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Calculus 1 : Calculus

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Example Questions

Example Question #138 : How To Find Rate Of Change

The rates of change of the width and length of a rectangle are $3e^{-t}$ and $6e^{-t}$ respectively. If the intial widths and lengths are, respectively and , what is the rate of change of the rectangle's area at time t=ln(3) ?

Possible Answers:

Correct answer:

Explanation:

The rates of change of the width and length of a rectangle are $3e^{-t}$ and $6e^{-t}$ , which is to say

$\frac{dw(t)}{dt}=3e^{-t}$

$\frac{dl(t)}{dt}=6e^{-t}$

These equations can be used to find formulae for the width and length by taking the integral of each:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$w(t)=-3e^{-t}+C_w$

$l(t)=-6e^{-t}+C_l$

These constants of integration can be found by using the initial conditions:

$w(0)=-3e^{0}+C_w=4$

-3+C_w=4

C_w=7

$l(0)=-6e^{0}+C_l=8$

-6+C_l=8

C_l=14

$w(t)=-3e^{-t}+7$

$l(t)=-6e^{-t}+14$

Now the area of a rectangle is given by the formula

A=wl

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

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

$\frac{dA(t)}{dt}=(7-3e^{-t})(6e^{-t})+(14-6e^{-t})(3e^{-t})$

$\frac{dA(ln(3))}{dt}=(7-3e^{-ln(3)})(6e^{-ln(3)})+(14-6e^{-ln(3)})(3e^{-ln(3)})$

$\frac{dA(ln(3))}{dt}=24$

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Example Question #139 : How To Find Rate Of Change

The rate of change of the sides of a square is 5-3sin(t) . What is the rate of change of the area at time $t=\pi$ if the the sides have an initial length of ?

Possible Answers:

$2+2\pi$

$20+20\pi$

$2+5\pi$

$20+50\pi$

$5+5\pi$

Correct answer:

$20+50\pi$

Explanation:

The rate of change of the sides of a square is 5-3sin(t) , which is to say

$\frac{ds(t)}{dt}=5-3sin(t)$ .

A formula for the length of the sides themselves can be found by integrating this equation with respect to time:

$\int a +bsin(ct)=at-\frac{bcos(ct)}{c}+C;(a,b,c,C:constants)$

s(t)=5t+3cos(t)+C

To find this constant of integration, use the initial condition given

s(0)=5(0)+3cos(0)+C=8

3+C=8

C=5

s(t)=5t+3cos(t)+5

Now, the area of a square is given by the equation

A=s^2

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

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

$\frac{dA(t)}{dt}=2(5t+3cos(t)+5)(5-3sin(t))$

$\frac{dA(\pi)}{dt}=2(5\pi+3cos(\pi)+5)(5-3sin(\pi))$

$\frac{dA(\pi)}{dt}=20+50\pi$

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Example Question #131 : How To Find Rate Of Change

The rate of change of the radius of a circle is given by the equation . What is the rate of growth of the circle's area at time $t=2\pi$ if the radius has an initial length of ?

Possible Answers:

$18\pi+24\pi^2$

$3\pi+4\pi^2$

$18+24\pi$

$6\pi+8\pi^2$

$3+4\pi$

Correct answer:

$18\pi+24\pi^2$

Explanation:

The rate of change of the radius of a circle is given by the equation which is to say

$\frac{dr(t)}{dt}=2+cos(t)$

The equation for the length of the radius can be found by taking the integral with respect to time:

$\int (a+bcos(ct))dt=at+\frac{bsin(ct)}{c}+C;(a,b,c,C:constants)$

r(t)=2t+sin(t)+C

This constant of integration can be found by using the initial condition:

r(0)=2(0)+sin(0)+C=3

C=3

r(t)=2t+sin(t)+3

The area of a circle is given by the equation

$A=\pi r^2$

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

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

$\frac{dA(t)}{dt}=2\pi (2t+sin(t)+3)(2+cos(t))$

$\frac{dA(2\pi)}{dt}=2\pi (2(2\pi)+sin(2\pi)+3)(2+cos(2\pi))$

$\frac{dA(2\pi)}{dt}=18\pi+24\pi^2$

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Example Question #3051 : Calculus

The rate of change of the radius of a sphere is given by the equation $8e^{-2t}$ . If the initial length of the radius is , what is the rate of change of the sphere's volume at time t=ln(2) ?

Possible Answers:

$49\pi$

$98\pi$

$392\pi$

$56\pi$

$8\pi$

Correct answer:

$392\pi$

Explanation:

The rate of change of the radius of a sphere is given by the equation $8e^{-2t}$ . Written in mathematical terms

$\frac{dr(t)}{dt}=8e^{-2t}$

The formula for the length of the radius can be found by taking the integral of this function with respect to time

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$r(t)=-4e^{-2t}+C$

This constant of integration can be found by utilizing the initial condition

$r(0)=-4e^{0}+C=4$

-4+C=4

C=8

$r(t)=8-4e^{-2t}$

The volume of a sphere is given by the equation

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

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

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

$\frac{dV(t)}{dt}=4 \pi (8-4e^{-2t})^2 (8e^{-2t})$

$\frac{dV(ln(2))}{dt}=4 \pi (8-4e^{-2ln(2)})^2 (8e^{-2ln(2)})$

$\frac{dV(ln(2))}{dt}=392\pi$

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Example Question #142 : Rate Of Change

The rate of change of the radius of a sphere is $9e^{-t}$ . If the sphere has an initial radius of , what is the rate of change of the sphere's surface area at time t=ln(3) ?

Possible Answers:

$288\pi$

$18\pi$

$36\pi$

$144\pi$

$72\pi$

Correct answer:

$288\pi$

Explanation:

To say that the rate of change of the radius of a sphere is $9e^{-t}$ means

$\frac{dr(t)}{dt}=9e^{-t}$

The equation for the length of the radius can be found by integrating this equation with respect to time:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$r(t)=-9e^{-t}+C$

The constant of integration can be found by utilizing the initial condition:

$r(0)=-9e^{0}+C=6$

-9+C=6

C=15

$r(t)=15-9e^{-t}$

The surface area of a sphere is given by the equation

$A=4\pi r^2$

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

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

$\frac{dA(t)}{dt}=8\pi (15-9e^{-t}) (9e^{-t})$

$\frac{dA(ln(3))}{dt}=8\pi (15-9e^{-ln(3)}) (9e^{-ln(3)})$

$\frac{dA(ln(3))}{dt}=288\pi$

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Example Question #143 : Rate Of Change

The rate of growth of the sides of a cube is $16e^{-4t}$ . If the cube has an initial volume of , what is the rate of growth of the surface area at time $t=ln(\sqrt{2})$ ?

Possible Answers:

336

672

112

Correct answer:

336

Explanation:

Saying that the rate of growth of the sides of a cube is $16e^{-4t}$ , means in mathematical terms

$\frac{ds(t)}{dt}=16e^{-4t}$

The length of the sides at any time can be found by integrating this equation with respect to time:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$s(t)=-4e^{-4t}+C$

To find this constant of integration, we'll need the value for the length of the sides at some known point of time. We're told the initial volume; we can use this to find the initial lengths of the cube's sides:

V_0=s_0^3=64

s_0=4

$s(0)=-4e^{-0}+C=4$

-4+C=4

C=8

$s(t)=8-4e^{-4t}$

The surface area of a cube is given by the equation

A=6s^2

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

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

$\frac{dA(t)}{dt}=12(8-4e^{-4t})(16e^{-4t})$

$\frac{dA(ln(\sqrt{2}))}{dt}=12(8-4e^{-4ln(\sqrt{2})})(16e^{-4ln(\sqrt{2})})$

$\frac{dA(ln(\sqrt{2}))}{dt}=336$

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Example Question #144 : Rate Of Change

The rate of change of the sides of a square is $6e^{-2t}$ . If the sides have initial lengths of , what is the rate of growth of the square's area at time $t=ln(\sqrt{2})$ ?

Possible Answers:

$\frac{9}{4}$

$\frac{25}{4}$

$\frac{25}{8}$

$\frac{75}{4}$

$\frac{3}{4}$

Correct answer:

$\frac{75}{4}$

Explanation:

We're told that the rate of growth of the sides of the square is $6e^{-2t}$

Which is to say $\frac{ds(t)}{dt}=6e^{-2t}$

The function for the sides can thus be found by integrating this function:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$s(t)=-3e^{-2t}+C$

To find this constant of integration, use the initial condition

s(0)=-3^0+C=4

-3+C=4

C=7

$s(t)=-3e^{-2t}+7$

Now, consider the area of a square

A=s^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:

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

$\frac{dA(t)}{dt}=2(-3e^{-2t}+7)(6e^{-2t})$

$\frac{dA(ln(2))}{dt}=2(-3e^{-2ln(2)}+7)(6e^{-2ln(2)})$

$\frac{dA(ln(2))}{dt}=\frac{75}{4}$

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Example Question #3051 : Calculus

Find the derivative of the following function.

f(x)=3x^4+2x^2-17x+15

Possible Answers:

12x^3+4x^2-17

3x^4+2x^2-17x+15

12x^3+2x^2-17x+15

12x^3+4x-17

12x^3+4x

Correct answer:

12x^3+4x-17

Explanation:

Because this function can be written in four seperate terms of addition (3x^4, 2x^2, -17x, 15) we are allowed to take the derivative of each component individually.

Using the power rule $(\frac{\mathrm{d} }{\mathrm{d} x}(x^n)=nx^{n-1})$ , we can derive each component individually to give us our answer,

12x^3+4x-17 (the 15 disappears because it is just a constant)

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Example Question #231 : Rate

Find the average rate of change of the function given by

f(x)=3x^2+sin(x) from x=0 to $x=\pi$

Possible Answers:

$3\pi^2$

$3\pi$

$-3\pi$

Correct answer:

$3\pi$

Explanation:

The average rate of change $\Delta f$ is given by

$\Delta f=\frac{f(b)-f(a)}{b-a}$ , where and are the boundaries between which we're trying to determine the average.

For this function, $b=\pi$ and a=0

f(x)=3x^2+sin(x)

Plugging into the equation, the average rate of change $\Delta f$ can be given by:

$\Delta f=\frac{f(\pi)-f(0)}{\pi-0}$

$\Delta f=\frac{(3(\pi )^2+sin(\pi))-(3(0)^2+sin(0))}{\pi-0}=\frac{3\pi^2}{\pi}=3\pi$

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Example Question #141 : Rate Of Change

A spherical balloon is being filled with air. What is the surface area of the sphere at the instance the rate of growth of the volume is three times the rate of growth of the radius?

Possible Answers:

$\frac{9}{\pi}$

$3\pi$

Correct answer:

Explanation:

Let's begin by writing the equations for the volumeof a sphere with respect to the sphere's radius:

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

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, let's solve for a radius that satisfies it the problem statement: What is the surface area of the sphere at the instance the rate of growth of the volume is three times the rate of growth of the radius?

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

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

$r=\sqrt{\frac{3}{4\pi}}$

Now to solve for the surface area:

$A=4\pi r^2$

$A=4\pi (\sqrt{\frac{3}{4\pi}})^2$

A=3

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #138 : How To Find Rate Of Change

Example Question #139 : How To Find Rate Of Change

Example Question #131 : How To Find Rate Of Change

Example Question #3051 : Calculus

Example Question #142 : Rate Of Change

Example Question #143 : Rate Of Change

Example Question #144 : Rate Of Change

Example Question #3051 : Calculus

Example Question #231 : Rate

Example Question #141 : Rate Of Change

All Calculus 1 Resources

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