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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3481 : 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 one half times the rate of growth of the circumference?

Possible Answers:

$\frac{\sqrt{2}}{4}$

$\frac{1}{2}$

$\frac{1}{8}$

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

$\frac{1}{4}$

Correct answer:

$\frac{1}{2}$

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 one half times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{2})2\pi \frac{dr}{dt}$

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

$r =\sqrt{\frac{\frac{1}{2}}{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

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Example Question #3482 : 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 $\frac{2}{49}$ times the rate of growth of the circumference?

Possible Answers:

$\frac{1}{14}$

$\frac{2}{7}$

$\frac{1}{28}$

$\frac{4}{7}$

$\frac{1}{7}$

Correct answer:

$\frac{1}{7}$

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 $\frac{2}{49}$ times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(\frac{2}{49})2\pi \frac{dr}{dt}$

$r^2=(\frac{2}{49})\frac{1}{2}$

$r =\sqrt{\frac{1}{49}}=\frac{1}{7}$

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Example Question #3483 : 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 1250 times the rate of growth of the circumference?

Possible Answers:

$50\sqrt{2}$

$25\sqrt{2}$

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 1250 times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(1250)2\pi \frac{dr}{dt}$

$r^2=(1250)\frac{1}{2}$

$r =\sqrt{625}=25$

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Example Question #3484 : 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 686 times the rate of growth of the circumference?

Possible Answers:

$7\sqrt{14}$

$7\sqrt{7}$

$14\sqrt{7}$

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 #3485 : 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 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 #3486 : 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 294 times the rate of growth of the circumference?

Possible Answers:

$9\sqrt{7}$

$3\sqrt{7}$

$14\sqrt{3}$

$7\sqrt{3}$

$7\sqrt{7}$

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:

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

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

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

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Example Question #3487 : 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 90 times the rate of growth of the circumference?

Possible Answers:

$15\sqrt{3}$

$25\sqrt{3}$

$5\sqrt{3}$

$9\sqrt{5}$

$3\sqrt{5}$

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:

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

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

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

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Example Question #3488 : 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 104 times the rate of growth of the circumference?

Possible Answers:

$8\sqrt{13}$

$13\sqrt{6}$

$2\sqrt{13}$

$4\sqrt{13}$

$13\sqrt{2}$

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:

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

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

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

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Example Question #3489 : 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 198 times the rate of growth of the circumference?

Possible Answers:

$11\sqrt{6}$

$3\sqrt{11}$

$11\sqrt{3}$

$6\sqrt{11}$

$9\sqrt{11}$

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:

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

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

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

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Example Question #2461 : Functions

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:

$22\sqrt{6}$

$11\sqrt{6}$

$11\sqrt{3}$

$6\sqrt{11}$

$22\sqrt{3}$

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:

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

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

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

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3481 : Calculus

Example Question #3482 : Calculus

Example Question #3483 : Calculus

Example Question #3484 : Calculus

Example Question #3485 : Calculus

Example Question #3486 : Calculus

Example Question #3487 : Calculus

Example Question #3488 : Calculus

Example Question #3489 : Calculus

Example Question #2461 : Functions

All Calculus 1 Resources

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