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Calculus 1 : Calculus

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Example Questions

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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Example Question #131 : How To Find Rate Of Change

The radius of a sphere is given with respect to time by the equation $r(t)=6-4e^{-\frac{1}{2}t}$ .

What is the rate of expansion of the sphere's volume at time $t=\ln(9)$ ?

Possible Answers:

$\frac{1568}{3}\pi$

$\frac{1568}{243}\pi$

$\frac{1568}{81}\pi$

$\frac{1568}{27}\pi$

$\frac{1568}{9}\pi$

Correct answer:

$\frac{1568}{27}\pi$

Explanation:

The volume of a sphere is given by the equation

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

Since the radius is given as

$r(t)=6-4e^{-\frac{1}{2}t}$

The volume can be expressed as:

$V(t)=\frac{4}{3}\pi(6-4e^{-\frac{1}{2}t})^3$

The rate of change of the volume can then, in turn, by taking the derivative of each side of the equation with respect to time.

Now, we could expand the function outwards, actually cubing it; however this is not necessary. We need only utilize the following derivative rules:

$d(u^n)=nu^{n-1}du;(n=constant)$

d(e^u)=e^udu

These rules hold even when u is a complex function; just don't assume du is 1!

Following these rules, we can find the derivative:

$\frac{dV(t)}{dt}=\frac{4}{3}\pi(3)(2e^{-\frac{1}{2}t})(6-4e^{-\frac{1}{2}t})^2$

$\frac{dV(t)}{dt}=4\pi(2e^{-\frac{1}{2}t})(6-4e^{-\frac{1}{2}t})^2$

At time $t=\ln(9)$ :

$\frac{dV(\ln(9))}{dt}=4\pi(2e^{-\frac{1}{2}(\ln(9))})(6-4e^{-\frac{1}{2}(\ln(9))})^2$

$\frac{dV(ln(9))}{dt}=\frac{1568}{27}\pi$

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Example Question #132 : How To Find Rate Of Change

The legs of a right triangle are given by the formulas $a(t)=6-3e^{-2t}$ , $b(t)=5-e^{-3t}$ .

What is the rate of change of the triangle's hypotenuse at time $t=\ln(2)$ ?

Possible Answers:

2.89

4.77

14.99

12.35

1.35

Correct answer:

1.35

Explanation:

The hypotenuse of a triangle can be found in terms of the triangle's legs using the Pythagorean Theorem:

c^2=a^2+b^2

$c=\sqrt{a^2+b^2}$

This equation also enables us to find the rate of change of the hypotenuse; we need only take the derivative with respect to time.

The legs are given as $a(t)=6-3e^{-2t}$ and $b(t)=5-e^{-3t}$ .

Plugging this into the equation for the hypotenuse, we find

$c(t)=\sqrt{(6-3e^{-2t})^2+(5-e^{-3t})^2}$

Now, to take the derivative of this equation, we'll utilize the following rules:

d(e^u)=du(e^u)

$d(u^n)=n(u^{n-1})du; (n:\uptext{constant})$

Note that u can represent a complicated function; the du term should not be dismissed! Consideration of it is known as the chain rule.

With these rules in mind, let's consider the derivative. First, let's rewrite the hypotenuse equation in terms of exponents rather than the square root symbol:

$c(t)=[(6-3e^{-2t})^2+(5-e^{-3t})^2]^{\frac{1}{2}}$

Now let's take the derivative:

$\frac{dc(t)}{dt}=\frac{1}{2}\frac{2(6e^{-2t})(6-3e^{-2t})+2(3e^{-3t})(5-e^{-3t})}{[(6-3e^{-2t})^2+(5-e^{-3t})^2]^{\frac{1}{2}}}$

Notice how the chain rule was utilized for the functions inside of the square root as well, for example the first term: $\frac{d}{dt}((6-3e^{-2t})^2)=2(6e^{-2t})(6-3e^{-2t})$

It may seem complicated, but being mindful of order will serve well here.

Now, the rate of change of the hypotenuse at time $t=\ln(2)$ is

$\frac{dc(\ln(2)))}{dt}=\frac{1}{2}\frac{2(6e^{-2(\ln(2))})(6-3e^{-2(\ln(2))})+2(3e^{-3(\ln(2))})(5-e^{-3(\ln(2))})}{[(6-3e^{-2(\ln(2))})^2+(5-e^{-3(\ln(2))})^2]^{\frac{1}{2}}}$

$\frac{dc(\ln(2)))}{dt}=1.35$

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Example Question #221 : Rate

A circle is expanding in a pulsating manner. The radius is given by the function $r(t)=t^2+\cos(t)+4$

What is the rate of growth of the area at time $t=\frac{\pi}{2}$ ?

Possible Answers:

44.52

108.93

87.03

23.26

11.15

Correct answer:

87.03

Explanation:

The area of a circle, in terms of its radius, is given by the function:

$A=\pi r^2$

Since the radius is given by the formula

$r(t)=t^2+\cos(t)+4$

The area in turn can be expressed as

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

We're looking for the rate of change of this area now; this can be found by taking the derivative of each side of the equation with respect to time.

To do perform this derivative recall the following derivative properties:

$d(u^n)=n(u^{n-1})du$

$d(\cos(u))=-du(\sin(u))$

Where can represent a complex function.

Following these rules, the rate of change of the area can be found to be

$\frac{dA(t)}{dt}=2\pi (2t-\sin(t))(t^2+\cos(t)+4)$

At time $t=\frac{\pi}{2}$

$\frac{dA(\frac{\pi}{2})}{dt}=2\pi (2(\frac{\pi}{2})-\sin(\frac{\pi}{2}))((\frac{\pi}{2})^2+\cos(\frac{\pi}{2})+4)$

$\frac{dA(\frac{\pi}{2})}{dt}=87.03$

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Example Question #134 : How To Find Rate Of Change

The sides of a cube are given by the formula $s(t)=5-2e^{-2t}$

What is the rate of change of the cube's surface area at time $t=\ln(4)$ ?

Possible Answers:

$\frac{117}{2}$

$\frac{117}{32}$

$\frac{117}{8}$

$\frac{117}{4}$

$\frac{117}{16}$

Correct answer:

$\frac{117}{8}$

Explanation:

The surface area of a cube is given by the formula

A=6s^2

For sides expressed by the formula

$s(t)=5-2e^{-2t}$

The area equation becomes

$A(t)=6(5-2e^{-2t})^2$

To find the rate of change of the surface area, take the derivative of each side of equation with respect to time. We'll utilize the following rules:

$d(u^n)=nu^{n-1}du; (n:constant)$

d(e^u)=e^udu

Note that u can be a complex function.

Thus the derivative is

$\frac{dA(t)}{dt}=12(4e^{-2t})(5-2e^{-2t})$

At time $t=\ln(4)$

$\frac{dA(\ln(4))}{dt}=12(4e^{-2(\ln(4))})(5-2e^{-2(\ln(4))})$

$\frac{dA(\ln(4))}{dt}=\frac{117}{8}$

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Example Question #135 : How To Find Rate Of Change

The radius of a sphere is currently , though it begins to expand a rate of 0.9 . What is the rate of growth of the sphere's surface area?

Possible Answers:

$3.24\pi$

$10.8 \pi$

$21.6\pi$

$43.2\pi$

$6.48\pi$

Correct answer:

$21.6\pi$

Explanation:

The surface area of a sphere is given by the formula

$A=4\pi r^2$

We're told that r=3 and $\frac{dr}{dt}=0.9$

Now, to find the rate of change of the surface area, take the derivative of the area equation with respect to time:

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

$\frac{dA}{dt}=8\pi (3)(0.9)$

$\frac{dA}{dt}=21.6\pi$

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Example Question #136 : How To Find Rate Of Change

The rate of change of a circle's radius is given by the equation $10e^{-\frac{1}{2}t}$ . The circle's radius has an initial value of . What is the rate of growth of the circle's area at time t=ln(4) ?

Possible Answers:

$25\pi$

$160 \pi$

$100\pi$

$50\pi$

$80 \pi$

Correct answer:

$160 \pi$

Explanation:

We're given a rate of change for the circle's radius

$\frac{dr(t)}{dt}=10e^{-\frac{1}{2}t}$

The function for the radius can be found by integrating this function

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

$r(t)=-20e^{-\frac{1}{2}t}+C$

To find the constant of integration, use the initial condition

$r(0)=-20e^{0}+C=6$

-20+C=6

C=26

$r(t)=-20e^{-\frac{1}{2}t}+26$

Now, the formula for a circle's area is

$A=\pi r^2$

To find the rate of change, take the derivative with respect to time:

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

$\frac{dA(t)}{dt}=2\pi (26-20e^{-\frac{1}{2}t})(10e^{-\frac{1}{2}t})$

$\frac{dA(t)}{dt}=2\pi (26-20e^{-\frac{1}{2}(ln(4))})(10e^{-\frac{1}{2}(ln(4))})$ $\frac{dA(t)}{dt}=160 \pi$

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Example Question #221 : Rate

The rate of change of the sides of a square is $5-5e^{-t}$ . If the square has an initial area of , what is the rate of change of the area at time t=ln(5) ?

Possible Answers:

40+40ln(5)

16+40ln(5)

8+40ln(5)

16+16ln(5)

8+5ln(5)

Correct answer:

16+40ln(5)

Explanation:

We're told that the rate of change of the square's sides is $5-5e^{-t}$ which is to say

$\frac{ds(t)}{dt}=5-5e^{-t}$

The formula for the sides themselves can be found by integrating this equation

$\int (a+e^{bt})dt=at+\frac{e^{bt}}{b}+C;(a,b,C:constants)$

$s(t)=5t+5e^{-t}+C$

To find this constant of integration, we need to know the value of the equation at some point of time. We do not know the initial value of the sides, but we know the initial value of the area which enables us to calculate this:

A_0=s_0^2=36

s_0=6

$s(0)=5(0)+5e^{0}+C=6$

5+C=6

C=1

$s(t)=5t+5e^{-t}+1$

Now let's return to the area 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:

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

$\frac{dA(t)}{dt}=2(5t+5e^{-t}+1)(5-5e^{-t})$

$\frac{dA(ln(5))}{dt}=2(5ln(5)+5e^{-ln(5)}+1)(5-5e^{-ln(5)})$

$\frac{dA(ln(5))}{dt}=16+40ln(5)$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #128 : Rate Of Change

Example Question #129 : Rate Of Change

Example Question #130 : Rate Of Change

Example Question #131 : How To Find Rate Of Change

Example Question #132 : How To Find Rate Of Change

Example Question #221 : Rate

Example Question #134 : How To Find Rate Of Change

Example Question #135 : How To Find Rate Of Change

Example Question #136 : How To Find Rate Of Change

Example Question #221 : Rate

All Calculus 1 Resources

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