A 3-meter ladder is leaning against a wall and its base is being pulled away from the wall at a rate of 20 cm/s. Find the rate the top of the ladder is falling when the base is 2.5 meters from the wall. The image shows the setup:

Formalizing the information given, we want to find $\frac{dy}{dt}$ when $x=2.5 \text{ m}$ given $\frac{dx}{dt}=0.2 \text{ m/s}$. (Note the conversion to meters.) Also, using the pythagorean theorem we have $$x^2+y^2=3^2$$ This is the equation we need to take the derivative of with respect to $t$.

Using Leibniz's notation (instead of writing $x$ and $y$ as functions of $t$), this equation says $$2x \frac{dx}{dt}+2y\frac{dy}{dt}=0$$ We want to solve this for $\frac{dy}{dt}$.

Here, we have $$\frac{dy}{dt}=-\frac{x\space \frac{dx}{dt}}{y}$$

To evaluate this we need $x$, $\frac{dx}{dt}$, and $y$, which we don't have yet, but we can solve for it given $x=2.5$.

We can ignore the negative value since $y$ is a distance, therefore $y=\frac{\sqrt{11}}{2}$. This finally gives us:

This gives $$\frac{dy}{dt}=-0.3015 \text{ m/s}$$

The value is negative since the distance $y$ is actually decreasing ($y$ is going down to 0.) Leaving off the **.n()** we get a value interms of a number multiplied by $\sqrt{11}$. Give this a try. To get an exact value we would need to evaluate:
$-\frac{\frac{5}{2}\cdot \frac{1}{5}}{\sqrt{11}}$.

**Question:** What happens as $x\rightarrow 3$? Hmmmm.