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Wize AP Calculus (BC) Textbook > Applications of Differentiation: Contextual

Related Rates

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Related Rates

A related rates problem involves quantities which vary with time and equations linking these quantities. Typically, the value of these quantities are given at a certain time as well as their rates of change. Often, one rate is missing and has to be found.

Procedure for Solving Related Rates

  1. Draw a picture.
  2. Identify the meaningful variables and constants with their respective units (put them on the picture if applicable).
  3. Identify the known and unknown rates of change.
  4. Find an equation linking the variables found in 2.
  5. Use implicit differentiation to differentiate both sides of the equation found in 4.
  6. Substitute the known quantities and rates of change in the equation found in 5.
  7. Solve 6 for the desired rate of change and answer the question.
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Example: Related Rates

A 10 ft10 \text{ ft} ladder is resting against a wall with the base of the ladder initially 5 ft5 \text{ ft} from the wall. If the bottom gets pushed towards the wall at a constant rate of 0.5 ft/s0.5 \text{ ft/s}, how fast is the top of the ladder moving up the wall after 8 seconds?

1) Draw a Picture

2) Identify the meaningful variables and constants with their respective units (put them on the picture if applicable).

3) Identify the known and unknown rates of change

x′(t)=−.5 ft/sx'\left(t\right)=-.5 \text{ ft/s}

y′(t)=?y'(t)=?

4) Find an equation linking the variables found in 2)
x2+y2=102⇒y=z2−x2=100−x2\begin{array}{c} x^2+y^2 = 10^2\\ \Rightarrow y = \sqrt{z^2-x^2} = \sqrt{100-x^2} \end{array}

x(t)=x(0)+x′t⇒x(t)=5−0.5t⇒x(8)=5−4=1\begin{array}{c} x(t) = x(0)+x't\\ \text{} \\ \Rightarrow x(t)= 5-0.5t \\ \text{} \\ \Rightarrow x(8) = 5-4 = 1 \end{array}

x=1⇒y=100−12=99x=1\Rightarrow y=\sqrt{100-1^2}=\sqrt{99}

5) Use implicit differentiation to differentiate both sides of the equation found in 4)
ddt[x2+y2=100]  ⟹  2xx′+2yy′=0  ⟹  y′=−2xx′2y\begin{array}{c} \dfrac{d}{dt}\left[x^2+y^2=100\right]\\ \text{} \\ \implies 2xx'+2yy'=0 \\ \text{} \\ \implies y' = \dfrac{-2xx'}{2y} \end{array}

6) Substitute the known quantities and rates of change into the equation in 5)
−(1)(−0.5)99=1299 ft/s\dfrac{-(1)(-0.5)}{\sqrt{99}} =\boxed{\dfrac{1}{2\sqrt{99}} \text{ ft/s} }

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Example: Related Rates

If a tree trunk adds 1/41/4 of an feet to its diameter and 11 foot to its height each year, how rapidly is its volume changing when its diameter is 33 feet and its height is 5050 feet (assume that the tree trunk is a circular cylinder).

1) Draw a Picture (see video)

2) Identify the meaningful variables and constants with their respective units (put them on the picture if applicable).
Let r=radius of the trunkr=\text{radius of the trunk}

h=height of the trunkh=\text{height of the trunk}

3) Identity the known and unknown rates of change
Given drdt=12(14)=18 ft/year\displaystyle\frac{dr}{dt}=\frac{1}{2}\left(\frac{1}{4}\right)=\frac{1}{8}\text{ ft/year}


and dhdt=1 ft/year\displaystyle \frac{dh}{dt}=1\text{ ft/year}

dVdt=?\displaystyle \frac{dV}{dt}=?

4) Find an equation linking the variables found in 2)
The volume of the trunk is

V=πr2h V=\pi r^2 h

5) Use implicit differentiation to differentiate both sides of the equation found in 4)
dVdt=2πrhdrdt+πr2dhdt\frac{dV}{dt} =2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}

6) Substitute the known quantities and rates of change into the equation in 5)

=2π(32)(50)(18)+π(32)2(1)=2\pi\left(\frac{3}{2}\right)(50)\left(\frac{1}{8}\right)+\pi\left(\frac{3}{2}\right)^2(1)

=150π8+9π4=21π  ft3/year=\frac{150\pi}{8}+\frac{9\pi}{4} =\boxed{21\pi \ \text{ ft}^3/\text{year}}

The radius rr of a pizza plate increases at a rate of .02.02 cm/min placed in an oven. At what rate is the area of the plate increasing when r=40r=40 cm?

A rocket rising straight up from a level field by an observer 600600 feet from the point of launch. At the moment the observer's angle of elevation is π4 \frac{\pi}{4} , the angle is increasings at a rate of .1.1 rads/minute. How fast is the rocket rising at that moment?
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A man running straight at a rate of 1010 ft/sec along a level street passes under a vertically rising balloon when it is 5050 feet high. Assuming that the balloon is rising at a constant rate of 1.51.5 ft/sec, how fast is the distance between the balloon and the man increasing 22 seconds later?