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Calculus 1 : Calculus

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Example Questions

Example Question #18 : How To Find Rate Of Change

Boat leaves a port at noon traveling mph . At the same time, boat leaves the port traveling east at mph . At what rate is the distance between the two boats changing at $3\; p.m.$ ?

Possible Answers:

$\frac{20\sqrt{13}}{3}}$ mph

mph

$\frac{5\sqrt{13}}{39}$ mph

$10\sqrt{13}$ mph

Correct answer:

$10\sqrt{13}$ mph

Explanation:

This scenario describes a right triangle where the hypotenuse is the distance between the two boats. Let denote the distance boat is from the port, denote the distance boat is from the port, denote the distance between the two boats, and denote the time since they left the port. Applying the Pythagorean Theorem we have,

D(t)^2=x(t)^2+y(t)^2 .

Implicitly differentiating this equation we get

$2D(t)\frac{dD}{dt}=2x(t)\frac{dx}{dt}+2y(t)\frac{dy}{dt}$ .

We need to find $\frac{dD}{dt}$ when t=3 .

We are given

$\frac{dx}{dt}=30, \frac{dy}{dt}=20$ which tells us

$x(3)=3\cdot 30=90, y(3)=3\cdot 20=60,$

$D(3)=\sqrt{90^2+60^2}=30\sqrt{13}$ .

Plugging this in we have

$2\cdot 30\sqrt{13}\frac{dD}{dt}|_{t=3}=2\cdot90\cdot 30+2\cdot60\cdot20$ .

Solving we get

$\frac{dD}{dt}|_{t=3}=10\sqrt{13}$ .

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Example Question #11 : Rate Of Change

Find $\frac{dV}{dt}$ if the radius of a spherical balloon is increasing at a rate of $2\,cm$ per second.

Possible Answers:

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

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

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

$\frac{dV}{dt}=8\pi r^2$

Correct answer:

$\frac{dV}{dt}=8\pi r^2$

Explanation:

The volume function, in terms of a radius , is given as

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

The change in volume over the change in time, or

$\frac{dV}{dt}$ is given as

$\frac{dV}{dt} = \frac{d}{dt}[V(r)]$

and by implicit differentiation, the chain rule, and the power rule,

$\frac{d}{dt}[V(r)]=\frac{d}{dt}[\frac{4}{3}\pi r^3]=4\pi r^2 \frac{dr}{dt}$ .

Setting $\frac{dr}{dt}=2$ we get

$\frac{d}{dt}[V(r)]= 4\pi r^2(2)=8\pi r^2$ .

As such,

$\frac{dV}{dt} =8\pi r^2$ .

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Example Question #2931 : Calculus

Find the rate of change of a function f(x)=5x+11 from x_1=2 to x_2=3 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

f(x_1)=f(2)=5(2)+11=10+11=21

f(x_2)=f(3)=5(3)+11=15+11=26

Plugging our values into the average rate of change formula, we get:

$\frac{(26-21)}{(3-2)}=\frac{5}{1}=5$ .

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Example Question #102 : Rate

Find the rate of change of a function f(x)=7x-8 from x_1=5 to x_2=6 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

f(x_1)=f(5)=7(5)-8=35-8=27

f(x_2)=f(6)=7(6)-8=42-8=34

Plugging our values into the average rate of change formula, we get:

$\frac{(34-27)}{(6-5)}=\frac{7}{1}=7$ .

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Example Question #1903 : Functions

Find the rate of change of a function f(x)=10x+12 from x_1=7 to x_2=9 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

f(x_1)=f(7)=10(7)+12=70+12=82

f(x_2)=f(9)=10(9)+12=90+12=102

Plugging our values into the average rate of change formula, we get:

$\frac{(102-82)}{(9-7)}=\frac{20}{2}=10$ .

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Example Question #21 : Rate Of Change

A spherical balloon is inflating at rate of 0.1 cubic meters per second. Find the rate of change of the surface area of the balloon (in square meters per second) when the balloon has a radius of 1 meter.

Possible Answers:

$\frac{1}{40\pi }$

$\frac{1}{5}$

$-\frac{1}{5}$

$8\pi$

Correct answer:

$\frac{1}{5}$

Explanation:

The volume of a sphere is given by

$V=\frac{4\pi }{3}r^3$ , where r is the radius of the sphere.

The surface area of a sphere is given by $A=4\pi r^2$ (r is the radius of the sphere). The rate of change of the volume of the sphere is found by taking the first derivative of the function for volume:

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ , where $\frac{dr}{dt}$ is the change of the radius with respect to time.

This derivative was found by using the power rule

$\frac{d}{dt}x^n=nx^{n-1}\frac{dx}{dt}$ .

The rate of change of the surface area of the sphere is found the same way, instead using the function for surface area:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ .

The balloon's volume is increasing at a constant rate of 0.1 cubic meters per second when the radius of the balloon is 1 meter, thus plugging in these values into the $\frac{dV}{dt}$ equation gives us the following:

$0.1 \frac{m^3}{sec}=4\pi (1^2)\frac{dr}{dt}$ . Solving for $\frac{dr}{dt}$ , we get $\frac{dr}{dt}=\frac{1}{40\pi }$ meters per second. Since we are solving for the rate of change of surface area of the balloon, we plug this value into its equation along with the radius we want (1 m). Solving for the rate of change of the surface area we get:

$\frac{dA}{dt}=8\pi (1)\left(\frac{1}{40\pi }\right)=\frac{1}{5}$ square meters per second.

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Example Question #21 : Rate Of Change

The spherical bottom of a snowman is melting such that its volume is decreasing at a rate of $3\frac{cm^3}{hr}$ .

What is the rate of change of the radius, to the nearest whole number, when the radius is 24cm ?

Possible Answers:

$18\frac{cm}{hr}$

$18.8\frac{cm}{hr}$

$19\frac{cm}{hr}$

$20\frac{cm}{hr}$

Correct answer:

$19\frac{cm}{hr}$

Explanation:

First, find the rate of increase in the radius $\left(\frac{dr}{dt}\right)$ :

$\frac{3\frac{cm^3}{hr}}{24cm}=\frac{1}{8}\frac{cm^2}{hr}$

Next, take the derivative of the area of a circle ( $A=\pi\cdot r^2$ ) by using the Power Rule ( $x^n=nx^{n-1}$ ):

$\frac{dA}{dt}=(2)(\pi)(r)\left(\frac{dr}{dt}\right)$

Lastly, plug in the givens and simplify:

$\frac{dA}{dt}=2\pi(24)\left(\frac{1}{8}\right)=6\pi\approx 18.85 \approx19\frac{cm}{hr}$

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Example Question #111 : Rate

For which of the following functions is the average rate of change along the interval [-2, 2] equal to the instentaneous rate of change at x=0 ?

f(x)=x

$g(x)=x^{2}$

$h(x)=x^{3}$

Possible Answers:

$f(x)\textup{\textup{ only}}$

$f(x)\textup{ and }h(x)$

$f(x) \textup{\textup{ and }} g(x)$

$f(x),g(x),\textup{and }h(x)$

Correct answer:

$f(x) \textup{\textup{ and }} g(x)$

Explanation:

The average rate of change of a function f(x) on an interval [a,b] is given by

$\frac{\Delta f(x)}{\Delta x}=$ $\frac{f(b)-f(a)}{b-a}$

while the instantaneous rate of change at a is given by

$\frac{\mathrm{d} }{\mathrm{d} x}f(a)$ .

Finding each of these on the interval [-2, 2] for the equations given:

$f'(0)=1, \frac{\Delta f(x)}{\Delta x}=1$

$g'(0)=0, \frac{\Delta g(x)}{\Delta x}=0$

$h'(0)=0, \frac{\Delta h(x)}{\Delta x}=4$

We can see that the two rates are equal for f(x) and g(x), but not h(x).

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Example Question #1906 : Functions

Find $\frac{dy}{dx}$ for the equation 8y^2 - 4xy + 3 = 12

Possible Answers:

x + 1

y + x(1 + 4y)

$\frac{x+y}{4y}$

$\frac{y}{4y - x}$

$\frac{x-y}{x + 2y}$

Correct answer:

$\frac{y}{4y - x}$

Explanation:

To evaluate, first take the derivatives of both sides of the equation. This gives

$16y\frac{dy}{dx} - 4x\frac{dy}{dx} - 4y = 0$

By factoring and simple algebraic simplification, the equation becomes

$\frac{dy}{dx} = \frac{y}{x - 4y}$

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Example Question #111 : Rate

A spherical balloon with radius r = 4in is being inflated with air. Its radius is expanding at a rate of $2 \frac{in}{min}$ . What is the change in volume of the sphere (in $\frac{in^3}{min}$ )?

Possible Answers:

$25\pi \frac{in^3}{min}$

$64\pi \frac{in^3}{min}$

$80\pi \frac{in}{min}$

$256\pi \frac{in^3}{min}$

$128\pi \frac{in^3}{min}$

Correct answer:

$128\pi \frac{in^3}{min}$

Explanation:

The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$

To find the change in volume, differentiate the volume with respect to time.

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

Simply plug in the values given in the problem to find the rate of change of the volume.

$\frac{dV}{dt} = 4\pi(4)^2(2) = 128\pi$ $\frac{in^3}{min}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #18 : How To Find Rate Of Change

Example Question #11 : Rate Of Change

Example Question #2931 : Calculus

Example Question #102 : Rate

Example Question #1903 : Functions

Example Question #21 : Rate Of Change

Example Question #21 : Rate Of Change

Example Question #111 : Rate

Example Question #1906 : Functions

Example Question #111 : Rate

All Calculus 1 Resources

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