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Calculus 1 : How to find rate of change

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Example Question #31 : 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:

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

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

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

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

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

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 #33 : 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:

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

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

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

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

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

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 #34 : 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:

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

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

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

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

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

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 #35 : 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:

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

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

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

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

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

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 #36 : 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:

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

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

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

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

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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Example Question #37 : Rate Of Change

A 13 foot ladder is leaning against the wall. The bottom of the ladder is initially 10 feet away from the wall and being pushed toward the wall at a rate of $\frac{1}{2}\frac{ft}{sec}$ . How fast is the top of the ladder moving after 10 seconds?

Possible Answers:

$\frac{1}{4}\frac{ft}{sec}$

Not enough information given.

$-\frac{5}{24}\frac{ft}{sec}$

$\frac{5}{24}\frac{ft}{sec}$

$-\frac{1}{4}\frac{ft}{sec}$

Correct answer:

$\frac{5}{24}\frac{ft}{sec}$

Explanation:

In this question we are asked to find how fast the top of the ladder is going down. To solve this quesiton, we must first realize that the ladder forms a triangle with the wall and the ground.

By realizing this we then simply take the derivative of the Pythagorean Theorem,

x^2+y^2=13^2 using the power rule,

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

The derivative of any constant is 0, therefore the derivative of the Pythagorean Theorem becomes

2x dx+2ydy=0 .

We must now find the values for and in order to solve for . We know that , the distance of the bottom of the ladder to the wall is intially 10 feet and is being pushed toward the wall at a rate of $\frac{1}{2}\frac{ft}{sec}$ for 10 seconds, therefore

$x=10-\frac{1}{2}(10)=5 ft$ .

To solve for we simply plug back into the Pythagorean Theorem in the beginning, obtaining 5^2+y^2=13^2 . Solving for , we find that y=12 . Plugging back into derivative of the Pythagorean Theorem, we know the values for x,y,dx making it easy to solve for . Note that is negative in this equation because is decreasing not increasing.

$2(5)\left(-\frac{1}{2}\right)+2(12)dy=0$

-5+24 dy=0

$dy=\frac{5}{24}\frac{ft}{sec}$

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Example Question #38 : Rate Of Change

A cup with a shape of an inverted cone that is being filled up with water has a base radius of 3 inches and height 5 inches. If the water is being poured into the cup at a rate of $1\frac{in^{3}}{second}$ find the rate at which the water level of the cup is rising when the water is 4 inches deep.

Possible Answers:

$\frac{1}{2.14\pi}\frac{in^3}{sec}$

$\frac{1}{2.12\pi}\frac{inc^3}{sec}$

$\frac{1}{2\pi}\frac{in^3}{sec}$

Not enough information is given.

$\frac{1}{9.54\pi}\frac{in^3}{sec}$

Correct answer:

$\frac{1}{2.12\pi}\frac{inc^3}{sec}$

Explanation:

To solve this equation, we must first know that the equation for the volume of a cone is

$V=\frac{1}{3}\pi r^2 h$

where is the base radius of the cone and is the height of the cone.

We are asked to find the rate of which the water level of the cup changes after the water height in the cone is equal to 4 inches, therefore we take the derivative of the volume equation with respect to time.

To take the derivative of the volume equation, we must use the power rule

$\frac{d}{dx}x^n=nx^{n-1}$ as well as the product rule (uv)'=u'v+uv' .

The derivative comes out to be

$\frac{dV}{dt}=\frac{\pi}{3}\left(r^2\frac{dh}{dt}+2rh\frac{dr}{dt}\right)$ .

We know that

$\frac{dV}{dt}=1\frac{in^3}{sec},r=2.4 in, h=4in,\frac{dr}{dt}=\frac{3}{5}\frac{dh}{dt}$

based off of the information given to us in the beginning; now we must just solve for $\frac{dh}{dt}$ .

Note that $r,\frac{dr}{dt}$ different then expected due to the ratio of the cone.

This ratio is

$\frac{h}{2r}=\frac{5}{6}$ or $r=\frac{3}{5}h$ .

$1\frac{in^3}{sec}=\frac{\pi}{3}\left(2.4^2\frac{dh}{dt}+\frac{3}{5}\frac{dh}{dt}\right)$

$\frac{3}{6.36\pi}\frac{in^3}{sec}=\frac{dh}{dt}=\frac{1}{2.12\pi}\frac{inc^3}{sec}$

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Example Question #39 : Rate Of Change

A spherical snowball stuck into an oven and is melting at a rate such that its volume is decreasing by $2\frac{cm^3}{sec}$ . When the snowball has a radius of 10 centimeters, how fast is the diameter of the snowball changing? Assume only the surface of the snowball is melting at a uniform rate.

Possible Answers:

Not enough information is given.

$\frac{1}{200\pi}\frac{cm}{sec}$

$\frac{1}{300\pi}\frac{cm}{sec}$

$\frac{1}{100\pi}\frac{cm}{sec}$

$\frac{1}{400\pi}\frac{cm}{sec}$

Correct answer:

$\frac{1}{100\pi}\frac{cm}{sec}$

Explanation:

In order to solve this question, we must first know that the volume of a sphere is defined as

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

We are asked to find the rate of which the diameter of the snowball changes when it is melting. Therefore we must take the derivative of the volume equation with respect to time. To do this we will need to utilize the power rule

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

The derivative becomes

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$ and since we know that $r=10 cm, \frac{dV}{dt}=2\frac{cm^3}{sec}$ , we simply plug the variables into the equation and solve for $\frac{dr}{dt}$ .

$2\frac{cm^3}{sec}=4\pi (10cm)^2\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{1}{200\pi}\frac{cm}{sec}$

However we are asked to find the rate of which the diamater of the snowball changes, therefore we multiply the change of rate of the radius by 2 and obtain that the diameter changes at a rate of $\frac{1}{100\pi}\frac{cm}{sec}$ .

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Example Question #40 : Rate Of Change

A car 40 miles east of point A starts driving north at a rate of 30 miles per hour. What is the rate at which the distance between the car and point A is changing after one hour?

Possible Answers:

19mph

18pmh

17mph

20mph

None of the above.

Correct answer:

18pmh

Explanation:

In order to solve this question, we must first see that a triangle is being formed between the car and point A. Therefore a Pythagorean theorem derivative can be used to solve this problem. The Pythagorean therorem

x^2+y^2=z^2 can be derived with respect to time in using the power rule

$\frac{d}{dx}x^n=nx^{n-1}$ to obtain

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$ .

We know from the information given to us that

$x=40 miles,y=30miles, \frac{dy}{dt}=30 mph,\frac{dx}{dt}=0$ ,

therefore simply plugging in and solving for $\frac{dz}{dt}$ will give us the answer to this problem. However, we must still solve for . can be solved by plugging in the values of and back into the Pythagorean theorem. is equal to 30 miles because after one hour that is how far the car has traveled.

(30)^2+(40)^2=z^2

z=50 miles

Now we simply plug everything into the derivative equation and solve for $\frac{dz}{dt}$

$2(40miles)(0)+2(30miles)(30mph)=2(50miles)\frac{dz}{dt}$

$\frac{dz}{dt}=\frac{1800}{100}mph=18 mph$

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Calculus 1 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #31 : Rate Of Change

Example Question #32 : Rate Of Change

Example Question #33 : Rate Of Change

Example Question #34 : Rate Of Change

Example Question #35 : Rate Of Change

Example Question #36 : Rate Of Change

Example Question #37 : Rate Of Change

Example Question #38 : Rate Of Change

Example Question #39 : Rate Of Change

Example Question #40 : Rate Of Change

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