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Calculus 1 : How to find rate of change

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Example Questions

Example Question #121 : Rate Of Change

The sides of a cube are expanding at a rate of 0.5 . If the sides have length , what is the rate of growth of the surface area of the cube?

Possible Answers:

1.5

Correct answer:

Explanation:

The surface area of a cube is given in terms of its sides as

A=6s^2

This relation can be used to also relate rates of change. This can be done by taking the derivative of each side of the equation with respect to time:

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

We are given the rate of change of the sides $\frac{ds}{dt}=0.5$ and the lengths of the sides s=4 . Thus the rate of growth of the surface area is

$\frac{dA}{dt}=12(4)(0.5)$

$\frac{dA}{dt}=24$

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Example Question #122 : Rate Of Change

A trapezoid is changing in shape. The parallel sides have lengths and , and the height is . If the short and long sides are growing at a rate of 0.2 and 0.3 , and the height is constant, what is the rate of growth of the area?

Possible Answers:

0.5

Correct answer:

Explanation:

The area of a trapezoid is given by the formula

$A=\frac{1}{2}(a+b)h$

Let a represent the short side and b the long side.

For this problem, while not necessary, it may make things easier to expand this equation

$A=\frac{1}{2}ah+\frac{1}{2}bh$

Now, we can relate rates of change by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=\frac{1}{2}a\frac{dh}{dt}+\frac{1}{2}h\frac{da}{dt}+\frac{1}{2}b\frac{dh}{dt}+\frac{1}{2}h\frac{db}{dt}$

Now recall that the height is constant. This means that $\frac{dh}{dt}=0$ , which shortens the equation

$\frac{dA}{dt}=\frac{1}{2}h\frac{da}{dt}+\frac{1}{2}h\frac{db}{dt}$

$\frac{dA}{dt}=\frac{1}{2}h(\frac{da}{dt}+\frac{db}{dt})$

$\frac{dA}{dt}=\frac{1}{2}(4)(0.2+0.3)$

$\frac{dA}{dt}=1$

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Example Question #123 : Rate Of Change

A rectangular prism has the following dimensions; w=2;l=3;h=8 . If the width and length are growing at a rate of 0.1 and the height is constant, what is the rate of growth of the prism's surface area?

Possible Answers:

3.6

2.1

5.4

0.02

4.2

Correct answer:

4.2

Explanation:

The surface area of a rectangular prism is given by the formula:

A=2wl+2wh+2lh

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

$\frac{dA}{dt}=2w\frac{dl}{dt}+2l\frac{dw}{dt}+2w\frac{dh}{dt}+2h\frac{dw}{dt}+2l\frac{dh}{dt}+2h\frac{dl}{dt}$

Now, we've been given some rates of change, namely $\frac{dw}{dt}=\frac{dl}{dt}=0.1$ . Since the height is constant $\frac{dh}{dt}=0$ , and the equation reduces to

$\frac{dA}{dt}=2w\frac{dl}{dt}+2l\frac{dw}{dt}+2h\frac{dw}{dt}+2h\frac{dl}{dt}$

Plugging in our other known values w=2;l=3;h=8 , we can solve for the rate of change of the surface area

$\frac{dA}{dt}=2(2)(0.1)+2(3)(0.1)+2(8)(0.1)+2(8)(0.1)$

$\frac{dA}{dt}=0.4+0.6+1.6+1.6$

$\frac{dA}{dt}=4.2$

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Example Question #124 : Rate Of Change

The parallel sides of a cube are expanding at rates 0.3 , 0.4 , and 0.5 . If the cube has a volume of , what is the rate of growth of this volume?

Possible Answers:

14.8

0.12

0.06

19.2

Correct answer:

19.2

Explanation:

Although we are told that the sides of the cube have different rates of growth (meaning it won't stay a cube for long), at the time of this problem, since the shape is a cube it means all of the sides are equal:

s_1=s_2=s_3=s

The volume of a cube is given by the expression:

V=s^3

We'll be using this to find the current lengths of the sides

s^3=64

s=4

Now, rewrite the volume equation in terms of the three sides:

V=s_1s_2s_3

The rate of change of the volume can be found by taking the derivative of each side of the equation with respect to time. Remember to take the derivative of the right side with respect to each variable:

$\frac{dV}{dt}=s_2s_3\frac{ds_1}{dt}+s_1s_3\frac{ds_2}{dt}+s_1s_2\frac{ds_3}{dt}$

With the three rates of change 0.3 , 0.4 , and 0.5 and the known length of the sides, the rate of change is

$\frac{dV}{dt}=(4^2)(0.3)+(4^2)(0.4)+(4^2)(0.5)$

$\frac{dV}{dt}=19.2$

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Example Question #125 : Rate Of Change

The rate of change of a sphere's radius is given by the equation $5e^{-\frac{1}{2}t}$ . If the sphere has an initial radius of , what volume will the sphere will approach if given unlimited time to expand?

Possible Answers:

$\frac{2744}{3}\pi$

$234\pi$

$\frac{10976}{3}\pi$

$2744\pi$

Correct answer:

$\frac{10976}{3}\pi$

Explanation:

For this problem we're given the initial radius of a sphere and the rate of change of this radius. It will be useful to find an equation for the radius itself, however.

The rate, we are told, is $\frac{dr(t)}{dt}=5e^{-\frac{1}{2}t}$ .

To find an expression for the radius, we need to integrate this function.

$\int e^{at}dt=\frac{e^{at}}{a}+C; [a,C:constants]$

$r(t)=-10e^{-\frac{1}{2}t}+C$

To find C, the constant of integration, utilize the initial condition:

$r(0)=-10e^{-\frac{1}{2}(0)}+C=4$

C-10=4

C=14

So our radius expression is

$r(t)=14-10e^{-\frac{1}{2}t}$

Now to find the maximum radius, take the limit as time goes towards infinity:

$lim_{t\rightarrow \infty }14-10e^{-\frac{1}{2}t}=14-0=14$

The volume of a sphere is given by the equation:

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

Therefore, the maximum volume is

$V_{max}=\frac{4}{3}\pi 14^3$

$V_{max}=\frac{10976}{3}\pi$

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Example Question #121 : Rate Of Change

The sides of an expanding rectangle can be expressed with the functions $w(t)=2e^{2t}$ , l(t)=5+e^t . What is the rate of growth of the rectangle's area?

Possible Answers:

$4e^{3t}$

$20e^{2t}+6e^{3t}$

$20e^{2t}+4e^{2t^2}$

$20e^{2t}$

$4e^{2t^2}$

Correct answer:

$20e^{2t}+6e^{3t}$

Explanation:

The area of a rectangle in relation to its sides is given by the formula

A=wl

This rule holds even when the the sides are expressed in more complicated terms such as those of this problem.

Since $w(t)=2e^{2t}$ and l(t)=5+e^t

$A(t)=2e^{2t}(5+e^t)$

$A(t)=10e^{2t}+2e^{3t}$

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

Recall that the rule of derivation for an exponential:

$d(e^{u})=du(e^u)$ , where is any function in the exponent.

For this problem, the derivative is then

$\frac{dA(t)}{dt}=20e^{2t}+6e^{3t}$

This is the formula for the rate of growth of the rectangle's area.

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Example Question #121 : Rate Of Change

The rate of change of the sides of a square is 2.5 . If the square has an area of , what is the rate of change of the area?

Possible Answers:

6.25

3.12

7.5

Correct answer:

Explanation:

The area of a square is given in terms of its sides as

A=s^2

For an area of

s^2=9

s=3

The area equation can also be used to find rates of change; take the derivative of each side of the equation with respect to time:

A=s^2

$\frac{dA}{dt}=2s\frac{ds}{dt}$

We're told that $\frac{ds}{dt}=2.5$

Therefore

$\frac{dA}{dt}=2(3)(2.5)$

$\frac{dA}{dt}=15$

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Example Question #128 : Rate Of Change

Let $g(x)=\pi e^{-2x}$ , h(y) = -3y and $f(y) = \cos(-y)$ .

If it is known that g(x)h(y)=f(y) ,

where is related to by an unknown function,

find an expression for $\frac{dy}{dx}$ .

Possible Answers:

$\frac{dy}{dx}= \frac{6\pi e^{-2x}y}{3\pi e^{-2x}-\sin y}$

$\frac{dy}{dx}=\frac{6\pi e^{-2x}y}{-\sin y}$

$\frac{dy}{dx}=0$

Not enough information is given to find an expression for $\frac{dy}{dx}$ .

Correct answer:

$\frac{dy}{dx}= \frac{6\pi e^{-2x}y}{3\pi e^{-2x}-\sin y}$

Explanation:

Since it is known that g(x)h(y)=f(y) , differentiate implicitly on both sides w.r.t. (it is also helpful to notice that, since cosine is an even function, $\cos(-y)=\cos (y)$ :

$-3\pi (-2e^{-2x}y+e^{-2x}\frac{dy}{dx})=-\sin y\frac{dy}{dx}$ USE PRODUCT RULE

$6\pi e^{-2x}y-3\pi e^{-2x}\frac{dy}{dx}=-\sin y\frac{dy}{dx}$ FACTOR IN THE CONSTANT

$6\pi e^{-2x}y = (3\pi e^{-2x}-\sin y)\frac{dy}{dx}$ COMBINE $\frac{dy}{dx}$ TERMS

$\frac{dy}{dx}= \frac{6\pi e^{-2x}y}{3\pi e^{-2x}-\sin y}$ DIVIDE

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Example Question #129 : Rate Of Change

A sphere is fixed inside of a cube, such that it is completely snug. If the sides of the cube, which have length , begin to grow at a rate of 0.4 , what is the rate of growth of the volume of the sphere?

Possible Answers:

$57.6\pi$

$0.003 \pi$

$0.021\pi$

$28.8\pi$

$115.2\pi$

Correct answer:

$28.8\pi$

Explanation:

The volume of sphere, in terms of its radius, is defined as

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

However, in the case of the problem, we're given the lengths of the sides of a cube in which the sphere fits. Since the outside of the sphere is touching the sides walls of the cube, the length of the cube is the diameter of the sphere:

s=d=2r

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

$r=\frac{6}{2}=3$

Furthermore, the rate of growth of a sphere's radius will be half the rate of growth of it's diameter

$\frac{dr}{dt}=\frac{1}{2}\frac{ds}{dt}$

$\frac{dr}{dt}=\frac{1}{2}(0.4)=0.2$

Now returning to the volume equation

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

The rate of growth can be found by taking the derivative with respect to time:

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

$\frac{dV}{dt}=4\pi (6^2)(0.2)$

$\frac{dV}{dt}=28.8\pi$

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Example Question #130 : Rate Of Change

The sides of an equilateral triangle are growing at a rate of 0.5 . If the sides of length $2\sqrt{3}$ , what is the rate of growth of the triangle's area?

Possible Answers:

2.25

1.5

$\frac{\sqrt{3}}{8}$

$\frac{\sqrt{3}}{16}$

Correct answer:

1.5

Explanation:

The area of an equilateral triangle, in terms of its side, is

$A=\frac{\sqrt{3}}{4}s^2$

We're told that the length of a side is $2\sqrt{3}$ and the rate of change is 0.5 . Now, to utilize the fact, let's find an equation for the rate of change of the area. Find this by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=\frac{\sqrt{3}}{2}s\frac{ds}{dt}$

$\frac{dA}{dt}=\frac{\sqrt{3}}{2}(2\sqrt{3})(0.5)$

$\frac{dA}{dt}=1.5$

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Calculus 1 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #121 : Rate Of Change

Example Question #122 : Rate Of Change

Example Question #123 : Rate Of Change

Example Question #124 : Rate Of Change

Example Question #125 : Rate Of Change

Example Question #121 : Rate Of Change

Example Question #121 : Rate Of Change

Example Question #128 : Rate Of Change

Example Question #129 : Rate Of Change

Example Question #130 : Rate Of Change

All Calculus 1 Resources

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