Related rates with similar triangles

Warm-up

Three quick questions from earlier lessons.

Discover

A 6-ft person walks away from a 15-ft lamp. With $x$ the distance to the lamp and $s$ the shadow's length, two right triangles share the angle at the shadow's tip. Which equation links $s$ and $x$ ?

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

Key idea

Similar triangles remove an extra variable.

When a problem has more changing quantities than rates you know, use a fixed ratio (similar triangles, a cone's shape) to write one quantity in terms of another before differentiating.

Worked example

6-ft person, 15-ft lamp, walking at 3 ft/s. How fast does the shadow grow?

6x + 6s = 15s

, so

s = \tfrac{2}{3}x

Cross-multiply 6/s = 15/(x + s).

s' = \tfrac{2}{3}x'

Differentiate.

s' = 2

ft/s

x' = 3.

Your turn, guided

Your turn: a cone tank whose radius is half the depth: $V = \tfrac{\pi}{12}h^3$ . Differentiate.

\dfrac{dV}{dt} = \dfrac{\pi}{4}h^2 \cdot \dfrac{dh}{dt}

Chain rule: 3h²/12 = h²/4.

Practice ★

Put the steps for finding $\dfrac{dA}{dt}$ when $r = 3$ in order.
- Draw a picture and name the changing quantities
- Write an equation linking them: $A = \pi r^2$
- Differentiate with respect to $t$ : $A' = 2\pi r \, r'$
- Substitute $r = 3$ and the known rate

A person 6 ft tall walks away from a 13-ft street lamp at 4 ft/s. How fast is the length of their shadow growing? (ft/s, exact)
A person 5 ft tall walks away from a 18-ft street lamp at 4 ft/s. How fast is the length of their shadow growing? (ft/s, exact)

Watch out!

Practice ★★

Water flows at 9 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 4$ m? (m/min, exact)
Water flows at 3 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 3$ m? (m/min, exact)

Exit ticket

Three questions to check you have got it.

Water flows at 9 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 4$ m? (m/min, exact)
Water flows at 3 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 3$ m? (m/min, exact)
Water flows at 7 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 4$ m? (m/min, exact)

Summary card

Similar triangles

Use the fixed ratio to write one variable in terms of another.
Differentiate with respect to $t$ .
Substitute the moment's values.

Watch out: the big triangle's base runs from the lamp to the shadow's tip.

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