Math · Calculus I · Worked example

Solve a sliding ladder problem

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

x^{2} + y^{2} = 100, \frac{d x}{d t} = 1

Name the changing quantities

Let x be the distance from the wall to the base and y the height of the top, both in feet. The ladder is the hypotenuse of a right triangle, so x² + y² = 10² at every moment.

x^{2} + y^{2} = 100

Write what is known and what is wanted

The base moves away at dx/dt = 1 ft/s. We want dy/dt at the instant when x = 6.

\frac{d x}{d t} = 1, \frac{d y}{d t} = ? when x = 6

Differentiate with respect to t

Differentiate both sides with respect to time. By the chain rule, x² gives 2x·dx/dt and y² gives 2y·dy/dt; the constant 100 gives 0.

2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0

Find y at that instant

When x = 6, the equation x² + y² = 100 gives y² = 64, so y = 8. A height is positive, so the negative root does not apply.

6^{2} + y^{2} = 100 ⟹ y = 8

Substitute and solve

Put x = 6, y = 8 and dx/dt = 1 into the differentiated equation: 12 + 16·dy/dt = 0, so dy/dt = −12/16 = −3/4.

2 (6) (1) + 2 (8) \frac{d y}{d t} = 0 ⟹ \frac{d y}{d t} = - \frac{3}{4}

\begin{array}{c} 2 (6) (1) + 2 (8) \frac{d y}{d t} = 0 \\ \frac{d y}{d t} = - \frac{3}{4} \end{array}

Interpret the sign

The rate is negative because the height is decreasing: the top slides down at 0.75 ft/s. The graph of the top, y = √(100 − x²), has slope −3/4 at (6, 8), the same number, because dy/dt = (dy/dx)(dx/dt) and dx/dt = 1.

Result

The top slides down at 0.75 ft/s: dy/dt = −3/4 ft/s when the base is 6 ft from the wall.

Your turn

For the same ladder, how fast is the top sliding down when the base is 8 ft from the wall?

Show the answer and explanation

4/3 ≈ 1.33 ft/s.

At x = 8, y = √(100 − 64) = 6. Then 2(8)(1) + 2(6)·dy/dt = 0, so dy/dt = −16/12 = −4/3. The top falls at about 1.33 ft/s, faster than before.

2 (8) (1) + 2 (6) \frac{d y}{d t} = 0 ⟹ \frac{d y}{d t} = - \frac{4}{3}

\begin{array}{c} 2 (8) (1) + 2 (6) \frac{d y}{d t} = 0 \\ \frac{d y}{d t} = - \frac{4}{3} \end{array}

Keep exploring

In Graph, the tangent at (6, 8) has slope −3/4. The curve gets steeper as x grows, so the same 1 ft/s at the base makes the top fall faster and faster.

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OpenStax Calculus Volume 1 — Related rates

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