MHB MaleahP's question at Yahoo Answers regarding related rates

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The discussion revolves around a calculus problem involving related rates, where a boat is being pulled towards a dock. The key variables defined include the horizontal distance from the dock, the length of the rope, and the vertical height difference between the pulley and the boat. By applying the Pythagorean theorem and implicit differentiation, the rate at which the boat approaches the dock is derived. When evaluated at a distance of 7 meters from the dock, the boat approaches at approximately -1.01 m/s. This solution effectively demonstrates the application of calculus in real-world scenarios.
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Here is the question:

Calc homework help please!?

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.
 
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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:

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