# MaleahP's question at Yahoo! Answers regarding related rates

#### MarkFL

Staff member
Here is the question:

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 7 m from the dock? (Round your answer to two decimal places.)
I have posted a link there to this thread so the OP can view my work.

#### MarkFL

Staff member
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:

$$\displaystyle x^2+h^2=\ell^2$$

Implicitly differentiating with respect to time $t$, we find:

$$\displaystyle 2x\frac{dx}{dt}=2\ell\frac{d\ell}{dt}$$

We are asked to find $$\displaystyle \frac{dx}{dt}$$, which describes the rate of change of the boat's position. So, solving for this quantity, we find:

$$\displaystyle \frac{dx}{dt}=\frac{\ell}{x}\frac{d\ell}{dt}$$

Since we are given a value of $x$ at which to evaluate this, we may use:

$$\displaystyle \ell=\sqrt{x^2+h^2}$$

to get our formula in terms of know values:

$$\displaystyle \frac{dx}{dt}=\frac{\sqrt{x^2+h^2}}{x}\frac{d\ell}{dt}$$

Now, using the given data:

$$\displaystyle x=7\text{ m},\,h=1\text{ m},\,\frac{d\ell}{dt}=-1\frac{\text{m}}{\text{s}}$$

we find:

$$\displaystyle \left.\frac{dx}{dt} \right|_{x=7}=-\frac{\sqrt{7^2+1^2}}{7}\frac{\text{m}}{\text{s}}=-\frac{5}{7}\sqrt{2}\frac{\text{m}}{\text{s}} \approx-1.01\frac{\text{m}}{\text{s}}$$