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Lecture Comments (6)

0 answers

Post by Mohamed Al Mohannadi on January 31, 2017

does y = 2x ?

2 answers

Last reply by: Richard Gregory
Thu Nov 14, 2013 9:54 PM

Post by Richard Gregory on November 14, 2013

In example 5. Should Z be 1/3 not 3 given that the area not the perimeter is increasing 3 times faster than the diagonal. Therefore the perimeter would be changing at a rate of 1/3 in proportion to the area?

1 answer

Last reply by: Akshay Tiwary
Sun May 5, 2013 1:26 AM

Post by James Xie on January 23, 2013

So for example 6, did we really need to find the radius given that 4(pi)r^2 = the circle's area?

Related Rates

  • Rate of change of 2 variables with respect to a 3rd variable
    • Find mathematically relationship between variables
    • Differentiate
    • Use appropriate substitutions
  • To avoid confusion: keep big picture clear
  • Organization is key!

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 = πr2
  • [d/dt] A = [d/dt] πr2
  • [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 [(m3)/s] from an initial volume V0 = 100 m3. 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] πr3
  • 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 = V0 + [dV/dt] t
  • V = 100 m3 − 10 [(m3)/s] 5 s = 50 m3
  • We subtracted above because the problem said the volume was decreasing. Now we plug in to the volume formula and solve for d.
  • 50 m3 = [4/3] π([d/2])3
  • [(150 m3)/(4 π)] = ([d/2])3
  • [d/2] = 3√{[(150 m3)/(4 π)]}
d = 2 3√{[150/(4 π)]} m
A hose is pouring water at a rate of 3 [(m3)/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 r2)
  • We need a formula that relates the volume of a cylinder and its height.
  • V = πh r2
  • 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 r2
  • [dV/dt] = πr2 [dh/dt]
  • [dh/dt] = [dV/dt] [1/(πr2)]
  • [dh/dt] = 3 [(m3)/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)/(height2)]. A 70 inch person is increasing their BMI by 1 [lb/(in2)] every year while maintaining a constant height. At what rate is their weight changing?
  • [d/dt] BMI = [d/dt] [(703  weight)/(height2)]
  • [d/dt] BMI = [703/(height2)] [d/dt] weight
  • [d/dt] weight = [d/dt] BMI [(height2)/703]
  • [d/dt] weight = 1 [lb/(in2  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/(V2)] [dV/dt]
Water pressure at a depth is given by P = P0 + ρg h. At the surface of water, P0 = 105 [kg/(s2  m)], ρ = 1000 [kg/(m3)], g = 9.8 [m/(s2)], and h is the height or depth of the water. If pressure is increasing at a rate of 9800 [kg/(s3  m)], how fast are you descending?
  • [d/dt] P = [d/dt] P0 + [d/dt] ρg h
  • [dP/dt] = ρg [dh/dt]
  • [dh/dt] = [([dP/dt])/(ρg)]
  • [dh/dt] = [(9800 [kg/(s3  m)])/(1000 [kg/(m3)] 9.8 [m/(s2)])]
[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 = l3
  • [d/dt] V = [d/dt] l3
  • [dV/dt] = 3  l2 [d/dt] l
  • [dV/dt] = 3 (4  m)2 2 [m/s]
[dV/dt] = 96 [(m3)/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.


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
  • Related Rates 0:06
    • Finding Rate of Change: Organization & Big Picture
  • Example 2: Area of a Circle 1:17
  • Example 3: Spherical Volume Expanding 4:19
  • Example 4: Traveling Problem 7:57
  • Example 5: Square Increase 12:37
  • Example 6: Standard Related Rates Problem 16:59
  • Example 7: Standard Related Rates Problem 19:49