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Calculus 1 : Rate of Change

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Example Question #51 : How To Find Rate Of Change

The edges of a cube extend a rate of $1.5\frac{in}{s}$ . If the cube has a volume of 216in^3 , what is the rate of the growth of the surface area?

Possible Answers:

$18\frac{in^2}{s}$

$108\frac{in^2}{s}$

$216\frac{in^2}{s}$

$72\frac{in^2}{s}$

$36\frac{in^2}{s}$

Correct answer:

$108\frac{in^2}{s}$

Explanation:

Surface area of a cube is given by the formula:

A=6l^2

Therefore, deriving each side with respect to time relates the growth of surface area to the growth of the edges:

$\frac{dA}{dt}=12l\frac{dl}{dt}$

The only current unknown now is the length of an edge, which can be found since the volume of the cube is known:

l^3=216in^3

l=6in

Therefore, the growth of the surface area is:

$\frac{dA}{dt}=12(6in)\left(1.5\frac{in}{s}\right)=108\frac{in^2}{s}$

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Example Question #52 : How To Find Rate Of Change

A cylinder with a radius of two inches and a height of eight inches is enlongating at a rate of half an inch per second. If the thickness is maintained constant, what is the rate of growth of the surface area?

Possible Answers:

$6\pi \frac{in^2}{s}$

$\pi \frac{in^2}{s}$

$0.5\pi \frac{in^2}{s}$

$2\pi \frac{in^2}{s}$

$4\pi \frac{in^2}{s}$

Correct answer:

$2\pi \frac{in^2}{s}$

Explanation:

The surface area of a cylinder is given by the formula:

$A=2\pi r^2+2\pi r h$

Therefore the rate of growth over time is:

$\frac{dA}{dt}=2\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt}+2\pi r\frac{dh}{dt}$

Since the cylinder doesn't grow thicker, $\frac{dr}{dt}=0$

Therefore

$\frac{dA}{dt}=2\pi r \frac{dh}{dt}=2\pi(2in)\left(0.5 \frac{in}{s}\right)=2\pi \frac{in^2}{s}$ .

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Example Question #53 : How To Find Rate Of Change

A sphere with a volume of $36\pi$ is expanding at a rate of $0.9 \pi$ . What is the rate of growth of the surface area?

Possible Answers:

$3\pi$

$\frac{3}{5}\pi$

$\frac{12}{5}\pi$

$15\pi$

$\frac{6}{5}\pi$

Correct answer:

$\frac{3}{5}\pi$

Explanation:

To begin, find the radius of the sphere and the radius' rate of change, since these terms are used when describing the surface area. Working with volume:

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

$\frac{4}{3}\pi r^3 = 36\pi$

r=3

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

$\frac{dr}{dt}=\frac{1}{36\pi}(0.9\pi)=\frac{1}{40}$

Now, to work with surface area:

$A=4\pi r^2$

$\frac{dA}{dt}=8\pi r \frac{dr}{dt} = 24\pi \left(\frac{1}{40}\right)=\frac{3}{5}\pi$

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Example Question #54 : How To Find Rate Of Change

The amount of money in an account is given by the formula: $f(t)= 10000(e^{{0.02}t})dollars$

What will be the rate of growth in five years time?

Possible Answers:

$202.01\frac{dollars}{year}$

$20.2\frac{dollars}{year}$

$2020.1\frac{dollars}{year}$

$40.4\frac{dollars}{year}$

$404.02\frac{dollars}{year}$

Correct answer:

$202.01\frac{dollars}{year}$

Explanation:

The rate of growth of money in the account can be found by taking the time derivative of the amount:

$r(t)= \frac{d}{dt}(10000(e^{{0.02}t}))=200e^{{.002}t}\frac{dollars}{year}$

Therefore, the rate of growth in five years will be:

$r(t)=200e^{.01}\frac{dollars}{year}=202.01\frac{dollars}{year}$

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Example Question #55 : How To Find Rate Of Change

A leaky trough is ten feet long with isosceles triangle cross sections. The cross sections have a base of two feet and a height of two feet six inches. The trough is being filled with water at one cubic foot per minute. However, it is also leaking at a rate of two cubic feet per minute.

When the depth of the water is one foot five inches, how fast is the water level falling?

Possible Answers:

$\frac{3}{34}\ ft/min$

$-\frac{3}{34}\ ft/min$

$-\frac{1}{12}\ ft/min$

$2 \ ft/min$

Correct answer:

$-\frac{3}{34}\ ft/min$

Explanation:

You know the net volume is decreasing at a rate of -1 ft/min by adding the rates 1 (being added) and -2(leaking from the trough). However, the question asks what the rate of change of the height is. The equation V=1/2blh (because the cross sections are triangles; the trough is a prism) relates height to volume.

The length (l) is a constant 10 feet, and the base needs to be written in terms of something we know the rate of change. Because the cross sections are triangles, the sides are proportional.

Therefore, $\frac{b}{2}=\frac{h}{2.5}$ and b=0.8h.

After plugging the known values into the volume equation,

$V=\frac{1}{2}*10*0.8h*h$ or V=4h^2 .

Then differentiate both sides to relate the rates of change.

$\frac{dV}{dt}=8h\frac{dh}{dt}$ .

Finally, plug in the known values for the rate of change of volume(dV/dt) -1ft/min and the instantaneous height (1 ft 5 in = 17/12 ft).

$\\ -1=8\left(\frac{17}{12} \right )\frac{dh}{dt}\\ \\-1=\frac{34}{3}\frac{dh}{dt} \\ \\=-\frac{3}{34}=\frac{dh}{dt}$

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Example Question #56 : How To Find Rate Of Change

An swimming pool with dimensions 50 meters wide, 50 meters long, and 5 meters deep is being filled up with water. If the height of the water is rising at a rate of 0.1 meters per minute, how fast is the volume of the pool changing? Assume the pool is being filled uniformly from the bottom up.

Possible Answers:

$300\frac{m^3}{min}$

$250\frac{m^3}{min}$

$200\frac{m^3}{min}$

$150\frac{m^3}{min}$

None of the above.

Correct answer:

$250\frac{m^3}{min}$

Explanation:

To solve this problem, we must realize that the volume for the swimming pool can be expressed as V=l*w*h , where is the length, is the width, and is the height. In order to find the rate at which the volume of the swimming pool changes at, we must take the derivative of the volume with respect to time.We also know that l=w , therefore the volume equation becomes V=l^2*h . To take the derivative of the equation, we must be familiar with the power rule, $\frac{d}{dx}x^n=nx^{n-1}$ , and the product rule $\frac{d}{dt}u*v=u'v+uv'$ . By applying those rules in taking the derivative of the volume equation, we find that the derivative becomes $\frac{dV}{dt}=2l*\frac{dl}{dt}*h+l^2*\frac{dh}{dt}$ . Since we know that $l=50,\frac{dh}{dt}=0.1, h=5,\frac{dl}{dt}=0$ , we can plug in and solve for $\frac{dV}{dt}$ easily. Note that $\frac{dl}{dt}=0$ because the pool is being filled uniformly from the bottom up, this means that the length of water in the pool is constant, only the height of the water is changing.

$\frac{dV}{dt}=2*(50)*0*5+50*50*0.1=250\frac{m^3}{min}$

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Example Question #57 : How To Find Rate Of Change

The length of a rectangle is 50 feet and the width is 40 feet at a certain moment. If the length is increasing at 2 feet per second and the width is decreasing at 0.5 feet per second, how fast is the area of the rectangle changing at the same moment?

Possible Answers:

None of the above.

$60\frac{ft}{sec}$

$55\frac{ft}{sec}$

$50\frac{ft}{sec}$

$60\frac{ft}{sec}$

Correct answer:

$55\frac{ft}{sec}$

Explanation:

In order to solve this problem, we must first know that the area for a rectangle is A=l*w . In order to find the rate at which the area of the rectangle changes with respect to time, we simply take the derivative of the formula for the area of a rectangle with respect to time. In order to do so, we must apply the product rule,

$\frac{d}{dt}u*v=u'v+uv'$ .

Applying the product rule, the derivative becomes

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

Because we know that

$l=50,w=40,\frac{dw}{dt}=-0.5,\frac{dl}{dt}=2$ , we simply plug those into the equation to solve for $\frac{dA}{dt}$ .

$\frac{dA}{dt}=-0.5*50+2*40=55\frac{ft}{sec}$

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Example Question #58 : How To Find Rate Of Change

A circle's circumference is increasing at a rate of 0.6 feet per second. If the diameter of the circle is 80 feet at a moment, how fast is the area of the circle increasing at the same moment?

Possible Answers:

$27\frac{ft}{sec}$

$24 \frac{ft}{sec}$

$26 \frac{ft}{sec}$

None of the above.

$25 \frac{ft}{sec}$

Correct answer:

$24 \frac{ft}{sec}$

Explanation:

In order to solve this problem, we must know that the circumfrence of a circle is equivalent to $C=2\pi r$ .

Therefore by taking the derivative with respect to time, we obtain

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

Given that we know that

$\frac{dC}{dt}=0.6$ , we can solve for $\frac{dr}{dt}$ .

$0.6=2\pi\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{3}{10\pi}$

In order to solve for how fast the area of the circle changes, we must know that the area of a circle is defined as $A=\pi r^2$ .

Taking the derivative with respect to time, we obtain

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

We know that $r=40,\frac{dr}{dt}=\frac{3}{10\pi}$ , so by plugging in all the variables, we can solve for $\frac{dA}{dt}$ .

$\frac{dA}{dt}=2\pi(40)\left(\frac{3}{10\pi}\right)$

$\frac{dA}{dt}=24 \frac{ft}{sec}$ .

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Example Question #59 : How To Find Rate Of Change

Find the rate of change of the function f(t) = tan(t) + 4t^2 - 6 .

Possible Answers:

sec^2(t) + 8t

sin^2(t) + 4t^2

tan(t) +4t

sec^2(t) + 4t^2

cos(t) - 8t

Correct answer:

sec^2(t) + 8t

Explanation:

The rate of change of a function is its derivative.

Recall the following rules of differentiation to help solve this problem.

Power Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Differentiation rule for tangent:

$\frac{\mathrm{d} }{\mathrm{d} t}(atan(bt)) = absec^2(bt)$

Therefore, the rate of change of f(t) , by the power rule and the differentiation rule for tangent, is

f'(t) = sec^2(t) + 8t .

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Example Question #60 : How To Find Rate Of Change

The sides of a square shrink at a rate of $0.01\frac{in}{s}$ . What is the rate of growth of the square if its sides have lengths of 10in ?

Possible Answers:

$0.0002\frac{in^2}{s}$

$0.1\frac{in^2}{s}$

$-0.0001\frac{in^2}{s}$

$-0.2\frac{in^2}{s}$

$-0.1\frac{in^2}{s}$

Correct answer:

$-0.2\frac{in^2}{s}$

Explanation:

The area of a square is given by the formula:

A=l^2

The rate of growth of the area can be related to the rate of growth of sides by differentiating each side with respect to time:

$\frac{dA}{dt}=\frac{d}{dt}l^2=2l\frac{dl}{dt}$

Therefore, the rate of growth of the square is:

$\frac{dA}{dt}=2(10in)(-0.01\frac{in}{s})=-0.2\frac{in^2}{s}$

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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 #51 : How To Find Rate Of Change

Example Question #52 : How To Find Rate Of Change

Example Question #53 : How To Find Rate Of Change

Example Question #54 : How To Find Rate Of Change

Example Question #55 : How To Find Rate Of Change

Example Question #56 : How To Find Rate Of Change

Example Question #57 : How To Find Rate Of Change

Example Question #58 : How To Find Rate Of Change

Example Question #59 : How To Find Rate Of Change

Example Question #60 : How To Find Rate Of Change

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