x=5[/latex]) into the equation of variation.

[latex]y=\dfrac{9.3}{{5}}[/latex]

[latex]y=1.86[/latex]

Therefore, if [latex]y[/latex] varies inversely with [latex]x[/latex], and [latex]y=4.65[/latex] when [latex]x=2[/latex], then [latex]y=1.86[/latex] when [latex]x[/latex] is 5.

4) The quantity [latex]y[/latex] varies inversely with [latex]x[/latex]. If [latex]y=4.8[/latex] when [latex]x=55[/latex], find [latex]y[/latex] when [latex]x[/latex] is 8.

4a) Find the constant of variation.

4b) Find the equation of variation.

4c) Find y when [latex]x=8[/latex].

5) The quantity [latex]y[/latex] varies inversely with [latex]x[/latex]. If [latex]y=11.25[/latex] when [latex]x=4[/latex], find [latex]y[/latex] when [latex]x[/latex] is 6.

5a) Find the constant of variation.

5b) Find the equation of variation.

5c) Find y when [latex]x=6[/latex].

6) The time it takes to put up a fence varies inversely with the number of people working on the fence. Let [latex]x[/latex] be the number of people working on the fence, and let [latex]y[/latex] time it takes to complete the fence, in hours. Suppose your neighbor’s fence had 5 people working on it and it took 14 hours. You know that you will have 4 people to work on your fence of the same length. You are wondering how long it will take to complete your fence.

6a) Find the constant of variation for this scenario.

6b) What does that constant of variation represent in this scenario?

6c) Find the equation of variation for this scenario.

6d) Find the amount of time that it will take 4 people to finish your fence.

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

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solving direct and inverse variation problems

Inverse Variation: Practice Problems

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If $$ y $$ varies inversely as $$ x $$ and $$ y = 14 $$ when $$ x = 20 $$ . What is the value of $$ y $$ when $$ x $$ is 5?

Write down the variation equation.

$$ y = \frac k x $$

Use $$ \red{y = 14} $$ and $$ \blue{x = 20} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{14} & = \frac k {\blue{20}}\\[6pt] \red{14}(\blue{20}) & = \frac k {\blue{20}}\cdot \blue {20}\\[6pt] 280 & = \frac k {\cancelred{20}}\cdot \cancelred{20}\\[6pt] k & = 280 \end{align*} $$

The updated variation equation is $$ y = \frac{280} x $$

Determine the value of $$ y $$ when $$ \blue{x = 5} $$ .

$$ \begin{align*} y & = \frac{280}{\blue x}\\[6pt] y & = \frac{280}{\blue 5} = 56 \end{align*} $$

$$ y = 56 $$ when $$ x = 5 $$ .

Suppose $$ s $$ varies indirectly as $$ t $$, and $$ s(1.5)=25 $$ . Determine the value of $$ s(2.25) $$ .

$$ s = \frac k t $$

Use $$ \red{s = 25} $$ and $$ \blue{t = 1.5} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{25} & = \frac k {\blue{1.5}}\\[6pt] \red{25}(\blue{1.5}) & = \frac k {\blue{1.5}}\cdot \blue{1.5}\\[6pt] 27.5 & = \frac k {\cancelred{1.5}}\cdot \cancelred{1.5}\\[6pt] k & = 37.5 \end{align*} $$

The updated variation equation is $$ s = \frac{37.5} t $$

Determine the value of $$ s $$ when $$ \blue{t = 2.25} $$ .

$$ \begin{align*} s & = \frac{37.5}{\blue t}\\[6pt] s & = \frac{37.5}{\blue{2.25}} = \frac{3750}{225} = \frac{50} 3 = 16\,\frac 2 3 \approx 16.667 \end{align*} $$

$$ s = \frac{50} 3 $$ when $$ t = 2.25 $$ .

Suppose $$ y $$ varies inversely as $$ x $$, and $$ y = 0.025 $$ when $$ x = 2 $$ . What is the value of $$ x $$ when $$ y = 0.4 $$ ?

Use $$ \red{y = 0.025} $$ and $$ \blue{x = 2} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{0.025} & = \frac k {\blue 2}\\[6pt] \red{(0.025)}\blue 2 & = \frac k {\blue 2} \cdot \blue 2\\[6pt] 0.05 & = \frac k {\cancelred 2} \cdot \cancelred 2\\[6pt] k & = 0.05 = \frac 1 {20} \end{align*} $$

The updated variation equation is $$ y = \frac{0.05} x = \frac{1/20} x = \frac 1 {20x} $$

Determine the value of $$ x $$ when $$ \red{y = 0.4} $$ .

$$ \begin{align*} \red{0.4} & = \frac 1 {20x}\\[6pt] \red{0.4} \cdot (20x) & = 1\\[6pt] 8x & = 1\\[6pt] x & = \frac 1 8 = 0.125 \end{align*} $$

$$ x = 0.125 $$ when $$ y = 0.4 $$ .

Suppose $$ I $$ is inversely proportional to $$ R $$ and when $$ R = 200 $$ , $$ I = 35 $$ . Determine the value of $$ R $$ when $$ I = 100 $$ .

$$ I = \frac k R $$

Use $$ \red{I = 35} $$ and $$ \blue{R = 200} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{35} & = \frac k {\blue{200}}\\[6pt] \red{35}(\blue{200}) & = \frac k {\blue{200}}\cdot \blue{200}\\[6pt] 7000 & = \frac k {\cancelred{200}}\cdot \cancelred{200}\\[6pt] k & = 7000 \end{align*} $$

The updated variation equation is $$ I = \frac{7000} R $$ .

Determine the value of $$ R $$ when $$ \red{I = 100} $$ .

$$ \begin{align*} \red{100} & = \frac{7000} R\\[6pt] 100R & = 7000\\[6pt] R & = \frac{7000}{100} = 70 \end{align*} $$

$$ R = 70 $$ when $$ I = 100 $$ .

Suppose $$ y $$ varies inversely as the square-root of $$ x $$. If $$ y = 8 $$ when $$ x = 49 $$ , determine the value of $$ y $$ when $$ x = 121 $$

$$ y = \frac k {\sqrt x} $$

Use $$ \red{y = 8} $$ and $$ \blue{x = 49} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red 8 & = \frac k {\sqrt{\blue{49}}}\\[6pt] \red 8 & = \frac k 7\\[6pt] \red 8 (7) & = \frac k 7\cdot 7\\[6pt] 56 & = \frac k {\cancelred 7}\cdot \cancelred 7\\[6pt] k & = 56 \end{align*} $$

The updated variation equation is $$ y = \frac{56}{\sqrt x} $$

Determine the value of $$ y $$ when $$ \blue{x = 100} $$ .

$$ \begin{align*} y & = \frac{56}{\sqrt{\blue{100}}}\\[6pt] & = \frac{56}{10}\\[6pt] & = 5.6 \end{align*} $$

$$ y = 5.6 $$ when $$ x = 100 $$ .

Suppose $$ p $$ varies inversely as the cube of $$ q $$, and $$ p = 2 $$ when $$ q = 4 $$ . Determine the value of $$ q $$ when $$ p = 8 $$ .

$$ p = \frac k {q^3} $$

Use $$ \red{p = 2} $$ and $$ \blue{q = 4} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red 2 & = \frac k {\blue 4^3}\\[6pt] \red 2 & = \frac k {64}\\[6pt] \red 2 (64) & = \frac k {\cancelred{64}} \cdot \cancelred{64}\\[6pt] k & = 128 \end{align*} $$

The updated variation equation is $$ p = \frac{128}{q^3} $$ .

Determine the value of $$ q $$ when $$ \red{p = 8} $$ .

$$ \begin{align*} \red 8 & = \frac{128}{q^3}\\[6pt] 8q^3 & = 128\\[6pt] q^3 & = \frac{128} 8\\[6pt] & = 16\\[6pt] q & = \sqrt[3]{16} = 2\sqrt[3] 2 \end{align*} $$

$$ q = 2\sqrt[3] 2 $$ when $$ p = 8 $$ .

Word Problems

Speed and travel time are inversely related (the faster you go, the shorter the travel time). Suppose the daily trip to school normally takes 20 minutes when your speed is 30 miles per hour. How long should the trip take if you drive 45 miles per hour instead (and don't get pulled over for speeding)?

Define variables to describe travel time and speed, and state the equation relating them.

Let's use $$ t $$ for time (in minutes).

Let's use $$ s $$ for speed (in miles per hour).

The equation we need then is $$ s = \frac k t $$ .

Use $$ \red{s = 30} $$ and $$ \blue{t = 20} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{30} & = \frac k {\blue{20}}\\[6pt] \red{30}(\blue{20}) & = \frac k {\blue{20}}\cdot\blue{20}\\[6pt] 600 & = \frac k {\cancelred{20}}\cdot\cancelred{20}\\[6pt] k & = 600 \end{align*} $$

The updated variation equation is $$ s = \frac{600} t $$ .

Use the equation to determine the travel time for the trip when speed is 45 miles per hour.

$$ \begin{align*} \red{45} & = \frac{600} t\\[6pt] \red{45}t & = 600\\[6pt] t & = \frac{600}{\red{45}}\\[6pt] & = \frac{40} 3\\[6pt] & = 13\frac 1 3 \end{align*} $$

The trip will take $$ 13\frac 1 3 $$ minutes (or 13 minutes and 20 seconds) if you drive at 45 miles per hour.

The Ideal Gas Law from chemistry can be written as $$ P = \frac k V $$ , where $$ P $$ is the pressure (in pascals) and $$ V $$ is the volume of the container (in cubic meters).

Suppose that when a particular gas is stored in a 0.25 cubic meter container, and exerts 40 pascals of pressure. How much pressure will the gas exert if it were transferred to a container that only holds a volume of $$ 0.1 $$ cubic meters.

Use $$ \red{P = 40} $$ and $$ \blue{V = 0.25} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{P} & = \frac k {\blue{V}}\\[6pt] \red{40} & = \frac k {\blue{0.25}}\\[6pt] \red{40}(\blue{0.25}) & = \frac k {\blue{0.25}}\cdot\blue{0.25}\\[6pt] 10 & = \frac k {\cancelred{0.25}}\cdot\cancelred{0.25}\\[6pt] k & = 10 \end{align*} $$

So for this particular gas, $$ P = \frac{10} V $$

Use the equation to determine the pressure when $$ \blue{V = 0.1} $$ .

$$ P = \frac{10}{\blue{0.1}} = 100 $$

The gas exerts a pressure of 100 pascals in a 0.1 cubic meter container.

The time it takes a block of ice to melt varies indirectly with temperature. If a block of ice that is 1/8 cubic meters in volume requires 6 hours to melt when it is $$72 ^\circ F $$ , at what temperature will the block melt in only 2 hours?

Define variables to represent temperature and time and write down the inverse variation equation.

Let $$ t = $$ time (in hours)

Let $$ T = $$ temperature (in degrees Fahrenheit)

The inverse variation equation will be

$$ t = \frac k T $$

Use $$ \red{t = 6} $$ and $$ \blue{T = 72} $$ to find the value of $$ k $$, then write down the updated version of the inverse variation equation.

$$ \begin{align*} \red t & = \frac k {\blue T}\\[6pt] \red 6 & = \frac k {\blue{72}}\\[6pt] \red 6(\blue{72}) & = \frac k {\blue{72}}\cdot\blue{72}\\[6pt] 432 & = \frac k {\cancelred{72}}\cdot\cancelred{72}\\[6pt] k & = 432 \end{align*} $$

The inverse variation equation is now $$ t = \frac{432} T $$ .

Determine the temperature required to have the block of ice melt in only $$ \blue{t = 2} $$ hours.

$$ \begin{align*} \blue 2 & = \frac{432} T\\[6pt] \blue 2T & = 432\\[6pt] T & = \frac{432}{\blue 2}\\[6pt] & = 216 \end{align*} $$

The 1/8 cubic meter block of ice will melt in only 2 hours if the temperature is $$ 216 ^\circ F $$ .

Suppose 4 adults can build a shed in 16 hours. How much quicker would it be to have 5 adults building the shed?

Confirm this is an inverse variation problem.

We note that (in an ideal world) having more adults working on a particular project will mean the task takes less time.

Since the time decreases as the number of workers increase, we can model this as an inverse variation.

Define variables to represent the quantities involved, and write down the inverse variation equation.

Let $$ w = $$ the number of workers building the shed.

Let $$ t = $$ the number of hours it takes to complete the shed.

The inverse variation equation is

$$ t = \frac k w $$

Use $$ \red{t = 16} $$ and $$ \blue{w = 4} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red t & = \frac k {\red w}\\[6pt] \red{16} & = \frac k {\blue 4}\\[6pt] \red{16}(\blue 4) & = \frac k {\blue 4}\cdot \blue 4\\[6pt] 64 & = \frac k {\cancelred 4}\cdot \cancelred 4\\[6pt] k & = 64 \end{align*} $$

The updated variation equation is $$ t = \frac{64} w $$ .

Determine the length of time required for $$\blue{w = 5}$$ people to build the shed.

$$ t = \frac{64}{\blue 5} = 12\frac 4 5 = 12.8 $$

Determine the savings in time when 5 adults work on the shed versus only 4.

Note that the question didn't ask us, "How long does it take..." , but rather "How much quicker.." . This means we are interested in the difference between having 4 and 5 workers

$$ \mbox{Time saved} = (\mbox{time for 4 workers}) - (\mbox{time for 5 workers}) = 16 - 12.8 = 3.2 $$

Having 5 adults on the project instead of 4 will result in the shed be completed 3.2 hours more quickly (or 3 hours, 12 minutes more quickly).

The gravitational force (in newtons) between two objects is inversely proportional to square of the distance (in meters) between the centers of the objects.

Suppose the gravitational force between two particular objects is $$ 1.95\times 10^4 $$ newtons when they are separated by $$ 8.94\times 10^{12} $$ meters. If the gravitational force increases to $$ 4.1\times 10^6 $$ newtons, what has happened to the distance between the objects?

Define variables to represent the two quantities under investigation. Then write down the variation equation.

Let $$ F_g $$ represent the force due to gravity, measured in newtons.

Let $$ r $$ represent the distance between the centers of the objects, measured in meters.

The variation equation is $$ F_g = \frac k {r^2} $$ .

Use $$ \red{F_g = 1.95\times 10^4} $$ and $$ \blue{r = 8.94\times 10^{12}} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red{F_g} & = \frac k {\blue r^2}\\[6pt] \red{1.95\times 10^4} & = \frac k {(\blue{8.94\times 10^{12}})^2}\\[6pt] \red{1.95\times 10^4} & = \frac k {\blue{79.9236\times 10^{24}}}\\[6pt] \red{1.95\times 10^4}(\blue{79.9236\times 10^{24}}) & = \frac k {\blue{79.9236\times 10^{24}}}\cdot\blue{79.9236\times 10^{24}}\\[6pt] 1.5585102 \times 10^{30} & = \frac k {\cancelred{79.9236\times 10^{24}}}\cdot\cancelred{79.9236\times 10^{24}}\\[6pt] k & = 1.5585102 \times 10^{30}\\[6pt] & \approx 1.56 \times 10^{30} \end{align*} $$

The updated variation equation is $$ \displaystyle F_g = \frac{1.56\times 10^{30}}{r^2} $$ .

Determine the distance between the two objects if the force of gravity is $$ \red{F_g = 4.1\times 10^6} $$ newtons.

$$ \begin{align*} \red{4.1\times 10^6} & = \frac{1.56\times 10^{30}}{r^2}\\[6pt] (\red{4.1\times 10^6})r^2 & = 1.56\times 10^{30}\\[6pt] r^2 & = \frac{1.56\times 10^{30}}{\red{4.1\times 10^6}}\\[6pt] r^2 & = \frac{1.56}{4.1}\times \frac{10^{30}}{10^6}\\[6pt] & = 0.3805\times 10^{24}\\[6pt] r & = \pm\sqrt{0.3805\times 10^{24}}\\[6pt] & = 0.6169\times 10^{12}\\[6pt] & = 6.169 \times 10^{12} \end{align*} $$

The distance decreases to $$ 6.169\times 10^{12} $$ meters.

In the study of electricity, Ohm's Law says the electrical current (measured in amps) across a conductor is inversely proportional to electrical resistance (measured in ohms).

Suppose a particular circuit has a variable resistor that is currently set to 360 ohm, and this results in 1/3 amps of current. If the resistance were decreased to 80 ohms, what would the resulting change in current be?

Define variables to represent the quantities we are investigating, and then write down the appropriate variation equation.

Let $$ I = $$ electrical current, measured in amps.

Let $$ R = $$ electrical resistance, measured in volts.

The equation of variation is $$ I = \frac k R $$ .

Note: This is Ohm's Law, and the constant of proportionality represents the voltage on the circuit.

Use $$ \red{I = \frac 1 3} $$ and $$ \blue{R = 360} $$ to determine the value of $$ k $$. Then rewrite the variation equation.

$$ \begin{align*} \red I & = \frac k {\blue R}\\[6pt] \red{\frac 1 3} & = \frac k {\blue{360}}\\[6pt] \red{\frac 1 3}(\blue{360}) & = \frac k {\blue{360}}\cdot \blue{360}\\[6pt] 120 & = \frac k {\cancelred{360}}\cdot \cancelred{360}\\[6pt] k & = 120 \end{align*} $$

The variation equation is now $$ I = \frac{120} R $$ .

Use the equation to determine the current running through the circuit when the resistance drops to only $$ \blue{R = 80} $$ ohms.

$$ I = \frac{120}{\blue{80}} = \frac 3 2 $$

When the resistance decreases to 80 ohms, the current increases to 1.5 amps.

In economics, the basic Law of Demand tells us that as the price for a particular good (or service) increases, the demand for that good (or service) will decrease. This is an inverse variation relationship.

Suppose a new app is released for cell phones, and at a price of $$ \$4.99 $$ there are 3.2 million downloads each month. Later, the company that produced the app puts it on sale for $$ \$0.99 $$

(a) What will the accompanying change in the number of downloads per month be?

(b) If the wanted to phase out the app (for a newer, better one in development), and wanted to decrease the number of downloads per month to only $$ 500{,}000 $$ , what price should they charge for the app?

Define variables for the quantities we are investigating, then write down the appropriate variation equation.

Let $$ p = $$ the price, in dollars, for which the app is being sold.

Let $$ d = $$ the number of downloads per month, measured in millions of downloads per month.

So, $$ d = 1 $$ represents 1 million downloads per month.

The variation equation is $$ d = \frac k p $$ .

Use $$ \red{d = 3.2} $$ and $$ \blue{4.99} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red d & = \frac k {\blue p}\\[6pt] \red{3.2} & = \frac k {\blue{4.99}}\\[6pt] \red{3.2}(\blue{4.99}) & = \frac k {\blue{4.99}}\cdot \blue{4.99}\\[6pt] 15.968 & = \frac k {\cancelred{4.99}}\cdot \cancelred{4.99}\\[6pt] k & = 15.968 \end{align*} $$

The variation equation becomes $$ d = \frac{15.968} p $$ .

Use the variation equation to determine the number of downloads per month the company can expect from putting the app on sale for $$ \blue{p = \$0.99} $$ .

$$ d = \frac{15.968}{\blue{0.99}} = \frac{1596.8}{99} \approx 16.13 $$

The downloads should increase to approximately 16.13 million per month.

Use the variation equation to determine the price which will cause the number of downloads per month to decrease to only 5 hundred thousand ( $$ \red{d = 1/2} $$ ).

$$ \begin{align*} \red{\frac 1 2} & = \frac{15.968} p\\[6pt] \frac p 2 & = 15.968\\[6pt] p & = 31.936 \end{align*} $$

The price would need to be about $$ \$31.94 $$ to drive down the demand to only half a million.

(a) Demand will increase to about 16.13 million downloads per month.

(b) Increasing the price to about $$ \$31.94 $$ will drive down demand to $$ 500{,}000 $$ downloads per month.

When working with electrical circuits, it turns out that the electrical resistance is inversely proportional to the square of the current.

Suppose a particular circuit has a current measured at $$ 0.05 $$ amps with a resistance of 90 ohms.

(a) What will happen to the resistance if the current drops to only 0.01 amps?

(b) Suppose someone swaps out the 90-ohm resistor for one that is rated at only 60 ohms. What happens to the current?

Define variables for the quantities, then write down the variation equation.

Let $$ R = $$ the electrical resistance measured in ohms.

Let $$ I = $$ the electrical current measured in amps.

The variation equation is $$ R = \frac k {I^2} $$

Note: In a previous question about electrical circuits we mentioned the constant of proportionality was in fact the voltage on the circuit. This time, the constant of proportionality represents the electrical power (in watts).

Use $$ \red{R = 90} $$ and $$ \blue{I = 0.05} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red R & = \frac k {\blue I^2}\\[6pt] \red{90} & = \frac k {(\blue{0.05})^2}\\[6pt] \red{90}\cdot(\blue{0.0025}) & = \frac k {\blue{0.0025}}\cdot (\blue{0.0025})\\[6pt] 0.225 & = \frac k {\cancelred{0.0025}}\cdot (\cancelred{0.0025})\\[6pt] k & = 0.225 \end{align*} $$

The variation equation is now $$ \displaystyle R = \frac{0.225}{I^2} = \frac 9 {40I^2} $$

Determine what happens to the current if the resistance is changed to $$ \red{R = 60} $$ ohms.

$$ \begin{align*} \red{60} & = \frac 9 {40I^2}\\[6pt] \red{60}I^2 & = \frac 9 {40}\\[6pt] I^2 & = \frac 9 {2400}\\[6pt] I & = \sqrt{\frac 9 {2400}}\\[6pt] & = \frac 3 {20\sqrt 6}\\[6pt] & \approx 0.0612 \end{align*} $$

The current increases to $$ \frac 3 {20\sqrt 6} $$ or approximately 0.0612 amps.

Suppose an object is falling toward the earth, and it's speed is inversely proportional to the square root of the distance from the earth's surface.

Also, suppose the object's speed is 60 meters/second when it is 900 meters above the earth's surface. How close will the object be when it's velocity reaches 100 meters/second?

Define variables for the quantities we are interested in, then write down the variation equation.

$$ v = $$ velocity, in meters/second

$$ d = $$ distance from the earth's surface, in meters.

The variation equation is $$ \displaystyle v = \frac k {\sqrt d} $$ .

Use $$ \red{v = 60} $$ and $$ \blue{d =900} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red v & = \frac k {\sqrt{\blue d}}\\[6pt] \red{60} & = \frac k {\sqrt{\blue{900}}}\\[6pt] \red{60}(\blue{30}) & = \frac k {\blue{30}}\cdot\blue{30}\\[6pt] 1800 & = \frac k {\cancelred{30}}\cdot\cancelred{30}\\[6pt] k & = 1800 \end{align*} $$

The variation equation is now $$ v = \frac{1800}{\sqrt d} $$

Use the equation to determine the distance of the object from the earth's surface when it's velocity reaches $$ \red{v = 100} $$ meters per second.

$$ \begin{align*} \red{100} & = \frac{1800}{\sqrt d}\\[6pt] \sqrt d & = \frac{1800}{\red{100}} = 18\\[6pt] d & = 18^2 = 324 \end{align*} $$

The object is only 324 meters above the earth's surface when it's velocity reaches 100 meters per second.

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Direct and Inverse Variation Worksheets

Direct and inverse variation worksheets are designed for high schoolers and are divided into subtopics like identifying the type of variation by observing equations, graphs and tables, finding the constant of variation, and much more. The worksheets provide dual levels, level 1 deals with direct and inverse variations, while level 2 deals with direct, inverse, joint and combined variation. A prior knowledge of proportions will definitely be an added advantage. Kick-start your practice with our free worksheets!

Recognize Direct and Inverse Variation

The quintet multiple response pdf worksheets have exercises for learners to observe equations, graphs and recognize the type of variation as direct (linear graph) or inverse (rectangular hyperbola).

Direct and Inverse Variation - Equation

Equations representing the direct variation are in the form y = kx and inverse variation is in the form xy = k. Identify the type of variation in the equations featured in these printable worksheets. Also, find the constant of variation (k).

Direct and Inverse Variation - Table

State whether the table represents a direct variation or inverse variation or neither of them. Find the constant of variation as well.

Constant of Variation - Equation

These high school pdf worksheets feature exercises on finding the constant of variation (k) from the given equations, packed into two levels for easier navigation.

Constant of Variation with the given Values

Find the constant of variation (k) based on the values of the variables in the given statement. Validate your responses with our answer keys.

Write the Equation

Find the constant of variations and translate the statements into equations.

Missing Values

Navigate through the printable worksheets here to formulate an equation relating the variables to find the value of the missing entity.

Complete the Table

This set of pdf worksheets consist of exercises in tabular format. Find the constant of variation (k) and complete the table.

Direct and Inverse Variation Word Problems

Find 70+ word problems, where the student learns to apply the concept of variation to real-life scenarios.

(15 Worksheets)

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ACT Math : Direct and Inverse Variation

Study concepts, example questions & explanations for act math, all act math resources, example questions, example question #9 : variables.

The price of silver varies directly as the square of the mass. If 3.6 g of silver is worth $64.80, what is the value of 7.5 g of silver?

This is a direct variation problem of the form y = kx 2 The first set of data 3.6 g and $64.80 is used to calculate the proportionality constant, k. So 64.80 = k(3.6) 2 and solving the equation gives k = 5.

Now we move to the new data, 7.5 g and we get y = 5(7.5) 2 to yield an answer of $218.25.

$135.00 is the answer obtained if using proportions. This is an error because it does not take into consideration the squared elements of the problem.