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Calculus 1 : Calculus

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Example Questions

Example Question #81 : How To Find Rate Of Change

A right triangle has two legs of lengths 6in and 8 in . The longer leg is growing at a rate of $0.3\frac{in}{s}$ . What is the rate of change of the shorter leg if the hypotenuse is not currently changing length?

Possible Answers:

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

$-0.4\frac{in}{s}$

$-0.5\frac{in}{s}$

$-0.225\frac{in}{s}$

$-0.3\frac{in}{s}$

Correct answer:

$-0.4\frac{in}{s}$

Explanation:

The hypotenuse's length relation to the legs of a right triangle is shown by the Pythagorean Theorem:

a^2+b^2=c^2

This theorem can also be used to show how the rates of change of each parameter relate if we take the derivative of each side of the equation with respect tot ime:

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}$ or simply $a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}$

Treating as the smaller side and as the longer side, a= 6in , b=8in , $\frac{db}{dt}=0.3\frac{in}{s}$ . Since the hypotenuse is not changing length at the moment, $\frac{dc}{dt}=0$ . This allows us to plug in terms into the above equation:

$6in\frac{da}{dt}+8in(0.3\frac{in}{s})=c(0)=0$

(In this case, it was unnecessary to calculate that c=10in )

Solving for $\frac{da}{dt}$ :

$6in\frac{da}{dt}=-2.4\frac{in^2}{s}$

$\frac{da}{dt}=-0.4\frac{in}{s}$

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Example Question #81 : How To Find Rate Of Change

$z=3x^{2}y^{4}$ where , , and all change with respect time. When x=1 and y=3 , changes at rate of 2 units/s and changes at rate of 4 units/s . At what rate is changing?

Possible Answers:

$\frac{dz}{dt}=3$

$\frac{dz}{dt}=0$

$\frac{dz}{dt}=2268$

$\frac{dz}{dt}=812$

$\frac{dz}{dt}=57$

Correct answer:

$\frac{dz}{dt}=2268$

Explanation:

The rate at which changes is $\frac{dz}{dt}$ . The find $\frac{dz}{dt}$ , we will find the first derivative of the equation

$z=3x^{2}y^{4}$

Since and both change with respect to time, so we must use the product rule when differentiating.

The product rule is $\frac{d}{dt}(ab)=a\frac{d}{dt}(b)+b\frac{d}{dt}(a)$ .

The first derivative of is

$\frac{d}{dt}(z)=\frac{d}{dt}[3x^{2}y^{4}]=3\left [x^{2}\frac{d}{dt}(y^{4})+y^{4}\frac{d}{dt}(x^{2}) \right ]$

$\frac{dz}{dt}=3\left [x^{2}(4y^{3})\frac{dy}{dt}+y^{4}(2x)\frac{dx}{dt} \right ]$

$\frac{dz}{dt}=3\left [4x^{2}y^{3}\frac{dy}{dt}+2xy^{4}\frac{dx}{dt} \right ]$

We are given the following parameters

x=1 , y=3 , $\frac{dx}{dt}=2 units/s$ , and $\frac{dy}{dt}=4 units/s$ .

Subsituting these values into our derivative equation

$\frac{dz}{dt}=3\left [4x^{2}y^{3}\frac{dy}{dt}+2xy^{4}\frac{dx}{dt} \right ]$

$\frac{dz}{dt}=3\left [4(1)^{2}(3)^{3}(4)+2(1)(3)^{4}(2) \right ]$

$\frac{dz}{dt}=3\left [432+324 \right ]=2268$

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Example Question #1973 : Functions

Find the rate of change of a line connected by the points (0, 4) and (12, -7) .

Possible Answers:

$\frac{-11}{12}$

$\frac{1}{12}$

$\frac{-12}{7}$

Correct answer:

$\frac{-11}{12}$

Explanation:

To find the rate of change of a line connected by two points. We will use the following equation for slope.

$m=\frac{\Delta y}{\Delta x}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-7-4}{12-0}=\frac{-11}{12}$

The rate of change of the line is $\frac{-11}{12}$ .

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Example Question #1972 : Functions

A circle of radius r=3in is inscribed inside of a square with sides of length l=8in . If the radius of the circle is expanding at a rate of $0.3\frac{in}{s}$ , what is the rate of change of the sides such that the amount of area inscribed between the square and circle does not change?

Possible Answers:

$0.1125\pi\frac{in}{s}$

$0.3\frac{in}{s}$

$0.0375\frac{in}{s}$

$0.1125\frac{in}{s}$

$0.3\pi\frac{in}{s}$

Correct answer:

$0.1125\pi\frac{in}{s}$

Explanation:

The amount of area between the square and circle is given by the difference of the two individual areas, the larger and smaller:

$A_{diff}=l^2-\pi r^2$

It then holds that the rate of change of this difference in area can be found by taking the time derivative of each side of the equation:

$\frac{dA_{diff}}{dt}=2l\frac{dl}{dt}-2\pi r \frac{dr}{dt}$

We are told that the difference in area is not changing, which means that $\frac{dA_{diff}}{dt}=0$ . Therefore:

$2l\frac{dl}{dt}-2\pi r \frac{dr}{dt}=0$

$2(8in)\frac{dl}{dt}-2\pi(3in)(0.3\frac{in}{s})=0$

$8in\frac{dl}{dt}=0.9\pi\frac{in^2}{s}$

$\frac{dl}{dt}=0.1125\pi\frac{in}{s}$

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Example Question #91 : How To Find 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:

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

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

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

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

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

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 : 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:

2.20

7.99

6.18

12.36

15.98

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 #95 : How To Find Rate Of Change

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

0.201

1.039

$1.08\pi$

$1.039\pi$

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #81 : How To Find Rate Of Change

Example Question #81 : How To Find Rate Of Change

Example Question #1973 : Functions

Example Question #1972 : Functions

Example Question #91 : How To Find Rate Of Change

Example Question #1973 : Functions

Example Question #91 : Rate Of Change

Example Question #95 : How To Find Rate Of Change

Example Question #96 : How To Find Rate Of Change

Example Question #97 : How To Find Rate Of Change

All Calculus 1 Resources

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