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Calculus 1 : Functions

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Example Questions

Example Question #676 : Rate

A spherical balloon is being filled with air. What is the circumference of the sphere at the instance the rate of growth of the surface area is 232 times the rate of growth of the circumference?

Possible Answers:

$464\pi$ $\displaystyle 464\pi$

$116\pi$ $\displaystyle 116\pi$

$232\pi$ $\displaystyle 232\pi$

$58\pi$ $\displaystyle 58\pi$

$29\pi$ $\displaystyle 29\pi$

Correct answer:

$116\pi$ $\displaystyle 116\pi$

Explanation:

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

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

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

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. Now we can use the relation given in the problem statement, the rate of growth of the surface area is 232 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(232)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(232)2\pi \frac{dr}{dt}$

$r =\frac{232}{4}=58$ $\displaystyle r =\frac{232}{4}=58$

The circumference is then:

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

$C=116\pi$ $\displaystyle C=116\pi$

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Example Question #677 : Rate

A spherical balloon is being filled with air. What is the circumference of the sphere at the instance the rate of growth of the surface area is 192 times the rate of growth of the circumference?

Possible Answers:

$384\pi$ $\displaystyle 384\pi$

$768\pi$ $\displaystyle 768\pi$

$96\pi$ $\displaystyle 96\pi$

$1536\pi$ $\displaystyle 1536\pi$

$192\pi$ $\displaystyle 192\pi$

Correct answer:

$96\pi$ $\displaystyle 96\pi$

Explanation:

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

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

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

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. Now we can use the relation given in the problem statement, the rate of growth of the surface area is 192 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(192)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(192)2\pi \frac{dr}{dt}$

$r =\frac{192}{4}=48$ $\displaystyle r =\frac{192}{4}=48$

The circumference can then be found at this instant:

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

$C=96\pi$ $\displaystyle C=96\pi$

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Example Question #678 : Rate

A spherical balloon is being filled with air. What is the circumference of the sphere at the instance the rate of growth of the surface area is 1024 times the rate of growth of the circumference?

Possible Answers:

$4096\pi$ $\displaystyle 4096\pi$

$256\pi$ $\displaystyle 256\pi$

$512\pi$ $\displaystyle 512\pi$

$1024\pi$ $\displaystyle 1024\pi$

$128\pi$ $\displaystyle 128\pi$

Correct answer:

$512\pi$ $\displaystyle 512\pi$

Explanation:

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

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

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

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. Now we can use the relation given in the problem statement, the rate of growth of the surface area is 1024 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(1024)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(1024)2\pi \frac{dr}{dt}$

$r =\frac{1024}{4}=256$ $\displaystyle r =\frac{1024}{4}=256$

The circumference is then:

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

$C=512\pi$ $\displaystyle C=512\pi$

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Example Question #679 : Rate

A spherical balloon is being filled with air. What is the circumference of the sphere at the instance the rate of growth of the surface area is 3132 times the rate of growth of the circumference?

Possible Answers:

$783\pi$ $\displaystyle 783\pi$

$1566\pi$ $\displaystyle 1566\pi$

$1044\pi$ $\displaystyle 1044\pi$

$261\pi$ $\displaystyle 261\pi$

$522\pi$ $\displaystyle 522\pi$

Correct answer:

$1566\pi$ $\displaystyle 1566\pi$

Explanation:

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

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

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

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. Now we can use the relation given in the problem statement, the rate of growth of the surface area is 3132 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(3132)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(3132)2\pi \frac{dr}{dt}$

$r =\frac{3132}{4}=783$ $\displaystyle r =\frac{3132}{4}=783$

The circumference is then:

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

$C=1566\pi$ $\displaystyle C=1566\pi$

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Example Question #591 : Rate Of Change

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the radius when the radius is 17?

Possible Answers:

$1156\pi$ $\displaystyle 1156\pi$

1156 $\displaystyle 1156$

289 $\displaystyle 289$

$289\pi$ $\displaystyle 289\pi$

$578\pi$ $\displaystyle 578\pi$

Correct answer:

$1156\pi$ $\displaystyle 1156\pi$

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:

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

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

$\phi =4\pi(17)^2=1156\pi$ $\displaystyle \phi =4\pi(17)^2=1156\pi$

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Example Question #592 : Rate Of Change

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the radius when the radius is 34?

Possible Answers:

$1156\pi$ $\displaystyle 1156\pi$

$4624\pi$ $\displaystyle 4624\pi$

$1068\pi$ $\displaystyle 1068\pi$

$534\pi$ $\displaystyle 534\pi$

$136\pi$ $\displaystyle 136\pi$

Correct answer:

$4624\pi$ $\displaystyle 4624\pi$

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:

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

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

$\phi =4\pi(34)^2=4624\pi$ $\displaystyle \phi =4\pi(34)^2=4624\pi$

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Example Question #593 : Rate Of Change

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the radius when the radius is 28?

Possible Answers:

$3136\pi$ $\displaystyle 3136\pi$

$784\pi$ $\displaystyle 784\pi$

$448\pi$ $\displaystyle 448\pi$

$112\pi$ $\displaystyle 112\pi$

$6272\pi$ $\displaystyle 6272\pi$

Correct answer:

$3136\pi$ $\displaystyle 3136\pi$

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:

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

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

$\phi =4\pi(28)^2=3136\pi$ $\displaystyle \phi =4\pi(28)^2=3136\pi$

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Example Question #594 : Rate Of Change

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 $\frac{1}{34}$ $\displaystyle \frac{1}{34}$?

Possible Answers:

$\frac{8\pi}{289}$ $\displaystyle \frac{8\pi}{289}$

$\frac{4\pi}{289}$ $\displaystyle \frac{4\pi}{289}$

$\frac{2\pi}{289}$ $\displaystyle \frac{2\pi}{289}$

$\frac{\pi}{289}$ $\displaystyle \frac{\pi}{289}$

$\frac{16\pi}{289}$ $\displaystyle \frac{16\pi}{289}$

Correct answer:

$\frac{\pi}{289}$ $\displaystyle \frac{\pi}{289}$

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:

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

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

$\phi =4\pi(\frac{1}{34})^2=\frac{\pi}{289}$ $\displaystyle \phi =4\pi(\frac{1}{34})^2=\frac{\pi}{289}$

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Example Question #595 : Rate Of Change

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 $\frac{1}{60}$ $\displaystyle \frac{1}{60}$?

Possible Answers:

$\frac{2\pi}{225}$ $\displaystyle \frac{2\pi}{225}$

$\frac{\pi}{900}$ $\displaystyle \frac{\pi}{900}$

$\frac{\pi}{225}$ $\displaystyle \frac{\pi}{225}$

$\frac{\pi}{450}$ $\displaystyle \frac{\pi}{450}$

$\frac{4\pi}{225}$ $\displaystyle \frac{4\pi}{225}$

Correct answer:

$\frac{\pi}{900}$ $\displaystyle \frac{\pi}{900}$

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:

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

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

$\phi =4\pi(\frac{1}{60})^2=\frac{\pi}{900}$ $\displaystyle \phi =4\pi(\frac{1}{60})^2=\frac{\pi}{900}$

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Example Question #596 : Rate Of Change

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 $\frac{1}{71}$ $\displaystyle \frac{1}{71}$?

Possible Answers:

$\frac{4\pi}{5041}$ $\displaystyle \frac{4\pi}{5041}$

$\frac{8\pi}{5041}$ $\displaystyle \frac{8\pi}{5041}$

$\frac{2\pi}{5041}$ $\displaystyle \frac{2\pi}{5041}$

$\frac{\pi}{5041}$ $\displaystyle \frac{\pi}{5041}$

$\frac{\pi}{10082}$ $\displaystyle \frac{\pi}{10082}$

Correct answer:

$\frac{4\pi}{5041}$ $\displaystyle \frac{4\pi}{5041}$

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:

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

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

$\phi =4\pi(\frac{1}{71})^2=\frac{4\pi}{5041}$ $\displaystyle \phi =4\pi(\frac{1}{71})^2=\frac{4\pi}{5041}$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #676 : Rate

Example Question #677 : Rate

Example Question #678 : Rate

Example Question #679 : Rate

Example Question #591 : Rate Of Change

Example Question #592 : Rate Of Change

Example Question #593 : Rate Of Change

Example Question #594 : Rate Of Change

Example Question #595 : Rate Of Change

Example Question #596 : Rate Of Change

All Calculus 1 Resources

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