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Calculus 1 : Functions

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Example Questions

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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Example Question #21 : Rate Of Change

For the relation $\small x^2+y^4=2$ , compute $\small y'$ using implicit differentiation.

Possible Answers:

$\small y'=-\frac{x}{2y^3}$

$\small \small y'=-\frac{2x^4}{y}$

$\small \small y'=\frac{x}{2y^3}$

$\small \small y'=-\frac{x}{4y^3}$

Correct answer:

$\small y'=-\frac{x}{2y^3}$

Explanation:

To calculate $\small y'$ using implicit differentiation, we differentiate everything in the relation $\small x^2+y^4=2$ and solve for $\small y'$ :

$\small \frac{d}{dx}(x^2+y^4)=\frac{d}{dx}(2)$

$\small \small \implies 2x+4y^3\cdot y'=0$

$\small \small \iff 4y^3 y'=-2x$

$\small y'=-\frac{2x}{4y^3}=-\frac{x}{2y^3}$

So we have $\small y'=-\frac{x}{2y^3}$ .

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Example Question #22 : Rate Of Change

If b(t) models the number of books a library has a function of time (in days), at what rate was the library gaining or losing books when $t=\frac{3\pi}{4}$ ?

b(t)=6cos(x)+200x

Possible Answers:

The library was losing 19.576 books per day.

The library was gaining 19.576 books per day.

The library was gaining 195.76 books per day.

The library was losing 195.76 books per day.

Correct answer:

The library was gaining 195.76 books per day.

Explanation:

If b(t) models the rate at which a library buys and sells books, at what rate was the library gaining or losing books when $t=\frac{3\pi}{4}$ ?

b(t)=6cos(x)+200x

We need to find an instantaneous rate of change. Rates of change can be found with derivatives. Start by finding b'(t), and then using our given t value to find our final answer:

b(t)=6cos(x)+200x

b'(t)=-6sin(x)+200

$b'\left(\frac{3\pi}{4}\right)=-6sin\left(\frac{3\pi}{4}\right)+200=195.76$

So the library was gaining 195.76 books per day.

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Example Question #30 : How To Find Rate Of Change

Abel wants to fill a spherical ball with air to bring with him to the park. It has an initial volume of 450in^3 , and when Abel attaches the pumps, he sets it to inflate at a rate of $0.2\frac{in^3}{s}$ . How quickly does the point where the nozzle connects move from the center of the basketball?

Possible Answers:

$\frac{1}{180\pi\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

$\frac{3\pi}{180\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

$\frac{1}{180\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

$\frac{\pi}{180\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

$\frac{3}{180\pi\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

Correct answer:

$\frac{1}{180\pi\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

Explanation:

To solve this problem, first understand what it's asking. When it talks about the distance between the center of the basketball and the point of connection, it's talking about a radius, and when it asks about the movement of this point, it's talking about a change in the radius.

First, write an equation for the radius with respect to the volume:

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

$r=\sqrt[3]{\frac{3}{4\pi}V}$

Afterwards, derive each side with respect to time, and plug in known values.

$\frac{dr}{dt}=\frac{1}{3}(\frac{3}{4\pi})(\frac{3}{4\pi}V)^{-\frac{2}{3}}\frac{dV}{dt}$

$\frac{dr}{dt}=\frac{1}{4\pi \sqrt[\frac{2}{3}]{\frac{3}{4\pi}V}}\frac{dV}{dt}$

$\frac{dr}{dt}=\frac{1}{180\pi\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

Another approach is to derive the original volume equation:

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

$\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$

However, since $r=\sqrt[3]{\frac{3}{4\pi}V}$ , this also reduces to:

$\frac{dr}{dt}=\frac{1}{4\pi \sqrt[\frac{2}{3}]{\frac{3}{4\pi}V}}\frac{dV}{dt}$

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Example Question #31 : How To Find Rate Of Change

A cylinder is growing wider and wider. If it has a radius of ten inches, a height of forty inches, and its radius expands at a rate of two inches per second, what is the rate of growth of its volume?

Possible Answers:

$800\pi\frac{in^3}{s}$

$2000\pi\frac{in^3}{s}$

$1600\pi\frac{in^3}{s}$

$100\pi\frac{in^3}{s}$

$1800\pi\frac{in^3}{s}$

Correct answer:

$1600\pi\frac{in^3}{s}$

Explanation:

A cylinder's volume is defined in terms of its radius and height as:

$V=\pi r^2h$

Therefore, it's rate of volume expansion is found by taking the derivative of each side with respect to time:

$\frac{dV}{dt}=\frac{d}{dt}(\pi r^2h)=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}$

Since there's zero growth in height, the second term goes to zero, and we're left with:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}=2\pi(10in)(40in)(2\frac{in}{s})=1600\pi\frac{in^3}{s}$

What an unnatural cylinder.

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Example Question #32 : Rate Of Change

The surface area of a cone is given by the formula $A=\pi r^2 + \pi r\sqrt{h^2+r^2}$ where is the radius of the cone at its base, and is its height.

A cone with a radius of and a height of is widening, its radius growing at a rate of $0.2\frac{m}{s}$ , though its height remains fixed. What is the rate of growth of its surface area?

Possible Answers:

$0.32\pi\frac{m^2}{s}$

$5.12\pi\frac{m^2}{s}$

$1.28\pi\frac{m^2}{s}$

$2.56\pi\frac{m^2}{s}$

$0.64\pi\frac{m^2}{s}$

Correct answer:

$2.56\pi\frac{m^2}{s}$

Explanation:

To find the rate of growth of the area, take the time derivative of both sides of the surface area equation:

$\frac{dA}{dt}=\frac{d}{dt}(\pi r^2 + \pi r\sqrt{h^2+r^2})$

$\frac{dA}{dt}=2\pi r \frac{dr}{dt}+\pi\frac{dr}{dt}\sqrt{h^2+r^2}+\frac{\pi r^2\frac{dr}{dt}}{\sqrt{h^2+r^2}}+\frac{\pi rh\frac{dh}{dt}}{\sqrt{h^2+r^2}}$

Since the height does not change with time, the fourth term is eliminated, leaving:

$\frac{dA}{dt}=(2\pi r +\pi\sqrt{h^2+r^2}+\frac{\pi r^2}{\sqrt{h^2+r^2}})\frac{dr}{dt}$

$\frac{dA}{dt}=(2\pi (3m) +\pi\sqrt{(4m)^2+(3)^2}+\frac{\pi (3m)^2}{\sqrt{(4m)^2+(3m)^2}})0.2\frac{m}{s}$

$\frac{dA}{dt}=0.2\pi(6+5+1.8)\frac{m^2}{s}=2.56\pi\frac{m^2}{s}$

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Example Question #32 : How To Find Rate Of Change

The volume of a hot hair balloon decreases at a rate of $1\frac{m^3}{s}$ . What is the rate at which its radius decreases when the diameter of the balloon is 10m ?

Possible Answers:

$\frac{1}{400\pi}\frac{m}{s}$

$\frac{1}{50\pi}\frac{m}{s}$

$\frac{1}{100\pi}\frac{m}{s}$

$50\pi\frac{m}{s}$

$100\pi\frac{m}{s}$

Correct answer:

$\frac{1}{100\pi}\frac{m}{s}$

Explanation:

To solve this problem, first note the volume of the baloon as a function of its radius:

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

Now, derive both sides with respect to time, treating both volume and radius and functions of it:

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

The rate of change of the radius, $\frac{dr}{dt}$ , is what we are interested in, so rewrite the equation in terms of that:

$\frac{dr}{dt} =\frac{1}{4\pi r^2 }\frac{dV}{dt}$

The diameter of the baloon is 10m , so its radius is . $\frac{dV}{dt}$ is given as $1\frac{m^3}{s}$ , so the rate of change of the radius at this instant in time is:

$\frac{dr}{dt} =\frac{1}{4\pi (5m)^2 }\left(1\frac{m^3}{s}\right)=\frac{1}{100\pi}\frac{m}{s}$

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Example Question #31 : Rate Of Change

The sides of a square increase at a rate of $0.1 \frac{in}{s}$ . If the area of the square is 9 in^2 , what is its rate of change?

Possible Answers:

$0.01\frac{in^2}{s}$

$0.1\frac{in^2}{s}$

$0.3\frac{in^2}{s}$

$0.06\frac{in^2}{s}$

$0.6\frac{in^2}{s}$

Correct answer:

$0.6\frac{in^2}{s}$

Explanation:

Area of a square is given in terms of its sides as:

A=s^2

Deriving each side with respect of time allows us to find its rate of change:

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

The rate of change of the sides, $\frac{ds}{dt}$ , is given to us as $0.1 \frac{in}{s}$ . To find the length of the side, use the orignal formula:

A=s^2

9in^2=s^2

3in=s

Therefore:

$\frac{dA}{dt}=2(3in)\left(0.1\frac{in}{s}\right)=0.6\frac{in^2}{s}$

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Example Question #34 : How To Find Rate Of Change

A rectangle with an area of 6in^2 and a perimeter of 10in is expanding. The longer side increases at a rate of $0.3\frac{in}{s}$ and the shorter side increases at a rate of $0.4\frac{in}{s}$ . What is the rate of change of the rectangle's area?

Possible Answers:

$1.7\frac{in^2}{s}$

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

$0.72\frac{in^2}{s}$

$0.12\frac{in^2}{s}$

$3.6\frac{in^2}{s}$

Correct answer:

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

Explanation:

Begin with finding the short and long sides of the rectangle, which we may designate as and respectively. Since the perimeter and area are given, we may write a system of equations:

wl=6in^2

2w+2l=10in

Solvin for the two yields w=2in and l=3in .

Now, to find the rate of change of the area, derive the area equation:

A=wl

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

Since $dl =0.3\frac{in}{s}$ and $dw= 0.4\frac{in}{s}$

The answer becomes:

$\frac{dA}{dt}=2in(0.3)\frac{in}{s}+3in(0.4)\frac{in}{s}=1.8\frac{in^2}{s}$

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Example Question #35 : How To Find Rate Of Change

A physician determines that the concentration of a medicine in the bloodsteam, in milligrams per liter, t minutes after it is given is modeled by the equation

$c(t)=2te^{-t}$ .

At what rate is the concentration of the medicine in the bloodstream changing after 1 hour?

Possible Answers:

$\frac{3}{e} \frac{mg}{L\cdot min}$

$0 \frac{mg}{L\cdot min}$

$\frac{2}{e} \frac{mg}{L\cdot min}$

$\frac{-2}{e} \frac{mg}{L\cdot min}$

Correct answer:

$0 \frac{mg}{L\cdot min}$

Explanation:

The derivative produces a new function that describes the rate of change of the original, so first calculate the derivative of c(t) using the product rule.

$c'(t)=2e^{-t}-2te^{-t}$ .

To find the rate of change when t=1, substitute 1 for t in the derivative.

$c'(1)=2e^{-1}-2(1)e^{-1}=0 \frac{mg}{L\cdot min}$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #111 : Rate

Example Question #21 : Rate Of Change

Example Question #22 : Rate Of Change

Example Question #30 : How To Find Rate Of Change

Example Question #31 : How To Find Rate Of Change

Example Question #32 : Rate Of Change

Example Question #32 : How To Find Rate Of Change

Example Question #31 : Rate Of Change

Example Question #34 : How To Find Rate Of Change

Example Question #35 : How To Find Rate Of Change

All Calculus 1 Resources

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