Related rates

When two quantities are tied together by an equation, their rates of change are tied together too. This lesson shows you how to use that.

The idea

Picture a ladder leaning on a wall. Pull the bottom away and the top slides down. You never touch the top, yet its speed is fixed by how fast you pull the bottom. The ladder’s length is the link between the two.

Calculus gives us a way to turn that link into a number. If x and y both change with time t and satisfy an equation, then differentiating the equation with respect to t relates dx/dt to dy/dt.

Differentiate with respect to time, using the chain rule: d/dt of x² is 2x · dx/dt.

Try it: the sliding ladder

A 10 ft ladder leans on a wall. Drag the slider to move the foot, or press Play to pull it away at a steady speed. Watch what happens to the top.

y x length = 10 ft
x, foot from wall y, top height dx/dt (you set this) dy/dt (result)

How the speed ratio changes

This graph shows how many times faster the top moves than the foot, which is x/y. The dot follows your slider. The curve climbs without limit as the foot approaches 10 ft, which is why a ladder’s top gets faster and faster at the end.

0510 ft0×10×

The foot moves at a constant speed, but the top does not. Slide x toward 10 and see how fast the top drops. We can predict that exactly.

The five-step method

Every related rates problem follows the same path. Step through the ladder problem: the foot is 6 ft from the wall and moving at 2 ft/s. How fast is the top sliding?

1. Draw it and name the variables

Let x be the foot’s distance from the wall and y the top’s height. Both depend on t.

2. Write what you know and what you want

Known: dx/dt = 2 ft/s when x = 6. Wanted: dy/dt at that moment.

3. Find an equation linking the variables

The ladder, wall, and floor form a right triangle: x² + y² = 100. The 100 is constant, so it will vanish when we differentiate.

4. Differentiate with respect to t

2x · dx/dt + 2y · dy/dt = 0, so dy/dt = −(x/y) · dx/dt.

5. Substitute, then solve

At x = 6, the triangle gives y = 8. So dy/dt = −(6/8)(2) = −1.5 ft/s. The top slides down at 1.5 ft/s.

The most common mistake: plugging in numbers before differentiating. Here, y changes constantly. If you replace it with 8 first, its derivative becomes zero and the answer is wrong. Differentiate first, then substitute.

You can check the result in the simulator above: set x to 6 and the pull speed to 2. The readout for dy/dt should say −1.50.

Your turn

Six problems, from simple to tricky. Enter a number, check it, and open the hint or solution if you get stuck. Problems 4 and 5 add a new tool: similar triangles, which let you write one variable in terms of another before differentiating.

AP Calculus BC free response

These two questions are modeled on the multi-part free-response format. On the real exam you would show your setup, include units, and justify each step. Try writing your work on paper first, then enter the numeric answer to check it and open the solution to compare your work against the scoring notes.

On free response, points come from the setup as much as the final number: the equation, the differentiated equation with the chain rule, and the substitution, with units on the answer.
Timed mode locks the solutions until time runs out or you stop. Pick one question and work it on paper, as on the exam.