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Calculus 1 : Rate of Change

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Example Questions

Example Question #291 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's surface area to the rate of growth of its diagonal when its sides have length $3\sqrt{5}$ $\displaystyle 3\sqrt{5}$ ?

Possible Answers:

$12\sqrt{15}$ $\displaystyle 12\sqrt{15}$

$60\sqrt{3}$ $\displaystyle 60\sqrt{3}$

$12\sqrt{5}$ $\displaystyle 12\sqrt{5}$

$3\sqrt{15}$ $\displaystyle 3\sqrt{15}$

$6\sqrt{5}$ $\displaystyle 6\sqrt{5}$

Correct answer:

$12\sqrt{15}$ $\displaystyle 12\sqrt{15}$

Explanation:

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

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

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$ $\displaystyle \frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=4s\sqrt{3}$ $\displaystyle \phi=4s\sqrt{3}$

$\phi =12\sqrt{15}$ $\displaystyle \phi =12\sqrt{15}$

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Example Question #292 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its diagonal when its sides have length 1.81?

Possible Answers:

12.54 $\displaystyle 12.54$

3.13 $\displaystyle 3.13$

10.27 $\displaystyle 10.27$

2.08 $\displaystyle 2.08$

8.20 $\displaystyle 8.20$

Correct answer:

12.54 $\displaystyle 12.54$

Explanation:

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

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

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$ $\displaystyle \frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=4s\sqrt{3}$ $\displaystyle \phi=4s\sqrt{3}$

$\phi =12.54$ $\displaystyle \phi =12.54$

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Example Question #293 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its diagonal when its sides have length 7.82?

Possible Answers:

54.18 $\displaystyle 54.18$

60.84 $\displaystyle 60.84$

12.32 $\displaystyle 12.32$

13.55 $\displaystyle 13.55$

49.26 $\displaystyle 49.26$

Correct answer:

54.18 $\displaystyle 54.18$

Explanation:

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

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

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$ $\displaystyle \frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=4s\sqrt{3}$ $\displaystyle \phi=4s\sqrt{3}$

$\phi =54.18$ $\displaystyle \phi =54.18$

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Example Question #294 : Rate Of Change

A spherical balloon is being filled with air. What is the volume of the sphere at the instance the rate of growth of the volume is 1.65 times the rate of growth of the surface area?

Possible Answers:

218.993 $\displaystyle 218.993$

118.818 $\displaystyle 118.818$

232.724 $\displaystyle 232.724$

78.912 $\displaystyle 78.912$

150.533 $\displaystyle 150.533$

Correct answer:

150.533 $\displaystyle 150.533$

Explanation:

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

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

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

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 1.65 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(1.65)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(1.65)8\pi r \frac{dr}{dt}$

$\displaystyle r =(1.65)2$

r=3.3 $\displaystyle r=3.3$

Then to find the volume:

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

$V=\frac{4}{3}\pi (3.3)^3=150.533$ $\displaystyle V=\frac{4}{3}\pi (3.3)^3=150.533$

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Example Question #295 : Rate Of Change

A spherical balloon is being filled with air. What is the diameter of the sphere at the instance the rate of growth of the volume is 2.08 times the rate of growth of the surface area?

Possible Answers:

26.14 $\displaystyle 26.14$

4.16 $\displaystyle 4.16$

8.32 $\displaystyle 8.32$

13.07 $\displaystyle 13.07$

11.22 $\displaystyle 11.22$

Correct answer:

8.32 $\displaystyle 8.32$

Explanation:

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

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

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

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 2.08 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(2.08)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(2.08)8\pi r \frac{dr}{dt}$

$\displaystyle r =(2.08)2$

r=4.16 $\displaystyle r=4.16$

The diameter is thus

d=8.32 $\displaystyle d=8.32$

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Example Question #296 : Rate Of Change

A spherical balloon is being filled with air. What is the circumference of the sphere at the instance the rate of growth of the volume is 7.19 times the rate of growth of the surface area?

Possible Answers:

103.22 $\displaystyle 103.22$

40.18 $\displaystyle 40.18$

90.35 $\displaystyle 90.35$

14.38 $\displaystyle 14.38$

51.62 $\displaystyle 51.62$

Correct answer:

90.35 $\displaystyle 90.35$

Explanation:

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

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

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

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 7.19 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(7.19)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(7.19)8\pi r \frac{dr}{dt}$

$\displaystyle r =(7.19)2$

r=14.38 $\displaystyle r=14.38$

The circumference is then

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

C=90.35 $\displaystyle C=90.35$

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Example Question #297 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the surface area of the sphere at the instance the rate of shrinkage of the volume is 10.13 times the rate of shrinkage of the surface area?

Possible Answers:

2729.43 $\displaystyle 2729.43$

1289.52 $\displaystyle 1289.52$

6343.81 $\displaystyle 6343.81$

5158.09 $\displaystyle 5158.09$

4596.03 $\displaystyle 4596.03$

Correct answer:

5158.09 $\displaystyle 5158.09$

Explanation:

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

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

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

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of shrinkage of the volume is 10.13 times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(10.13)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(10.13)8\pi r \frac{dr}{dt}$

$\displaystyle r =(10.13)2$

r=20.26 $\displaystyle r=20.26$

Then to find the surface area:

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

$\displaystyle A=5158.09$

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Example Question #298 : Rate Of Change

A spherical balloon is being filled with air. What is the volume of the sphere at the instance the rate of growth of the volume is 128 times the rate of growth of the circumference?

Possible Answers:

$\frac{4096}{3}\pi$ $\displaystyle \frac{4096}{3}\pi$

$\frac{256}{3}\pi$ $\displaystyle \frac{256}{3}\pi$

$\frac{2048}{3}\pi$ $\displaystyle \frac{2048}{3}\pi$

$\frac{512}{3}\pi$ $\displaystyle \frac{512}{3}\pi$

$\frac{1024}{3}\pi$ $\displaystyle \frac{1024}{3}\pi$

Correct answer:

$\frac{2048}{3}\pi$ $\displaystyle \frac{2048}{3}\pi$

Explanation:

Begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

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

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

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now given the problem information, the rate of growth of the volume is 128 times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(128)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(128)2\pi \frac{dr}{dt}$

$r^2=(128)\frac{1}{2}$ $\displaystyle r^2=(128)\frac{1}{2}$

$r =\sqrt{64}$ $\displaystyle r =\sqrt{64}$

r=8 $\displaystyle r=8$

Then to find the volume:

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

$V=\frac{2048}{3}\pi$ $\displaystyle V=\frac{2048}{3}\pi$

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Example Question #299 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the circumference of the sphere at the instance the rate of shrinkage of the volume is seven times the rate of shrinkage of the circumference?

Possible Answers:

$7\pi\sqrt{2}$ $\displaystyle 7\pi\sqrt{2}$

$\pi\sqrt{14}$ $\displaystyle \pi\sqrt{14}$

$2\pi\sqrt{3}$ $\displaystyle 2\pi\sqrt{3}$

$\pi\sqrt{7}$ $\displaystyle \pi\sqrt{7}$

Correct answer:

$\pi\sqrt{14}$ $\displaystyle \pi\sqrt{14}$

Explanation:

Begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

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

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

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now given the problem information, the rate of shrinkage of the volume is seven times the rate of shrinkage of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(7)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(7)2\pi \frac{dr}{dt}$

$r^2=(7)\frac{1}{2}$ $\displaystyle r^2=(7)\frac{1}{2}$

$r =\sqrt{\frac{7}{2}}$ $\displaystyle r =\sqrt{\frac{7}{2}}$

The circumference is then

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

C=11.75 $\displaystyle C=11.75$

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Example Question #300 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the surface area of the sphere at the instance the rate of shrinkage of the volume is fifteen times the rate of shrinkage of the circumference?

Possible Answers:

$225\pi$ $\displaystyle 225\pi$

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

$25\pi$ $\displaystyle 25\pi$

$30\pi$ $\displaystyle 30\pi$

$700\pi$ $\displaystyle 700\pi$

Correct answer:

$30\pi$ $\displaystyle 30\pi$

Explanation:

Begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

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

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

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now given the problem information, the rate of shrinkage of the volume is fifteen times the rate of shrinkage of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(15)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(15)2\pi \frac{dr}{dt}$

$r^2=(15)\frac{1}{2}$ $\displaystyle r^2=(15)\frac{1}{2}$

$r =\sqrt{\frac{15}{2}}$ $\displaystyle r =\sqrt{\frac{15}{2}}$

The surface area is then:

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

$A=30\pi$ $\displaystyle A=30\pi$

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #291 : Rate Of Change

Example Question #292 : Rate Of Change

Example Question #293 : Rate Of Change

Example Question #294 : Rate Of Change

Example Question #295 : Rate Of Change

Example Question #296 : Rate Of Change

Example Question #297 : Rate Of Change

Example Question #298 : Rate Of Change

Example Question #299 : Rate Of Change

Example Question #300 : Rate Of Change

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