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!
So we let
$$ x= \text{the height of the wall at time}\ t = 12 \\ x'= \text{the rate at which the ladder is sliding down the wall} =\textbf{*what we are solving for*} \\ y= \text{the distance between the bottom of the ladder and the wall} =\ ? \\ y'= \text{the rate at which the ladder is sliding away from the wall} = 1 $$
So our initial equation that we want to differentiate is
$$x^2 + y^2 = 225$$
So we differentiate this and get
$$2xx' + 2yy' = 0$$
Then we substitute
$$ 2(12)x' + 2y(1) = 0 \\ 12x' + y = 0 $$
We know that we're solving for $x'$ but what about $y$? Well we know that at this instant the ladder (hypotenuse) will always be 15, and the height of the wall is 12. So we use the Pythagorean theorem to solve for y:
$$ 144 + y^2 = 225 \\ y = 9 $$
We plug this value back into our differentiated equation and we get our answer.
$$ 12x' + 9 = 0 \\ x' = -\frac{3}{4} $$
So the top of the ladder is sliding down the wall at $\frac{3}{4} m/s$ (this implies it is negative since you are saying "slides down") or you can say the rate the ladder is moving is $-\frac{3}{4} m/s$.
Hope that helps!