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Calculus 1 : Rate

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Example Questions

Example Question #91 : Rate Of Change

A rectangle of length and width is changing shape. The length is shrinking at a rate of $0.1\frac{m}{s}$ and the width is growing at a rate of $0.3\frac{m}{s}$ . At the moment the rectangle becomes a square, what will be the rate of change of its area?

Possible Answers:

$3\frac{m^2}{s}$

$1.0\frac{m^2}{s}$

$-1.8\frac{m^2}{s}$

$1.5\frac{m^2}{s}$

$1.8\frac{m^2}{s}$

Correct answer:

$1.5\frac{m^2}{s}$

Explanation:

The area of a rectangle is given in terms of its length and width by the formula:

A=lw

We are asked to find the rate of change of the rectangle when it is a square, i.e at the time that w=l , so we must find the unknown value of and at this moment. The width and length at any time can be found in terms of their starting values and rates of change:

$w(t)=6m+0.3\frac{m}{s}t$

$l(t)=8m-0.1\frac{m}{s}t$

When they're equal:

$6m+0.3\frac{m}{s}t=8m-0.1\frac{m}{s}t$

$0.4\frac{m}{s}t=2m$

t=5s

And at this time w=l=7.5m .

Now, going back to our original area equation

A=lw

We can take the derivative of each side with respect to time to find the rate of change:

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

$\frac{dA}{dt}=7.5m(0.3\frac{m}{s})+7.5m(-0.1\frac{m}{s})$

$\frac{dA}{dt}=7.5m(0.2\frac{m}{s})$

$\frac{dA}{dt}=1.5\frac{m^2}{s}$

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Example Question #1973 : Functions

Find f'(3) if $f(x)=-2x^{3}+2x^{2}+1$ .

Possible Answers:

The derivative does not exist at that point.

Correct answer:

Explanation:

To find f'(3) , we must first find the derivative and then plug in for .

To evaluate this derivative, we need the following formulae:

$\frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1, \frac{d}{dx}cx=c, \frac{d}{dx}(x^{n}))=nx^{n-1}$

$\frac{d}{dx}(cx^{n})=ncx^{n-1}$

$f'(x)=(-2*3)x^{3-1}+(2*2)x^{2-1}+0=-6x^{2}+4x$

Then plug in for into f'(x) :

$f'(3)=-6(3)^{2}+4(3)=-54+12=-42$

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Example Question #91 : How To Find Rate Of Change

The length of a rectangle is defined by the function l(t)=2+0.3t+0.1t^2 and the width is defined by the function w(t)=1+0.4t+0.2t^2

What is the rate of change of the rectangle's area at time t=4 ?

Possible Answers:

6.18

15.98

2.20

7.99

12.36

Correct answer:

15.98

Explanation:

The area of a rectangle is given by the function:

A=lw

For the definitions of the sides

l(t)=2+0.3t+0.1t^2

w(t)=1+0.4t+0.2t^2

A(t)=(2+0.3t+0.1t^2)(1+0.4t+0.2t^2)

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

$\frac{dA(t)}{dt}=(0.3+0.2t)(1+0.4t+0.2t^2)+(2+0.3t+0.1t^2)(0.4+0.4t)$

$\frac{dA(4)}{dt}=(1.1)(5.8)+(4.8)(2)$

$\frac{dA(4)}{dt}=15.98$

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Example Question #181 : Rate

The area of a circle is given by the function $A(t)=\pi(4+0.09t^2)$ . What is the rate of change of the radius at time t=6 ?

Possible Answers:

$1.08\pi$

1.08

$1.039\pi$

1.039

0.201

Correct answer:

0.201

Explanation:

The area of a circle is given by the function:

$A=\pi r^2$

This equation can be rewritten to define the radius:

$r=\sqrt{\frac{A}{\pi}}$

For the area function

$A(t)=\pi(4+0.09t^2)$

The radius is then

$r(t)=\sqrt{4+0.09t^2}$

$r(t)=(4+0.09t^2)^{\frac{1}{2}}$

The rate of change of the radius can be found by taking the derivative of each side of this equation with respect to time:

$\frac{dr(t)}{dt}=\frac{1}{2}\frac{0.18t}{(4+0.09t^2)^{\frac{1}{2}}}$

$\frac{dr(6)}{dt}=\frac{1}{2}\frac{0.18(6)}{(4+0.09(6)^2)^{\frac{1}{2}}}$

$\frac{dr(6)}{dt}=0.201$

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Example Question #96 : How To Find Rate Of Change

The rate of change of the area of a square is given by the function A(t)=9+4t^2 .

What is the rate of change of its sides at time t=2 ?

Possible Answers:

3.2

0.8

1.6

Correct answer:

1.6

Explanation:

The sides of a square and its area are related via the function

A=s^2

Rewriting the equation in terms of its sides gives

$s=\sqrt{A}$

For the area definition

A(t)=9+4t^2

The sides are then

$s(t)=\sqrt{9+4t^2}$

$s(t)=(9+4t^2)^{\frac{1}{2}}$

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

$\frac{ds(t)}{dt}=\frac{1}{2}\frac{8t}{(9+4t^2)^{\frac{1}{2}}}$

$\frac{ds(2)}{dt}=\frac{1}{2}\frac{8(2)}{(9+4(2)^2)^{\frac{1}{2}}}$

$\frac{ds(2)}{dt}=\frac{8}{(25)^{\frac{1}{2}}}$

$\frac{ds(2)}{dt}=1.6$

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Example Question #97 : How To Find Rate Of Change

The sides of a cube are defined by the function s(t)=3+2t+t^2 .

What is the rate of growth of the cube's volume at time t=3 ?

Possible Answers:

7776

2592

864

256

512

Correct answer:

7776

Explanation:

A cube's volume is defined in terms of its sides as follows:

V=s^3

For sides defined as

s(t)=3+2t+t^2

The volume is

V(t)=(3+2t+t^2)^3

The rate of change can be found by taking the derivative of the function with respect to time

$\frac{dV(t)}{dt}=3(2+2t)(3+2t+t^2)^2$

$\frac{dV(3)}{dt}=3(2+6)(3+6+9)^2$

$\frac{dV(3)}{dt}=3(8)(18)^2$

$\frac{dV(3)}{dt}=7776$

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Example Question #91 : How To Find Rate Of Change

The radius of a sphere is defined in terms of time as follows:

r(t)=1+4t+3t^2 .

What is the rate of change of the sphere's surface area at time t=2 ?

Possible Answers:

$512\pi$

$896\pi$

$1024\pi$

$448\pi$

$2688\pi$

Correct answer:

$2688\pi$

Explanation:

The surface area of a sphere is given by the function

$A=4\pi r^2$

For a radius defined as

r(t)=1+4t+3t^2

The surface area equation becomes

$A(t)=4\pi(1+4t+3t^2)^2$

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

$\frac{dA(t)}{dt}=8\pi(4+6t)(1+4t+3t^2)$

$\frac{dA(2)}{dt}=8\pi(16)(21)$

$\frac{dA(2)}{dt}=2688\pi$

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Example Question #99 : How To Find Rate Of Change

A circle's radius at any point in time is defined by the function r(t)=4+0.1t+0.02t^2 . What is the rate of change of the area at time t=3 ?

Possible Answers:

$3.018\pi$

$1.971\pi$

$1.209\pi$

$3.942\pi$

$2.456\pi$

Correct answer:

$1.971\pi$

Explanation:

The area of a circle is defined by its radius as follows:

$A=\pi r^2$

In the case of the given function for the radius r(t)=4+0.1t+0.02t^2

The area is thus

$A(t)=\pi(4+0.1t+0.02t^2)^2$

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

$\frac{dA(t)}{dt}=2\pi(0.1+0.04t)(4+0.1t+0.02t^2)$

$\frac{dA(3)}{dt}=2\pi(0.1+0.04(3))(4+0.1(3)+0.02(3)^2)$

$\frac{dA(3)}{dt}=2\pi(0.22)(4.48)$

$\frac{dA(3)}{dt}=1.971\pi$

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Example Question #3011 : Calculus

The legs of a right triangle are given by the formulas a(t)=4-t and b(t)=1+2t . What is the maximum area of the triangle?

Possible Answers:

5.0625

12.0625

8.750

4.0125

10.125

Correct answer:

5.0625

Explanation:

The area of a right triangle can be written in terms of its legs (the two shorter sides):

$A=\frac{1}{2}ab$

For sides a(t)=4-t and b(t)=1+2t , the area expression for this problem becomes:

$A(t)=\frac{1}{2}(4-t)(1+2t)$

$A(t)=\frac{1}{2}(-2t^2+7t+4)$

To find where this area has its local maxima/minima, take the derivative with respect to time and set the new equation equal to zero:

$A'(t)=\frac{1}{2}(-4t+7)$

$\frac{1}{2}(-4t+7)=0$

4t=7

$t=\frac{7}{4}$

t=1.75

At an earlier time t=1 , the derivative is postive, and at a later time t=2 , the derivative is negative, indicating that t=1.75 corresponds to a maximum.

$A_{max}=A(1.75)=\frac{1}{2}(-2(1.75)^2+7(1.75)+4)$

$A_{max}=5.0625$

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Example Question #101 : Rate Of Change

A spherical ball of chocolate is melting on a table at a rate of $0.001 \frac{m^3}{s}$ . What is the rate of change of the surface area of the chocolate when the radius of the chocolate ball is 0.01 m?

Possible Answers:

$-5 \frac{m^2}{s}$

$10\frac{m^2}{s}$

$0 \frac{m^2}{s}$

$-\frac{1}{5} \frac{m^2}{s}$

Correct answer:

$-\frac{1}{5} \frac{m^2}{s}$

Explanation:

To solve for the rate of change of the surface area of the chocolate ball, we must relate that rate to the rate of change of the radius, which we get from the given rate of change of volume:

$V=\frac{4}{3}\pi r^3 , \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

$A=4\pi r^2, \frac{dA}{dt}=8 \pi r \frac{dr}{dt}$

Now, using the rate of change of volume, solve for the rate of change of the radius:

$-.001=4\pi (.01)^2 \frac{dr}{dt}$

$\frac{dr}{dt}=-\frac{5}{2\pi}\frac{m}{s}$

Now, plug this into the equation for the rate of change of the surface area, using the given radius:

$\frac{dA}{dt}=8\pi (.01)\left(\frac{-5}{2\pi}\right)=-\frac{1}{5}\frac{m^2}{s}$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #91 : Rate Of Change

Example Question #1973 : Functions

Example Question #91 : How To Find Rate Of Change

Example Question #181 : Rate

Example Question #96 : How To Find Rate Of Change

Example Question #97 : How To Find Rate Of Change

Example Question #91 : How To Find Rate Of Change

Example Question #99 : How To Find Rate Of Change

Example Question #3011 : Calculus

Example Question #101 : Rate Of Change

All Calculus 1 Resources

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