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Calculus 1 : Calculus

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Example Questions

Example Question #2121 : Functions

The vertical axis of an ellipse increases twice as fast as its horizontal axis. How does the rate of change of the ellipse's area compare to that of the rate of change of its vertical axis when the vertical axis is an eighth its horizontal axis?

Possible Answers:

$\frac{17}{8}\pi v$

$\frac{5}{8}\pi v$

$\frac{17}{2}\pi v$

$4 \pi v$

$10 \pi v$

Correct answer:

$\frac{17}{2}\pi v$

Explanation:

Start by writing the expression for the area of an ellipse in terms of its vertical and horizontal axes:

$A=\pi vh$

The rate of change of the area can be found by taking the derivative of the equation with respect to time:

$\frac{dA}{dt}=\pi h\frac{dv}{dt}+\pi v\frac{dh}{dt}$

Now, we're told two things:

The vertical axis of an ellipse increases twice as fast as its horizontal axis: $\frac{dv}{dt}=2\frac{dh}{dt}$

The vertical axis is an eighth its horizontal axis: $v=\frac{1}{8}h$

Using this, rewrite the area equation in terms of its vertical axis: $\frac{dA}{dt}=\pi (8v)\frac{dv}{dt}+\pi v(\frac{1}{2}\frac{dv}{dt})$

$\frac{dA}{dt}=\frac{17}{2}\pi v\frac{dv}{dt}$

The rate of change of the area is $\frac{17}{2}\pi v$ times the rate of change of the rate of change of the vertical axis.

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Example Question #231 : How To Find Rate Of Change

The vertical axis of an ellipse increases five times as fast as its horizontal axis. How does the rate of change of the ellipse's area compare to that of the rate of change of its vertical axis when the vertical axis is half its horizontal axis?

Possible Answers:

$\frac{5}{2}\pi v$

$\frac{7}{10}\pi v$

$7 \pi v$

$\frac{11}{2}\pi v$

$\frac{11}{5}\pi v$

Correct answer:

$\frac{11}{5}\pi v$

Explanation:

Start by writing the expression for the area of an ellipse in terms of its vertical and horizontal axes:

$A=\pi vh$

The rate of change of the area can be found by taking the derivative of the equation with respect to time:

$\frac{dA}{dt}=\pi h\frac{dv}{dt}+\pi v\frac{dh}{dt}$

Now, we're told two things:

The vertical axis of an ellipse increases five times as fast as its horizontal axis: $\frac{dv}{dt}=5\frac{dh}{dt}$

The vertical axis is half its horizontal axis: $v=\frac{1}{2}h$

Using this, rewrite the area equation in terms of its vertical axis: $\frac{dA}{dt}=\pi (2v)\frac{dv}{dt}+\pi v(\frac{1}{5}\frac{dv}{dt})$

$\frac{dA}{dt}=\frac{11}{5}\pi v\frac{dv}{dt}$

The rate of change of the area is $\frac{11}{5}\pi v$ times the rate of change of the rate of change of the vertical axis.

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Example Question #239 : 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 surface area when the radius is 10?

Possible Answers:

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 regardless of what parameter is being considered. To find the ratio of the rates of changes of the volume and surface area, simply divide:

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

$\phi =\frac{10}{2}$

$\phi =5$

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Example Question #241 : 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 surface area when the radius is one half?

Possible Answers:

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{4}$

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 regardless of the considered parameter. To find the ratio of the rates of changes of the volume and surface area, divide:

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

$\phi =\frac{\frac{1}{2}}{2}$

$\phi =\frac{1}{4}$

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Example Question #242 : 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 circumference when the radius is 5?

Possible Answers:

Correct answer:

Explanation:

Let's 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. To find the ratio of the rates of changes of the volume and circumference, divide:

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

$\phi =2r^2$

$\phi =50$

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Example Question #243 : 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 circumference when the radius is 3?

Possible Answers:

Correct answer:

Explanation:

Let's 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. To find the ratio of the rates of changes of the volume and circumference, divide:

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

$\phi =2r^2$

$\phi =18$

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Example Question #244 : Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is 6?

Possible Answers:

$12\pi$

$24\pi$

$6\pi$

Correct answer:

Explanation:

Let's begin 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. To find the ratio of the rates of changes of the surface area and circumference, divide:

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

$\phi =4r$

$\phi =24$

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Example Question #245 : Rate Of Change

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the surface area of the sphere to the rate of loss of the circumference when the radius is 13?

Possible Answers:

$52\pi$

$104\pi$

$26\pi$

Correct answer:

Explanation:

Let's begin 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:

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

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

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

$\phi =4r$

$\phi =52$

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Example Question #246 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 16?

Possible Answers:

Correct answer:

Explanation:

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

V=s^3

A=6s^2

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

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

$3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$

$\phi=\frac{s}{4}$

$\phi =4$

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Example Question #247 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length 2?

Possible Answers:

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

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

V=s^3

A=6s^2

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

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

$3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$

$\phi=\frac{s}{4}$

$\phi =\frac{1}{2}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2121 : Functions

Example Question #231 : How To Find Rate Of Change

Example Question #239 : Rate Of Change

Example Question #241 : Rate Of Change

Example Question #242 : Rate Of Change

Example Question #243 : Rate Of Change

Example Question #244 : Rate Of Change

Example Question #245 : Rate Of Change

Example Question #246 : Rate Of Change

Example Question #247 : Rate Of Change

All Calculus 1 Resources

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