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Calculus 1 : How to find rate of change

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Example Question #3352 : Calculus

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is 65?

Possible Answers:

260 $\displaystyle 260$

16.25 $\displaystyle 16.25$

$\displaystyle 65$

32.5 $\displaystyle 32.5$

130 $\displaystyle 130$

Correct answer:

260 $\displaystyle 260$

Explanation:

Let's begin by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$C=2\pi r$ $\displaystyle C=2\pi r$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ $\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the surface area and circumference, divide:

$8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =4r$ $\displaystyle \phi =4r$

$\phi =4(65)=260$ $\displaystyle \phi =4(65)=260$

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Example Question #3353 : Calculus

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is $3\pi$ $\displaystyle 3\pi$?

Possible Answers:

$12\pi$ $\displaystyle 12\pi$

$6\pi$ $\displaystyle 6\pi$

$9\pi$ $\displaystyle 9\pi$

$15\pi$ $\displaystyle 15\pi$

$3\pi$ $\displaystyle 3\pi$

Correct answer:

$12\pi$ $\displaystyle 12\pi$

Explanation:

Let's begin by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$C=2\pi r$ $\displaystyle C=2\pi r$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ $\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

$8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =4r$ $\displaystyle \phi =4r$

$\phi =4(3\pi)=12\pi$ $\displaystyle \phi =4(3\pi)=12\pi$

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Example Question #3354 : Calculus

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the surface area of the sphere to the rate of loss of the circumference when the radius is $\frac{1}{16}$ $\displaystyle \frac{1}{16}$?

Possible Answers:

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\frac{1}{16}$ $\displaystyle \frac{1}{16}$

$\displaystyle 16$

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

$\displaystyle 4$

Correct answer:

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

Explanation:

Let's begin by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$C=2\pi r$ $\displaystyle C=2\pi r$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ $\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

$8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =4r$ $\displaystyle \phi =4r$

$\phi =4(\frac{1}{16})=\frac{1}{4}$ $\displaystyle \phi =4(\frac{1}{16})=\frac{1}{4}$

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Example Question #3355 : Calculus

Possible Answers:

1220 $\displaystyle 1220$

3660 $\displaystyle 3660$

915 $\displaystyle 915$

305 $\displaystyle 305$

1830 $\displaystyle 1830$

Correct answer:

3660 $\displaystyle 3660$

Explanation:

Let's begin by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$C=2\pi r$ $\displaystyle C=2\pi r$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ $\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

$8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =4r$ $\displaystyle \phi =4r$

$\phi =4(915)=3660$ $\displaystyle \phi =4(915)=3660$

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Example Question #3356 : Calculus

Possible Answers:

$\frac{\pi}{8}$ $\displaystyle \frac{\pi}{8}$

$\frac{2}{\pi}$ $\displaystyle \frac{2}{\pi}$

$\frac{\pi}{2}$ $\displaystyle \frac{\pi}{2}$

$\frac{32}{\pi}$ $\displaystyle \frac{32}{\pi}$

$\frac{16}{\pi}$ $\displaystyle \frac{16}{\pi}$

Correct answer:

$\frac{32}{\pi}$ $\displaystyle \frac{32}{\pi}$

Explanation:

Let's begin by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$C=2\pi r$ $\displaystyle C=2\pi r$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ $\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

$8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =4r$ $\displaystyle \phi =4r$

$\phi =4(\frac{8}{\pi})=\frac{32}{\pi}$ $\displaystyle \phi =4(\frac{8}{\pi})=\frac{32}{\pi}$

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Example Question #3357 : Calculus

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the radius when the radius is 0.5?

Possible Answers:

$\frac{1}{\pi}$ $\displaystyle \frac{1}{\pi}$

$\frac{1}{\pi^2}$ $\displaystyle \frac{1}{\pi^2}$

$\displaystyle 1$

$\pi$ $\displaystyle \pi$

$\pi^2$ $\displaystyle \pi^2$

Correct answer:

$\pi$ $\displaystyle \pi$

Explanation:

Let's begin by writing the equation for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$ $\displaystyle V=\frac{4}{3}\pi r^3$

The rate of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$ $\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and radius, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$

$\phi =4\pi r^2$ $\displaystyle \phi =4\pi r^2$

$\phi =4\pi(0.5)^2=\pi$ $\displaystyle \phi =4\pi(0.5)^2=\pi$

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Example Question #441 : How To Find Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length $\frac{1}{4}$ $\displaystyle \frac{1}{4}$?

Possible Answers:

$\displaystyle 4$

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

$\displaystyle 1$

$\frac{1}{16}$ $\displaystyle \frac{1}{16}$

$\displaystyle 16$

Correct answer:

$\frac{1}{16}$ $\displaystyle \frac{1}{16}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dA}{dt}=12s\frac{ds}{dt}$ $\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

$3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$

$\phi=\frac{s}{4}$ $\displaystyle \phi=\frac{s}{4}$

$\phi =\frac{(\frac{1}{4})}{4}=\frac{1}{16}$ $\displaystyle \phi =\frac{(\frac{1}{4})}{4}=\frac{1}{16}$

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Example Question #441 : How To Find Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 68?

Possible Answers:

272 $\displaystyle 272$

$\displaystyle 34$

$\displaystyle 17$

136 $\displaystyle 136$

8.5 $\displaystyle 8.5$

Correct answer:

$\displaystyle 17$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dA}{dt}=12s\frac{ds}{dt}$ $\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

$3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$

$\phi=\frac{s}{4}$ $\displaystyle \phi=\frac{s}{4}$

$\phi =\frac{68}{4}=17$ $\displaystyle \phi =\frac{68}{4}=17$

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Example Question #442 : How To Find Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 34?

Possible Answers:

$\frac{17}{2}$ $\displaystyle \frac{17}{2}$

$\displaystyle 34$

$\displaystyle 17$

$\displaystyle 68$

$\frac{17}{4}$ $\displaystyle \frac{17}{4}$

Correct answer:

$\frac{17}{2}$ $\displaystyle \frac{17}{2}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dA}{dt}=12s\frac{ds}{dt}$ $\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

$3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$

$\phi=\frac{s}{4}$ $\displaystyle \phi=\frac{s}{4}$

$\phi =\frac{34}{4}=\frac{17}{2}$ $\displaystyle \phi =\frac{34}{4}=\frac{17}{2}$

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Example Question #441 : How To Find Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length 96?

Possible Answers:

$\displaystyle 12$

$\displaystyle 60$

$\displaystyle 36$

$\displaystyle 24$

$\displaystyle 48$

Correct answer:

$\displaystyle 24$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dA}{dt}=12s\frac{ds}{dt}$ $\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

$3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$

$\phi=\frac{s}{4}$ $\displaystyle \phi=\frac{s}{4}$

$\phi =\frac{96}{4}=24$ $\displaystyle \phi =\frac{96}{4}=24$

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Calculus 1 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3352 : Calculus

Example Question #3353 : Calculus

Example Question #3354 : Calculus

Example Question #3355 : Calculus

Example Question #3356 : Calculus

Example Question #3357 : Calculus

Example Question #441 : How To Find Rate Of Change

Example Question #441 : How To Find Rate Of Change

Example Question #442 : How To Find Rate Of Change

Example Question #441 : How To Find Rate Of Change

All Calculus 1 Resources

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