Related rates
Problem 3.93 · medium
A 25 ft ladder leans against a vertical wall. The bottom slides away from the wall at 4 ft/s. How fast is the top sliding down the wall when the bottom is 7 ft from the wall?
- Let x be the distance from the wall to the bottom and y the height of the top. The ladder is the hypotenuse: x² + y² = 625.
- Differentiate with respect to t: 2x dx/dt + 2y dy/dt = 0, so dy/dt = -(x/y) dx/dt.
- \[ \left. \sqrt{625 - x^{2}} \right|_{\substack{ x=7 }} = 24 \]When x = 7, y = √(625 − 49).✓ Proved
- \[ \left. - \frac{v x}{y} \right|_{\substack{ x=7\\ y=24\\ v=4 }} = - \frac{7}{6} \]dy/dt = −(x/y)(dx/dt) with x = 7, y = 24, dx/dt = 4.✓ Proved
Answer \( - \frac{7}{6} \ \text{ft/s} \approx -1.167\ \text{ft/s} \)
Lines: 2 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the situation was stepped forward and back by a microsecond and the quantity differenced numerically |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/related_rates, checked 2026-09-26 with SymPy 1.14.0.