# Solving problem using related rates yields incorrect result

I've been trying for a while to figure out what I did wrong on this problem, help would be appreciated.

"A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20 ft away from the wall, how fast does the ladder move up the wall 5 sec after we start pushing?"

$$x$$ = distance from wall, $$y$$ = height of ladder on wall, $$h$$ = length of ladder $$\frac {dx}{dt} = -1, t = 5$$ $$x^2 + y^2 = h^2$$ $$x^2 +y^2 = 25^2$$ $$2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = \frac{-x\frac{dx}{dt}}{y}$$ Since the ladder is moving toward the wall at 1 ft/sec for 5 sec, $$x(t = 5) =20-5 = 15$$

$$y(t=5) = \sqrt{25^2 - 15^2}=20$$ Substitute variables in equation: $$\frac{dy}{dt} = \frac{-(15)(-1)}{20}$$ $$\frac{dy}{dt} = \frac{3}{4}$$ The issue is that if $$\frac{dy}{dt} = \frac{3}{4},$$ the ladder would have risen 3.75 ft. in 5 seconds instead of 5 feet it SHOULD have risen, which is given by: $$\Delta y = y(t=5) - y(t=0)$$ $$(\sqrt{25^2 - 15^2}) - (\sqrt{25^2 - 20^2})$$ $$15-20$$ $$5$$ Thank you for reading and any answers.

At the start you have $$y(0)=\sqrt {25^2-20^2}=15$$. You are correct that the top has risen $$5$$ feet in $$5$$ seconds, but that does not mean that it has risen steadily at $$1$$ ft/sec for the whole time. Because $$\frac {dx}{dt}=-1$$ you found $$\frac {dy}{dt}=\frac xy$$ At $$t=0$$ that means that $$\frac {dy}{dt}=\frac {20}{15}=\frac 43$$ and at $$t=5$$ that means $$\frac {dy}{dt}=\frac {15}{20}=\frac 34$$. There is no contradiction here.

• Oh, I assumed that the rate of change had to be constant for some reason. Thank you for your help! – Kebab Boy Mar 14 at 20:07

Keeping your notation, you have that $$x(t) = 20 - t, \quad y(t)=\sqrt{25^2-x(t)^2},$$

and so, $$y'(t) = \frac{-2 x'(t)x(t)}{2 \sqrt{25^2-x(t)^2}},$$

which means that $$y'(5)=\frac{15}{\sqrt{25^2-15^2}}=\frac{3}{4}$$

The vertical displacement does not have to match the horizontal one. They are just connected through the expression $$x(t)^2 + y(t)^2=25^2$$.