For more information, please see full course syllabus of College Calculus: Level I

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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

## Discussion

## Study Guides

## Practice Questions

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## Table of Contents

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### Related Rates

- You will be differentiating both sides of an equation with respect to .
- Many problems here are geometrical – I strongly advise drawing a sketch!
- Use variables for those quantities that vary with other quantities in the problem.
- Wait to substitute in specific variable values until after you have differentiated!

### Related Rates

The radius of a sphere is growing at a rate of 1 [m/s]. Find the rate at which the area is changing.

- A = πr
^{2} - [d/dt]A = [d/dt]πr
^{2} - [dA/dt] = π(2r) [dr/dt]

[dA/dt] = 2 πr (1 [m/s])

The radius of a sphere is growing at a rate of 1 [m/s]. Find the rate at which the circumference is changing.

- C = 2 πr
- [d/dt]C = [d/dt]2 πr
- [dC/dt] = 2 π[dr/dt]
- [dC/dt] = 2 π[dr/dt] 1 [m/s]

[dC/dt] = 2 π[m/s]

A sphere's volume is decreasing at a rate of 10 [(m

^{3})/s] from an initial volume V_{0}= 100 m^{3}. How big is the diameter after 5 seconds of decreasing?- How does the volume relate to the diameter?
- Volume of a sphere is [4/3] πr
^{3} - V = [4/3] π([d/2])
^{3} - Find the volume after 5 seconds. To find the current volume, we need to add the initial volume and the rate of change multiplied by time.
- V = V
_{0}+ [dV/dt] t - V = 100 m
^{3}− 10 [(m^{3})/s] 5 s = 50 m^{3} - We subtracted above because the problem said the volume was decreasing. Now we plug in to the volume formula and solve for d.
- 50 m
^{3}= [4/3] π([d/2])^{3} - [(150 m
^{3})/(4 π)] = ([d/2])^{3} - [d/2] =
^{3}√{[(150 m^{3})/(4 π)]}

d = 2

^{3}√{[150/(4 π)]} mA hose is pouring water at a rate of 3 [(m

^{3})/s] into a rigid cylindrical container. The container has a radius of 10 m. What is the rate at which the height of the water level rises? (Volume of a cylinder = πh r^{2})- We need a formula that relates the volume of a cylinder and its height.
- V = πh r
^{2} - The problem says its a rigid container, so we can treat the radius as a constant. Otherwise, we would have to use the product rule when taking the derivative.
- [d/dt]V = [d/dt]πh r
^{2} - [dV/dt] = πr
^{2}[dh/dt] - [dh/dt] = [dV/dt] [1/(πr
^{2})] - [dh/dt] = 3 [(m
^{3})/s] [1/(π(10 m)^{2})]

[dh/dt] = [3/(100π)] [m/s]

A locomotive's position is given by x(t) = 36 [m/s] t. A train car unhitches from the engine and begins to slow. Its position is given by x(t) = −20 [m/s] t. At what rate is the distance between the two trains growing?

- In this case, the trains are traveling along the same axis, so all we have to do here is add the speeds.
- v(t) = [d/dt]36 [m/s] t + [d/dt](−20 [m/s] t)

v(t) = 16 [m/s]

A person's Body Mass Index in imperial units is given by BMI = [(703 weight)/(height

^{2})]. A 70 inch person is increasing their BMI by 1 [lb/(in^{2})] every year while maintaining a constant height. At what rate is their weight changing?- [d/dt]BMI = [d/dt][(703 weight)/(height
^{2})] - [d/dt]BMI = [703/(height
^{2})] [d/dt]weight - [d/dt]weight = [d/dt]BMI [(height
^{2})/703] - [d/dt]weight = 1 [lb/(in
^{2}year)] [((70 in)^{2})/703] - [d/dt]weight = [4900/703] [lb/year]

The person's weight is increasing at a rate of about 7 lbs per year

The pressure of a gas in a container is given by P = [nRT/V] where n is the number of molecules, R is a constant, T is the temperature, and V is the volume. Given that the volume is constant, solve for the rate of change of pressure.

- [d/dt]P = [d/dt][nRT/V]
- [dP/dt] = [R/V] [d/dt]nT
- [dP/dt] = [R/V] (n [d/dt]T + T [d/dt]n)
- [dP/dt] = [R/V] (n [dT/dt] + T [dn/dt])

[dP/dt] = [R/V] (n [dT/dt] + T [dn/dt])

Using the pressure formula of the previous problem, find the rate of change of P with respect to time given that n and T are constants.

- [d/dt]P = [d/dt][nRT/V]
- [dP/dt] = nRT [d/dt][1/V]
- We can use the quotient rule, or treat it as a negative exponent.
- [dP/dt] = nRT [d/dt]V
^{−1} - [dP/dt] = nRT (−V
^{−2}) [d/dt]V

[dP/dt] = −[nRT/(V

^{2})] [dV/dt]Water pressure at a depth is given by P = P

_{0}+ ρg h. At the surface of water, P_{0}= 10^{5}[kg/(s^{2}m)], ρ = 1000 [kg/(m^{3})], g = 9.8 [m/(s^{2})], and h is the height or depth of the water. If pressure is increasing at a rate of 9800 [kg/(s^{3}m)], how fast are you descending?- [d/dt]P = [d/dt]P
_{0}+ [d/dt]ρg h - [dP/dt] = ρg [dh/dt]
- [dh/dt] = [([dP/dt])/(ρg)]
- [dh/dt] = [(9800 [kg/(s
^{3}m)])/(1000 [kg/(m^{3})] 9.8 [m/(s^{2})])]

[dh/dt] = 1 [m/s]

A cube initially has sides 1 m in length. The length of the sides is increasing at 2 [m/s]. At what rate is the volume increasing when the length is 4 m?

- V = l
^{3} - [d/dt]V = [d/dt]l
^{3} - [dV/dt] = 3 l
^{2}[d/dt]l - [dV/dt] = 3 (4 m)
^{2}2 [m/s]

[dV/dt] = 96 [(m

^{3})/s]*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Related Rates

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- What Are Related Rates? 0:08
- Lecture Example 1 0:35
- Lecture Example 2 5:25
- Lecture Example 3 11:54
- Additional Example 4
- Additional Example 5

0 answers

Post by Alan Delez on October 4, 2012

A bug is moving along the right side of the parabola y=x^2 at a rate such that its distance from the origin is increasing at 1cm/min. At what rates are the x- and y-coordinates of the bug increasing when the bug is at the point (2,4)?

0 answers

Post by Rebecca Kerns on May 5, 2010

example 4 does not finish, it just starts over when the instructor get to the rate of water.