Related Rates

Overview

Related rates problems involve finding the rate of change of one quantity in terms of the rate of change of another quantity. Both quantities change with respect to time.

Problem-Solving Strategy

1. Draw a diagram and label all quantities
2. Identify known rates (given) and unknown rates (to find)
3. Write an equation relating the quantities
4. Differentiate implicitly with respect to time $t$
5. Substitute known values and solve for the unknown rate
6. Include units in your answer

Common Formulas

Geometry

• Circle: $A = \pi r^2$, $C = 2\pi r$
• Sphere: $V = \frac{4}{3}\pi r^3$, $S = 4\pi r^2$
• Cylinder: $V = \pi r^2 h$
• Cone: $V = \frac{1}{3}\pi r^2 h$
• Pythagorean: $a^2 + b^2 = c^2$

Trigonometry

• $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$
• $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$
• $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$

Examples

Example 1: Expanding Circle

A stone dropped in water creates circular ripples. If the radius increases at 2 m/s, how fast is the area increasing when $r = 5$ m?

Given: $\frac{dr}{dt} = 2$ m/s, $r = 5$ m Find: $\frac{dA}{dt}$

$A = \pi r^2$

$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} = 2\pi(5)(2) = 20\pi \text{ m}^2/\text{s}$

Example 2: Ladder Problem

A 10 ft ladder leans against a wall. The bottom slides away at 1 ft/s. How fast is the top sliding down when the bottom is 6 ft from the wall?

Given: $\frac{dx}{dt} = 1$ ft/s, $x = 6$ ft, ladder = 10 ft Find: $\frac{dy}{dt}$

$x^2 + y^2 = 100$

When $x = 6$: $y = \sqrt{100 - 36} = 8$ ft

Differentiating:

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$2(6)(1) + 2(8)\frac{dy}{dt} = 0$

$\frac{dy}{dt} = -\frac{12}{16} = -\frac{3}{4} \text{ ft/s}$

The top slides down at $\frac{3}{4}$ ft/s.

Example 3: Filling a Cone

Water pours into a conical tank at 10 ft³/min. The tank has radius 5 ft and height 10 ft. How fast is the water level rising when $h = 6$ ft?

Given: $\frac{dV}{dt} = 10$ ft³/min Find: $\frac{dh}{dt}$ when $h = 6$

Similar triangles: $\frac{r}{h} = \frac{5}{10}$, so $r = \frac{h}{2}$

$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$

$\frac{dV}{dt} = \frac{3\pi h^2}{12} \cdot \frac{dh}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$

$10 = \frac{\pi(6)^2}{4} \cdot \frac{dh}{dt}$

$\frac{dh}{dt} = \frac{40}{36\pi} = \frac{10}{9\pi} \text{ ft/min} \approx 0.354 \text{ ft/min}$

Example 4: Moving Shadow

A 6 ft person walks away from a 15 ft lamppost at 4 ft/s. How fast is the tip of their shadow moving?

Let $x$ = distance from lamppost, $s$ = shadow length

Similar triangles:

$\frac{15}{x + s} = \frac{6}{s}$

$15s = 6x + 6s$

$9s = 6x$

$s = \frac{2}{3}x$

The tip of shadow is at distance $x + s = x + \frac{2}{3}x = \frac{5}{3}x$ from post.

$\frac{d(x + s)}{dt} = \frac{5}{3} \cdot \frac{dx}{dt} = \frac{5}{3}(4) = \frac{20}{3} \text{ ft/s}$

Example 5: Angle of Elevation

A balloon rises at 3 m/s. An observer 100 m away watches it rise. How fast is the angle of elevation changing when the balloon is 100 m high?

Let $h$ = height, $\theta$ = angle

$\tan(\theta) = \frac{h}{100}$

$\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dh}{dt}$

When $h = 100$: $\tan(\theta) = 1$, so $\theta = \frac{\pi}{4}$ and $\sec^2(\theta) = 2$

$2 \cdot \frac{d\theta}{dt} = \frac{1}{100}(3)$

$\frac{d\theta}{dt} = \frac{3}{200} = 0.015 \text{ rad/s}$