The speed of the top of a sliding ladder

Solution 1:

You can't use an area relation because we're not told what $\dfrac{dA}{dt}$ is. If you draw out the situation, we see that we can actually use a different relation to solve this problem.

The equation that you want to use comes directly from the Pythagorean theorem; in particular, if $x$ is the distance the base of the ladder is from the wall and $y$ is the the height of the ladder along the wall, then for the 5 m long ladder, we have that $x^2+y^2 = 25$.

Differentiating this will then give you the appropriate equations involving $\dfrac{dx}{dt}$ (i.e. the rate the ladder is sliding away from the wall) and $\dfrac{dy}{dt}$ (i.e. the rate the ladder is sliding down the wall). You can then plug in all the known values at the appropriate instant in time (when $x=3$, $y=\ldots$ [which can be found using the Pythagorean theorem], and $dx/dt = 0.4\text{ m/s}$) and then solve for $\dfrac{dy}{dt}$.

Hopefully this is enough information to help you get the correct solution. :-)

Solution 2:

First we use Pythagorean Theorem

$$5^2=x^2+y^2 \implies y = \sqrt{25-x^2}.$$

Next we use that we can write $x=0.4t=\frac{2}{5}t$: $$y=\sqrt{25-\left(\frac{2}{5}t\right)^2} = \sqrt{25-\frac{4}{25}t^2}.$$

Now we calculate the derivative (using the chain rule) $$ \frac{dy}{dt}= \frac{\frac{d}{dt}\left(25-\frac{4}{25}t^2\right)}{2\sqrt{25-\frac{4}{25}t^2}} = \frac{-\frac{8}{25}t}{2\sqrt{25-\frac{4}{25}t^2}}.$$

This can be simplified further, but it isn't necessary. Now for the last step: evaluating the derivative at $t=7.5$ (which comes from solving $0.4t=3$).

$$\frac{dy}{dt}|_{t=7.5} =\frac{-\frac{8}{25}\cdot 7.5}{2\sqrt{25-\frac{4}{25}\cdot7.5^2}}=-0.3. $$ So the conclusion is that the speed, when the bottom is $3\text{m}$ from the wall, is $-0.3 \text{ m}/\text{s}$.