Solving Related Rates Problems — AP® Calculus AB Notes

The last topic showed the key move: differentiate an equation with respect to time, and the chain rule links the rates. A full related rates problem wraps that move in a story. You must turn the story into a picture, the picture into an equation, and the answer back into words. The same five steps handle ladders, tanks, shadows and planes.

The five steps

Draw and label. Sketch the situation at a general moment. Give a letter to every quantity that changes; write numbers only on lengths that stay fixed.
Relate. Write an equation that links the quantities at every instant: Pythagoras, similar triangles, an area or volume formula. If it has more changing quantities than you have rates for, use a second relation to remove one.
Differentiate both sides with respect to $t$ .
Substitute the values at the instant asked about, with signs: shrinking or approaching means a negative rate.
Solve and answer with units, in a sentence.

The sliding ladder

A 10-foot ladder (crimson) against a wall with its foot x = 6 ft out and top y = 8 ft up, the foot sliding away at dx/dt = 2 ft/s and the top's rate dy/dt unknown; a gray ghost shows a later position. Beside it: x² + y² = 100, differentiated to 2x dx/dt + 2y dy/dt = 0, giving dy/dt = −3/2 ft/s at this instant. — **Figure 1.** A 10-foot ladder whose foot slides away at 2 ft/s. Relate, differentiate, then substitute the instant. LevlPrep original diagram.

Worked example 1. A 10-foot ladder leans against a wall. Its foot slides away from the wall at 2 feet per second. How fast does the top slide down when the foot is 6 feet from the wall?

Draw: $x$ = distance from wall to foot, $y$ = height of the top. The ladder's 10 feet never changes, so it stays a number.

Relate: $x^2 + y^2 = 100$ .

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

Substitute: at this instant $x = 6$ , so $y = \sqrt{100 - 36} = 8$ , and $\frac{dx}{dt} = 2$ : $2(6)(2) + 2(8)\frac{dy}{dt} = 0$ .

Solve: $\frac{dy}{dt} = -\frac{24}{16} = -\frac{3}{2}$ . The top slides down the wall at 1.5 feet per second.

The rates are not equal: the foot moves at 2 ft/s, the top at 1.5 ft/s. And they change with time. When the foot is 8 feet out, $y = 6$ and $\frac{dy}{dt} = -\frac{8}{3}$ ; the closer the top gets to the floor, the faster it falls.

Angles work too. With $\theta$ the angle between the ladder and the floor, $\cos\theta = \frac{x}{10}$ . Differentiating gives $-\sin\theta\frac{d\theta}{dt} = \frac{1}{10}\frac{dx}{dt}$ . At $x = 6$ , $\sin\theta = \frac{8}{10}$ , so $\frac{d\theta}{dt} = -\frac{1}{4}$ radian per second: the ladder flattens.

The cone tank: remove a variable first

An inverted cone tank, top radius 3 ft and depth 6 ft, holding water of depth h and surface radius r. Similar triangles give r = h/2, so V = πh³/12 and dV/dt = (πh²/4)·dh/dt. — **Figure 2.** An inverted cone 6 ft deep with a top radius of 3 ft. Similar triangles give $r = \frac{h}{2}$ for the water at every depth. LevlPrep original diagram.

Worked example 2. Water pours into this tank at 4 cubic feet per minute. How fast is the water level rising when the water is 2 feet deep?

Draw: the water is a cone of depth $h$ and surface radius $r$ , both changing.

Relate: $V = \frac{1}{3}\pi r^2 h$ . That has two changing lengths, but we know only $\frac{dV}{dt}$ . Similar triangles fix the shape: $\frac{r}{h} = \frac{3}{6}$ , so $r = \frac{h}{2}$ at every instant. Substitute it now, because it is always true: $V$ $= \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h$ $= \frac{\pi h^3}{12}$ .

Differentiate: $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ .

Substitute and solve: $4 = \frac{\pi(4)}{4}\frac{dh}{dt}$ , so $\frac{dh}{dt} = \frac{4}{\pi} \approx 1.27$ feet per minute.

Notice the difference between the two substitutions. $r = \frac{h}{2}$ holds at every instant, so it may go in before differentiating. $h = 2$ holds at one instant, so it must wait until after.

Shadows

Worked example 3. A 6-foot person walks away from a 15-foot lamppost at 5 ft/s. How fast does the tip of the shadow move?

Let $x$ be the distance from the post to the person and $s$ the shadow's length. The light ray makes two similar right triangles: $\frac{s}{6} = \frac{x + s}{15}$ , so $9s = 6x$ and $s = \frac{2x}{3}$ .

The tip is at distance $x + s = \frac{5x}{3}$ from the post. Differentiate: $\frac{d}{dt}(x + s)$ $= \frac{5}{3}\frac{dx}{dt}$ $= \frac{25}{3}$ ft/s, faster than the person walks.

Slips that cost points

Slip	Fix
Substituting $x = 6$ before differentiating	Differentiate the general relation first
Keeping two unknown rates	Use similar triangles or a given condition to remove one
"Decreasing at 10" entered as $+10$	Decreasing means a negative rate
Labeling a fixed length with a variable	Constants of the problem stay numbers
A bare number	Units and a sentence: "the top slides down at 1.5 ft/s"