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.
Best Answer
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.