Question

A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 10 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 14 ft/min, at what rate will the boat be approaching the dock when 110 ft of rope is out? The boat will be approaching the dock at?

Answer 1

Answer:

The boat will be approaching the dock at 14.06 ft/min

Step-by-step explanation:

The very first thing we need to do when solving this kind of problems is to draw a diagram that will represent the problem. (Diagram attached).

When looking at the diagram, we can see that we can build a model of the situation by using a right triangle, whose height will be constant while its base (x) and its hypotenuse (l) are variable. These are the variables we are going to work with (Check uploaded picture of the triangle).

Once we have our triangle correctly set up, we can use the pythagorean theorem to relate each of the sides of the triangle.:

Pythagorean theorem: $c^{2}=a^{2}+b^{2}$

So the equation for our problem is:

$l^{2}=x^{2}+10^{2}$

Since we need to find the horizontal speed of the boat, we will need to solve the equation for x, so we get:

$x^{2} =l^{2}-10^{2}\\\\x=\sqrt{l^{2}-10^{2}}$

We can now use this equation to find the horizontal velocity of the boat. Position is related to velocity in that the derivative of position with respect to time is the instantaneous velocity of the parcticle at a given time t.

$\frac{dx}{dt}=velocity$

so we need to take the derivative of the given equation. We start by rewritting the radical as a rational power, so we get:

$x=(l^{2}-100)^{\frac{1}{2}}$

and now we can use the chain rule to take the derivative of this function. We take the derivative of the outern function and multiply it by the derivative of the inner function, like this:

$dx=\frac{1}{2}(l^{2}-100)^{-\frac{1}{2}}(2l)dl$

now we can simplify the equation and divide both sides of it into dt so we get:

$\frac{dx}{dt}=\frac{l}{\sqrt{l^{2}-10^{2}} }\frac{dl}{dt}$

the dx/dt part will represent the horizontal velocity of the boat, while the dl/dt part will represent the velocity at which the rope is being pulled, so we have enough data to solve our problem

$\frac{dx}{dt}=\frac{l}{\sqrt{l^{2}-10^{2}} }\frac{dl}{dt}\\\\\frac{dx}{dt}=\frac{110}{\sqrt{110^{2}-10^{2}} }*14\\\\\frac{dx}{dt}=\frac{110}{\sqrt{12000} }*14$

which yields

$\frac{dx}{dt}=14.06ft/min$

So the boat will be approaching the dock at 14.06ft/min.