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Calculus 1 : Rate of Change

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Example Questions

Example Question #571 : 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 686 times the rate of growth of the circumference?

Possible Answers:

$14\sqrt{7}$

$7\sqrt{14}$

$7\sqrt{7}$

Correct answer:

$7\sqrt{7}$

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

$4\pi r^2 \frac{dr}{dt}=(686)2\pi \frac{dr}{dt}$

$r^2=(686)\frac{1}{2}$

$r =\sqrt{343}=7\sqrt{7}$

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Example Question #572 : 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 250 times the rate of growth of the circumference?

Possible Answers:

$5\sqrt{5}$

$10\sqrt{5}$

$25\sqrt{5}$

Correct answer:

$5\sqrt{5}$

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

$4\pi r^2 \frac{dr}{dt}=(250)2\pi \frac{dr}{dt}$

$r^2=(250)\frac{1}{2}$

$r =\sqrt{125}=5\sqrt{5}$

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Example Question #573 : 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 294 times the rate of growth of the circumference?

Possible Answers:

$7\sqrt{7}$

$14\sqrt{3}$

$3\sqrt{7}$

$9\sqrt{7}$

$7\sqrt{3}$

Correct answer:

$7\sqrt{3}$

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

$4\pi r^2 \frac{dr}{dt}=(294)2\pi \frac{dr}{dt}$

$r^2=(294)\frac{1}{2}$

$r =\sqrt{147}=7\sqrt{3}$

Report an Error

Example Question #573 : How To Find 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 90 times the rate of growth of the circumference?

Possible Answers:

$5\sqrt{3}$

$9\sqrt{5}$

$15\sqrt{3}$

$3\sqrt{5}$

$25\sqrt{3}$

Correct answer:

$3\sqrt{5}$

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

$4\pi r^2 \frac{dr}{dt}=(90)2\pi \frac{dr}{dt}$

$r^2=(90)\frac{1}{2}$

$r =\sqrt{45}=3\sqrt{5}$

Report an Error

Example Question #574 : How To Find 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 104 times the rate of growth of the circumference?

Possible Answers:

$13\sqrt{2}$

$13\sqrt{6}$

$4\sqrt{13}$

$2\sqrt{13}$

$8\sqrt{13}$

Correct answer:

$2\sqrt{13}$

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

$4\pi r^2 \frac{dr}{dt}=(104)2\pi \frac{dr}{dt}$

$r^2=(104)\frac{1}{2}$

$r =\sqrt{52}=2\sqrt{13}$

Report an Error

Example Question #575 : How To Find 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 198 times the rate of growth of the circumference?

Possible Answers:

$11\sqrt{6}$

$3\sqrt{11}$

$11\sqrt{3}$

$9\sqrt{11}$

$6\sqrt{11}$

Correct answer:

$3\sqrt{11}$

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

$4\pi r^2 \frac{dr}{dt}=(198)2\pi \frac{dr}{dt}$

$r^2=(198)\frac{1}{2}$

$r =\sqrt{99}=3\sqrt{11}$

Report an Error

Example Question #576 : How To Find 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 726 times the rate of growth of the circumference?

Possible Answers:

$11\sqrt{6}$

$11\sqrt{3}$

$6\sqrt{11}$

$22\sqrt{6}$

$22\sqrt{3}$

Correct answer:

$11\sqrt{3}$

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

$4\pi r^2 \frac{dr}{dt}=(726)2\pi \frac{dr}{dt}$

$r^2=(726)\frac{1}{2}$

$r =\sqrt{363}=11\sqrt{3}$

Report an Error

Example Question #3491 : 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 volume is 676 times the rate of growth of the circumference?

Possible Answers:

$26\sqrt{2}$

$\sqrt{13}$

$2\sqrt{13}$

$13\sqrt{2}$

$13\sqrt{26}$

Correct answer:

$13\sqrt{2}$

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

$4\pi r^2 \frac{dr}{dt}=(676)2\pi \frac{dr}{dt}$

$r^2=(676)\frac{1}{2}$

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

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Example Question #574 : 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 surface area is 804 times the rate of growth of the circumference?

Possible Answers:

536

134

402

201

Correct answer:

201

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 804 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(804)2\pi \frac{dr}{dt}$

$r =\frac{804}{4}=201$

Report an Error

Example Question #575 : 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 surface area is 728 times the rate of growth of the circumference?

Possible Answers:

364

182

416

Correct answer:

182

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 728 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(728)2\pi \frac{dr}{dt}$

$r =\frac{728}{4}=182$

Report an Error

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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 #571 : Rate Of Change

Example Question #572 : Rate Of Change

Example Question #573 : Rate Of Change

Example Question #573 : How To Find Rate Of Change

Example Question #574 : How To Find Rate Of Change

Example Question #575 : How To Find Rate Of Change

Example Question #576 : How To Find Rate Of Change

Example Question #3491 : Calculus

Example Question #574 : Rate Of Change

Example Question #575 : Rate Of Change

All Calculus 1 Resources

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