Here is the problem:
A ladder 15 metres high is propped up against a high wall. The bottom of the ladder slides away from the wall at a rate of $1\ {m/s}$. How fast is the top of the ladder sliding down the wall when in it is $12\ m$ off the ground?
From this I can state the few things:
\begin{equation} \text{Let}\ y(t) = \text{Height of wall at time}\: t \\ \text{Let}\ x(t) = \text{Distance between bottom of ladder and bottom of wall} \\ \text{We know}\ x'(t) = 1 \\ \text{We need to find}\ y'(t)\ \text{when}\ y(t) = 12 \end{equation}
In order to find this, we need to get $y'(t)$ in terms of $x'(t)$ and $y(t)$.
My first thought is that these form a right triangle, so I call upon Pythagoras:
$$(y(t))^2 + (x(t))^2 = 15^2$$
Then implicitly differentiate with respect to $t$:
\begin{align} & \frac{d}{dt}\bigg[(y(t))^2 + (x(t))^2\bigg] = \frac{d}{dt}\big[15^2\big] \\ \leadsto\quad & 2y'y + 2x'x = 0 \\ \leadsto\quad & y' = -{{x}\over{y}}x' \end{align}
But here is where I reach a dead end. We have $y'$ in terms of $y$ and $x'$, but there's also a $x$ in there - the bottom of the triangle - that we don't know.
How do I approach this?
Thanks in advance!
