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Calculus 1 : Rate of Change

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Example Questions

Example Question #3023 : Calculus

A 13 ft ladder is leaning against a wall. If the ladder is sliding away from the wall at 4 ft/s horizontally, how fast is the ladder sliding down the wall when it is 5 ft above the ground?

Possible Answers:

$\frac{dy}{dt}=0$

$\frac{dy}{dt}=-16 ft/s$

$\frac{dy}{dt}=-5 ft/s$

$\frac{dy}{dt}=-4ft/s$

$\frac{dy}{dt}=-9.6 ft/s$

Correct answer:

$\frac{dy}{dt}=-9.6 ft/s$

Explanation:

We will use the Pythagorean Theorem and its derivative to solve this problem.

The ladder leaning against the wall forms a right triangle. Right triangles are governed by the Pythagorean Theorem

$x^{2}+y^{2}=r^{2}$

where is the length of the ladder.

We will now take the derivative of the equation with respect to time or . Using the power and chain rules,

$\frac{d}{dt}(x^{2})=2x\frac{dx}{dt}$

Similarly, $\frac{d}{dt}(y^{2})=2y\frac{dy}{dt}$ and $\frac{d}{dt}(r^{2})=2r\frac{dr}{dt}$ , giving us the equation

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2r\frac{dr}{dt}$ , which is simplified to $x\frac{dx}{dt}+y\frac{dy}{dt}=r\frac{dr}{dt}$

Where $\frac{dx}{dt}$ is the rate the ladder is sliding away from the building and $\frac{dy}{dt}$ is the rate the ladder is sliding down the building. $\frac{dy}{dt}$ is always negative.

$\frac{dr}{dt}$ is the rate at which the length of the ladder is changing.

We now have two equations

$x^{2}+y^{2}=r^{2}$

and

$x\frac{dx}{dt}+y\frac{dy}{dt}=r\frac{dr}{dt}$

The length of the ladder cannot change, therefore, $\frac{dr}{dt}=0$ .

We are given the following parameters

$\frac{dx}{dt}=4ft/s$ , r=13ft , and y=5 ft

We need to find $\frac{dy}{dt}$ and . We will use the Pythagorean Theorem to solve for .

$x^{2}+y^{2}=r^{2}$

$x^{2}+(5)^{2}=(13)^{2}$

$x^{2}+25=169$

$x^{2}=169-25=144$

$x=\sqrt{144}=12$

We will use the derivative equation to find $\frac{dy}{dt}$

$x\frac{dx}{dt}+y\frac{dy}{dt}=r\frac{dr}{dt}$

$(12)(4)+(5)\frac{dy}{dt}=(13)(0)$

$48+(5)\frac{dy}{dt}=0$

$(5)\frac{dy}{dt}=-48$

$\frac{dy}{dt}=-48/5=-9.6 ft/s$

$\frac{dy}{dt}$ is our solution, the rate at which the ladder is sliding down the wall.

Please note, $\frac{dy}{dt}$ is negative because the ladder is sliding down the wall.

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Example Question #111 : How To Find Rate Of Change

A 25 ft ladder is leaning against a building. If the ladder is sliding down the building at 8 ft/s horizontally, how fast is the ladder sliding away from the wall when it is 15 ft away from the wall?

Possible Answers:

$\frac{dx}{dt}=20 ft/s$

$\frac{dx}{dt}=12 ft/s$

$\frac{dx}{dt}=8 ft/s$

$\frac{dx}{dt}=10.667 ft/s$

$\frac{dx}{dt}=1 ft/s$

Correct answer:

$\frac{dx}{dt}=10.667 ft/s$

Explanation:

We will use the Pythagorean Theorem and its derivative to solve this problem.

The ladder leaning against the wall forms a right triangle. Right triangles are governed by the Pythagorean Theorem

$x^{2}+y^{2}=r^{2}$

where is the length of the ladder.

We will now take the derivative of the equation with respect to time or . Using the power and chain rules,

$\frac{d}{dt}(x^{2})=2x\frac{dx}{dt}$

Similarly, $\frac{d}{dt}(y^{2})=2y\frac{dy}{dt}$ and $\frac{d}{dt}(r^{2})=2r\frac{dr}{dt}$ , giving us the equation

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2r\frac{dr}{dt}$ , which is simplified to $x\frac{dx}{dt}+y\frac{dy}{dt}=r\frac{dr}{dt}$

Where $\frac{dx}{dt}$ is the rate the ladder is sliding away from the building and $\frac{dy}{dt}$ is the rate the ladder is sliding down the building. $\frac{dy}{dt}$ is always negative.

$\frac{dr}{dt}$ is the rate at which the length of the ladder is changing.

We now have two equations

$x^{2}+y^{2}=r^{2}$

and

$x\frac{dx}{dt}+y\frac{dy}{dt}=r\frac{dr}{dt}$

The length of the ladder cannot change, therefore, $\frac{dr}{dt}=0$ .

We are given the following parameters

$\frac{dy}{dt}=-8ft/s$ , r=25ft , and x=15 ft

We need to find $\frac{dx}{dt}$ and . We will use the Pythagorean Theorem to solve for .

$x^{2}+y^{2}=r^{2}$

$(15)^{2}+y^{2}=(25)^{2}$

$y^{2}=(25)^{2}-(15)^{2}=625-225=400$

$y=\sqrt{400}=20$

We will use the derivative equation to find $\frac{dx}{dt}$ .

$x\frac{dx}{dt}+y\frac{dy}{dt}=r\frac{dr}{dt}$

$(15)\frac{dx}{dt}+(20)(-8)=(25)(0)$

$(15)\frac{dx}{dt}=160$

$\frac{dx}{dt}=160/15=10.667 ft/s$

$\frac{dx}{dt}$ is our solution, the rate at which the ladder is sliding away from the building.

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Example Question #111 : Rate Of Change

Find the rate of change of the line $y=7x^{2}-5x+2$ .

Possible Answers:

$\frac{dy}{dx}=7x-5$

$\frac{dy}{dx}=7x^2+5$

$\frac{dy}{dx}=14x+3$

$\frac{dy}{dx}=14x-5$

$\frac{dy}{dx}=8x$

Correct answer:

$\frac{dy}{dx}=14x-5$

Explanation:

The rate of change of a line is also known as the slope. We find the slope by taking the first derivative of an equation.

Using the power rule

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

the derivative is found to be

$\frac{d}{dx}(y)=\frac{d}{dx}[7x^{2}-5x+2]=14x-5$

$\frac{dy}{dx}=14x-5$

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Example Question #111 : Rate Of Change

Find the tangent to the line $y=x^{3}+12x^{2}-28$ at x=5 .

Possible Answers:

y=578

y=x

y=5x-578

y=195x-578

$y=3x^{2}+24x$

Correct answer:

y=195x-578

Explanation:

The tangent line of a curve at a point is also known as the instantaneous rate of change.

To find the tangent line:

(1)Find the derivative of the equation

(2) Evaluate the derivative at the given point. This will be the slope of the tangent line.

(3) Evaluate the initial equation at the given point. This will give us an ordered pair.

(4) Find the equation of the tangent line using the equation y=mx+b .

Below is the solution to our problem.

(1) $\frac{d}{dx}(y)=y'=\frac{d}{dx}[x^{3}+12x^{2}-28]=3x^{2}+24x$

(2) $y'(5)=3(5)^{2}+24(5)=75+120=195$

m=195

(3) $y(5)=(5)^{3}+12(5)^{2}-28=125+300-28=397$

(5,397)

(4) y=mx+b

397=195(5)+b

b=397-195(5)=397-975=-578

y=195x-578

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Example Question #3026 : Calculus

Find the tangent to the line $y=x^{2}-7x+16$ at x=2 .

Possible Answers:

y=5x+7

y=2x-7

y=-3x+12

y=x

y=-3x-6

Correct answer:

y=-3x+12

Explanation:

The tangent line of a curve at a point is also known as the instantaneous rate of change.

To find the tangent line:

(1)Find the derivative of the equation

(2) Evaluate the derivative at the given point. This will be the slope of the tangent line.

(3) Evaluate the initial equation at the given point. This will give us an ordered pair.

(4) Find the equation of the tangent line using the equation y=mx+b

Below is the solution to our problem.

(1) $\frac{d}{dx}(y)=y'=\frac{d}{dx}[x^{2}-7x+16]=2x-7$

(2) y'(2)=2(2)-7=4-7=-3

m=-3

(3) $y(2)=(2)^{2}-7(2)+16=4-14+16=6$

(2,6)

(4) y=mx+b

6=-3(2)+b

b=6+3(2)=12

y=-3x+12

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Example Question #112 : Rate Of Change

The sides of a rectangle are given by the formulas w(t)=5t^2+4 and l(t)=3t^2+2t+3 . What is the rate of change of the rectange's diagonal at time t=4 ?

Possible Answers:

95.35

47.68

42.56

21.28

47.71

Correct answer:

47.68

Explanation:

The length of the diagonal of a rectangle can be related to the perpendicular sides of the rectangle via the Pythagorean Theorem:

w^2+l^2=d^2

$d=\sqrt{w^2+l^2}$

This relationship can be extended to rates of change as well; differentiating each side of the equation with respect to time is an example of this. To find the rate of change of the diagonal, perform this derivative.

This derivative can be found by taking the derivative of the function with respect to w, and adding that to the derivative of the function taken with respect to l.

$\frac{dd(t)}{dt}=\frac{2w\frac{dw(t)}{dt}+2l\frac{dl(t)}{dt}}{2\sqrt{w^2+l^2}}$

Now recall our functions for w and l:

w(t)=5t^2+4

l(t)=3t^2+2t+3

At time t=4 :

w(4)=5(16)+4=84

l(4)=3(16)+2(4)+3=59

Now, find the value of their derivatives at time t=4 :

$\frac{dw(t)}{dt}=10t$ ; $\frac{dw(4)}{dt}=40$

$\frac{dl(t)}{dt}=6t+2$ ; $\frac{dl(4)}{dt}=6(4)+2=26$

With these values known, we can find the rate of change of the diagonal at this time:

$\frac{dd(t)}{dt}=\frac{2w\frac{dw(t)}{dt}+2l\frac{dl(t)}{dt}}{2\sqrt{w^2+l^2}}$

$\frac{dd(4)}{dt}=\frac{2(84)(40)+2(59)(26)}{2\sqrt{84^2+59^2}}$

$\frac{dd(4)}{dt}=47.68$

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Example Question #111 : Rate Of Change

The rate of change of the radius of a circle is 0.3 . What is the rate of change of the circle's area when the radius is ?

Possible Answers:

$4.2\pi$

$14\pi$

$0.09\pi$

$7\pi$

$2.1\pi$

Correct answer:

$4.2\pi$

Explanation:

The area of a circle is related to its radius by the formula:

$A=\pi r^2$

This relationship can also be used to relate rates of changes between these two parameters. Take the derivative of each side of the equation with respect to time:

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

We're told that the rate of change of the radius is 0.3 , which gives the value for $\frac{dr}{dt}$ . We're also given the radius; r=7 . This allows us to find the rate of change of the area:

$\frac{dA}{dt}=2\pi (7)(0.3)$

$\frac{dA}{dt}=4.2\pi$

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Example Question #111 : Rate Of Change

The rate of change of the horizontal and vertical sides of a square are changing at a rate of 0.2 and 0.4 respectively. If the sides have a length of , what is the rate of change of the area?

Possible Answers:

0.6

2.4

1.2

1.8

0.3

Correct answer:

1.8

Explanation:

The area of a square can be given in terms of its sides as

A=s^2

However, we're told that the horizontal and vertical sides have different rates of change. It would be helpful to rewrite the equation in light of this fact:

A=s_1s_2 .

Now, the rates of change of the parameters can be related by taking the derivative of each side of the equation with respect to time. Be sure to take the derivative with respect to each variable and sum the results.

$\frac{dA}{dt}=s_1\frac{ds_2}{dt}+s_2\frac{ds_1}{dt}$

Now, we can utilize our rates of change $\frac{ds_1}{dt}=0.2;\frac{ds_2}{dt}=0.4$ . Recall that s_1=s_2=3 :

$\frac{dA}{dt}=3(0.4)+3(0.2)$

$\frac{dA}{dt}=1.2+0.6$

$\frac{dA}{dt}=1.8$

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Example Question #3031 : Calculus

The rates of change of the short and long legs of a right triangle are 0.4 and 0.3 respectively. If the triangle's legs have lengths and , what is the rate of change of the hypotenuse?

Possible Answers:

0.25

0.5

0.48

0.24

Correct answer:

0.48

Explanation:

The hypotenuse of a right triangle can be related to the triangle's legs via the Pythagorean Theorem:

c^2=a^2+b^2

We're told that a=3;b=4 :

c^2=3^2+4^2

c^2=25

c=5

Now let's return to the Pythagorean Theorem:

c^2=a^2+b^2

We can use it to relate rates of change by taking the derivative of each side of the equation with respect to time:

2cc'=2aa'+2bb'

We're told that a'=0.4;b'=0.3 , so we can in turn solve for :

$c'=\frac{aa'+bb'}{c}$

$c'=\frac{3(0.4)+4(0.3)}{5}$

$c'=\frac{2.4}{5}$

c'=0.48

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Example Question #120 : Rate Of Change

The sides of a rectangle are and . If the rate of change of the area is 0.6 and the rate of change of the longer side is 0.1 , what is the rate of change of the shorter side?

Possible Answers:

0.03

0.6

0.13

-0.13

Correct answer:

0.03

Explanation:

The area of a rectangle in terms of its sides is

A=wl

For the sake of this problem, let denote the shorter side and the longer.

This formula can also be used to relate rates of change. We need only take the derivative of each side of the equation with respect to time. Be sure to take the derivative with respect to each variable:

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

We're given some rates of change in the problem statement. Primarily

$\frac{dA}{dt}=0.6$ and $\frac{dl}{dt}=0.1$

Combining this with the lengths w=3 and l=10 , we can find the rate of change of the smaller side:

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

$\frac{dw}{dt}=\frac{0.6-3(0.1)}{10}$

$\frac{dw}{dt}=0.03$

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3023 : Calculus

Example Question #111 : How To Find Rate Of Change

Example Question #111 : Rate Of Change

Example Question #111 : Rate Of Change

Example Question #3026 : Calculus

Example Question #112 : Rate Of Change

Example Question #111 : Rate Of Change

Example Question #111 : Rate Of Change

Example Question #3031 : Calculus

Example Question #120 : Rate Of Change

All Calculus 1 Resources

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