Hello MaleahP,
Let's define the following:
$x$ = the horizontal distance of the boat from the dock.
$\ell$ = the length of rope from the pulley to the bow of the boat.
$h$ = the constant vertical difference between the pulley and the bow.
These three quantities form a right triangle at any point in time for which $0<x$. So, by Pythagoras, we may state:
$$x^2+h^2=\ell^2$$
Implicitly differentiating with respect to time $t$, we find:
$$2x\frac{dx}{dt}=2\ell\frac{d\ell}{dt}$$
We are asked to find $$\frac{dx}{dt}$$, which describes the rate of change of the boat's position. So, solving for this quantity, we find:
$$\frac{dx}{dt}=\frac{\ell}{x}\frac{d\ell}{dt}$$
Since we are given a value of $x$ at which to evaluate this, we may use:
$$\ell=\sqrt{x^2+h^2}$$
to get our formula in terms of know values:
$$\frac{dx}{dt}=\frac{\sqrt{x^2+h^2}}{x}\frac{d\ell}{dt}$$
Now, using the given data:
$$x=7\text{ m},\,h=1\text{ m},\,\frac{d\ell}{dt}=-1\frac{\text{m}}{\text{s}}$$
we find:
$$\left.\frac{dx}{dt} \right|_{x=7}=-\frac{\sqrt{7^2+1^2}}{7}\frac{\text{m}}{\text{s}}=-\frac{5}{7}\sqrt{2}\frac{\text{m}}{\text{s}} \approx-1.01\frac{\text{m}}{\text{s}}$$
