related rates: A dinghy is pulled toward a dock by a rope...

[ihatecalc]

This question just makes NO sense to me:

A dinghy is pulled toward a dock by a rope from the bow through a ring of the dock 6 feet above the bow, as shown in the figure. Rhe rope is hauled in at the rate of 2 ft/sec.

a) How fast is the boat approaching the dock when 10 ft of rope are out?
b) St what rate is angle theta changing at that moment?

 

[galactus]

Where's our picture as mentioned in the problem?.

Draw a triangle from the ring to the bow and then to a point 6' below the ring. You have a righ triangle.

Let x=the distance from the dock to the bow
Let y=the hypoteneuse of the triangle, that is, the length of the rope from the ring to the bow

For part a, you want dx/dt given that dy/dt=-2 ft/sec.

Use Pythagoras. \(\displaystyle \L\\x^{2}+36=y^{2}\)

Differentiate: \(\displaystyle \L\\2x\frac{dx}{dt}=2y\frac{dy}{dt}\)

\(\displaystyle \L\\\frac{y}{x}\cdot\frac{dy}{dt}\).....[1]

Using Pythagoras and y=10, we find x=8.

Using [1], \(\displaystyle \L\\\frac{dx}{dt}=\frac{10}{8}(-2)=\frac{-5}{2} \;\ ft/sec\)

Now, you try an tackle the second part. OK?.

 

[ihatecalc]

is there a formula i'm supposed to use to find theta? i really think im too stupid to be taking AP calc.. lol

 

[galactus]

Yes, there is a formula, so to speak. Use your trig functions.

Try \(\displaystyle sin({\theta})\)


Look at the diagram. You know your trig functions, I hope.

You have \(\displaystyle \frac{dy}{dt}={-}2\)

You want \(\displaystyle \frac{d{\theta}}{dt}\) when y=10

\(\displaystyle sin({\theta})=\frac{6}{y}\)

Now, finish up?.