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Calculus 1 : Rate of Change

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Example Questions

Example Question #411 : 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 $\frac{3}{\pi}$ times the rate of shrinkage of the surface area?

Possible Answers:

$\frac{12}{\pi^2}$

$\frac{12}{\pi}$

$\frac{\pi}{12}$

$12\pi^2$

Correct answer:

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 $\frac{3}{\pi}$ times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{3}{\pi})8\pi r \frac{dr}{dt}$

$r =(\frac{3}{\pi})2$

$r=\frac{6}{\pi}$

The circumference of a sphere is:

$C=2\pi r$

$C=2\pi(\frac{6}{\pi})$

C=12

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Example Question #412 : 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 0.16 times the rate of growth of the surface area?

Possible Answers:

0.070

0.039

0.314

0.137

0.282

Correct answer:

0.137

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 0.16 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(0.16)8\pi r \frac{dr}{dt}$

r =(0.16)2

r=0.32

Finding the volume:

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

$V=\frac{4}{3}\pi (0.32)^3=0.137$

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Example Question #413 : Rate Of Change

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the volume is 12.5 times the rate of growth of the surface area?

Possible Answers:

$50\pi$

$75\pi$

$25\pi$

Correct answer:

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 12.5 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(12.5)8\pi r \frac{dr}{dt}$

r =(12.5)2

r=25

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Example Question #414 : Rate Of Change

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the volume is 98 times the rate of growth of the circumference?

Possible Answers:

$\frac{7}{2}$

$\sqrt{7}$

$\sqrt{\frac{7}{2}}$

Correct answer:

Explanation:

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

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

$C=2\pi r^$

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

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

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

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 98 times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(98)2\pi \frac{dr}{dt}$

$r^2=(98)\frac{1}{2}$

$r =\sqrt{\frac{98}{2}}=\sqrt{49}=7$

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Example Question #415 : 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 72 times the rate of growth of the circumference?

Possible Answers:

$144\pi$

$288\pi$

288

$576\pi$

144

Correct answer:

$288\pi$

Explanation:

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

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

$C=2\pi r^$

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

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

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

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 72 times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(72)2\pi \frac{dr}{dt}$

$r^2=(72)\frac{1}{2}$

$r =\sqrt{\frac{72}{2}}=\sqrt{36}=6$

Finding the volume at this time:

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

$V=\frac{4}{3}\pi (6)^3=288\pi$

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Example Question #416 : 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 338 times the rate of shrinkage of the circumference?

Possible Answers:

$13\pi$

$26\pi$

Correct answer:

$26\pi$

Explanation:

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

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

$C=2\pi r^$

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

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

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

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 338 times the rate of shrinkage of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(338)2\pi \frac{dr}{dt}$

$r^2=(338)\frac{1}{2}$

$r =\sqrt{\frac{338}{2}}=\sqrt{169}=13$

The circumference is then:

$C=2\pi r$

$C=2\pi (13)=26\pi$

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Example Question #417 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the diameterof the sphere at the instance the rate of shrinkage of the volume is 450 times the rate of shrinkage of the circumference?

Possible Answers:

Correct answer:

Explanation:

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

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

$C=2\pi r^$

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

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

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

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 450 times the rate of shrinkage of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(450)2\pi \frac{dr}{dt}$

$r^2=(450)\frac{1}{2}$

$r =\sqrt{\frac{450}{2}}=\sqrt{225}=15$

The diameter is then:

d=2r

d=30

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Example Question #3331 : Calculus

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the surface area is 312 times the rate of growth of the circumference?

Possible Answers:

19.5

$19.5\pi$

$39\pi$

Correct answer:

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now we can use the relation given in the problem statement, the rate of growth of the surface area is 312 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(312)2\pi \frac{dr}{dt}$

$r =\frac{312}{4}$

r=78

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Example Question #3332 : Calculus

A spherical balloon is being filled with air. What is the ciircumference of the sphere at the instance the rate of growth of the surface area is 56 times the rate of growth of the circumference?

Possible Answers:

$28\pi ^2$

$56\pi^2$

$56\pi$

$28\pi$

$14\pi^2$

Correct answer:

$28\pi$

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now we can use the relation given in the problem statement, the rate of growth of the surface area is 56 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(56)2\pi \frac{dr}{dt}$

$r =\frac{56}{4}=14$

Now to find the circumference:

$C=2\pi r$

$C=28\pi$

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Example Question #3333 : Calculus

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 surface area is 12 times the rate of shrinkage of the circumference?

Possible Answers:

$64\pi$

$24\pi$

$36\pi$

$18\pi$

$32\pi$

Correct answer:

$36\pi$

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now we can use the relation given in the problem statement, the rate of shrinkage of the surface area is 12 times the rate of shrinkage of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(12)2\pi \frac{dr}{dt}$

$r =\frac{12}{4}=3$

Then to find the surface area at this point in time:

$A=4\pi r^2$

$A=4\pi (3)^2=36\pi$

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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 #411 : Rate Of Change

Example Question #412 : Rate Of Change

Example Question #413 : Rate Of Change

Example Question #414 : Rate Of Change

Example Question #415 : Rate Of Change

Example Question #416 : Rate Of Change

Example Question #417 : Rate Of Change

Example Question #3331 : Calculus

Example Question #3332 : Calculus

Example Question #3333 : Calculus

All Calculus 1 Resources

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