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Calculus 1 : Rate of Change

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Example Questions

Example Question #3141 : Calculus

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 equal to the rate of growth of the circumference?

Possible Answers:

$\frac{\pi\sqrt{3}}{4}$

$\frac{2\pi}{3}$

$\frac{\pi\sqrt{3}}{3}$

$\frac{\pi\sqrt{2}}{3}$

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

Correct answer:

$\frac{\pi\sqrt{2}}{3}$

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:

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

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

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 equal to the rate of growth of the circumference, let's solve for a radius that satisfies it.

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

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

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

Returning to the volume:

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

$V=\frac{4}{3}\pi (\frac{\sqrt{2}}{2})^3$

$V=\frac{\pi\sqrt{2}}{3}$

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Example Question #231 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the surface of the sphere at the instance the rate of shrinkage of the volume is 1/16 times the rate of shrinkage of the surface area?

Possible Answers:

$16\pi$

$\frac{\pi}{16}$

$\frac{\pi}{4}$

$4\pi$

$\frac{\pi}{8}$

Correct answer:

$\frac{\pi}{16}$

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 1/16 times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{16})8\pi r \frac{dr}{dt}$

$r =(\frac{1}{16})2$

$r=\frac{1}{8}$

Returning to the surface area:

$A=4\pi r^2$

$A=4\pi (\frac{1}{8})^2$

$A=\frac{\pi}{16}$

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Example Question #232 : Rate Of Change

A spherical balloon is being filled with air. What is the cirumference of the sphere at the instance the rate of growth of the volume is $\pi$ times the rate of growth of the surface area?

Possible Answers:

$4\pi^2$

$8\pi$

$16\pi^2$

$\frac{1}{\pi}$

$2\pi$

Correct answer:

$4\pi^2$

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

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

$r =2\pi$

Circumference is given as:

$C=2\pi r$

$C=4\pi^2$

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Example Question #3143 : Calculus

The width of a rectangle increases an eighth as fast as its length. How does the rate of change of the rectangle's area compare to that of the rate of change of the width when the length of the rectangle is a sixteenth of its width?

Possible Answers:

$\frac{129}{16}w$

$\frac{9}{16}w$

$\frac{129}{8}w$

24w

Correct answer:

$\frac{129}{16}w$

Explanation:

Begin by writing the expression for the area of a rectangle:

A=wl

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

$\frac{dA}{dt}=w\frac{dl}{dt}+l\frac{dw}{dt}$

Now, we're told two things:

The width of a rectangle increases an eighth as fast as its length: $\frac{dw}{dt}=\frac{1}{8}\frac{dl}{dt}$

The length of the rectangle is a sixteenth of its width: $l=\frac{1}{16}w$

Using this, rewrite the area equation in terms of width:

$\frac{dA}{dt}=w(8\frac{dw}{dt})+(\frac{1}{16}w)\frac{dw}{dt}$

$\frac{dA}{dt}=\frac{129}{16}w\frac{dw}{dt}$

The rate of change of the area is $\frac{129}{16}w$ times the rate of change of the rate of change of the width.

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Example Question #234 : Rate Of Change

The width of a rectangle increases a fifth as fast as its length. How does the rate of change of the rectangle's area compare to that of the rate of change of the width when the length of the rectangle is threefold its width?

Possible Answers:

$\frac{16}{5}w$

$\frac{8}{15}w$

$\frac{16}{3}w$

15w

Correct answer:

Explanation:

Begin by writing the expression for the area of a rectangle:

A=wl

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

$\frac{dA}{dt}=w\frac{dl}{dt}+l\frac{dw}{dt}$

Now, we're told two things:

The width of a rectangle increases a fifth as fast as its length: $w=\frac{1}{5}l$

The length of the rectangle is threefold its width: l=3w

Using this, rewrite the area equation in terms of width:

$\frac{dA}{dt}=w(5\frac{dw}{dt})+(3w)\frac{dw}{dt}$

$\frac{dA}{dt}=8w\frac{dw}{dt}$

The rate of change of the area is times the rate of change of the rate of change of the width.

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Example Question #235 : Rate Of Change

The width of a rectangle increases four times as fast as its length. How does the rate of change of the rectangle's area compare to that of the rate of change of the width when the length of the rectangle is a fourth of its width?

Possible Answers:

16w

$\frac{5}{6}w$

$\frac{1}{16}w$

$\frac{1}{2}w$

Correct answer:

$\frac{1}{2}w$

Explanation:

Begin by writing the expression for the area of a rectangle:

A=wl

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

$\frac{dA}{dt}=w\frac{dl}{dt}+l\frac{dw}{dt}$

Now, we're told two things:

The width of a rectangle increases four times as fast as its length: $\frac{dw}{dt}=4\frac{dl}{dt}$

The length of the rectangle is a fourth of its width: $l=\frac{1}{4}w$

Using this, rewrite the area equation in terms of width:

$\frac{dA}{dt}=w(\frac{1}{4}\frac{dw}{dt})+(\frac{1}{4}w)\frac{dw}{dt}$

$\frac{dA}{dt}=\frac{1}{2}w\frac{dw}{dt}$

The rate of change of the area is $\frac{1}{2}w$ times the rate of change of the rate of change of the width.

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Example Question #236 : Rate Of Change

The vertical axis of an ellipse increases a third 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 a third its horizontal axis?

Possible Answers:

$9\pi v$

$6\pi v$

$\frac{10}{3}\pi v$

$3\pi v$

$\frac{2}{3}\pi v$

Correct answer:

$6\pi v$

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 a third as fast as its horizontal axis: $\frac{dv}{dt}=\frac{1}{3}\frac{dh}{dt}$

The vertical axis is a third its horizontal axis: $v=\frac{1}{3}h$

Using this, rewrite the area equation in terms of its vertical axis: $\frac{dA}{dt}=\pi (3v)\frac{dv}{dt}+\pi v(3\frac{dv}{dt})$

$\frac{dA}{dt}=6\pi v\frac{dv}{dt}$

The rate of change of the area is $6\pi v$ times the rate of change of the rate of change of the vertical axis.

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Example Question #237 : Rate Of Change

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:

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

$10 \pi v$

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

$4 \pi v$

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

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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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3141 : Calculus

Example Question #231 : Rate Of Change

Example Question #232 : Rate Of Change

Example Question #3143 : Calculus

Example Question #234 : Rate Of Change

Example Question #235 : Rate Of Change

Example Question #236 : Rate Of Change

Example Question #237 : Rate Of Change

Example Question #231 : How To Find Rate Of Change

Example Question #239 : Rate Of Change

All Calculus 1 Resources

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