Help with related rates problem (calculus I)

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May 3rd 2009, 11:22 AM
gtgamer140@yahoo.com

Help with related rates problem (calculus I)

Two boats are racing with constant speed toward a finish marker, boat A sailing from the south at 13mph and boat B approaching from the east. When equidistant from the marker the boats are 16 miles apart and the distance between them is decreasing at the rate mph. which boat will win the race.

I'm lost at this problem so any help will be much appreciated(Happy)
May 3rd 2009, 11:37 AM
TheEmptySet

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Quote:

Originally Posted by gtgamer140@yahoo.com

Two boats are racing with constant speed toward a finish marker, boat A sailing from the south at 13mph and boat B approaching from the east. When equidistant from the marker the boats are 16 miles apart and the distance between them is decreasing at the rate mph. which boat will win the race.

I'm lost at this problem so any help will be much appreciated(Happy)

P.S you are missing one piece of information... the rate that the distance between them is decreasing. Here is the setup.

First Always draw a diagram.

Attachment 11242

Then from the pythagorean theorem we know that

$x^2+y^2=z^2$ taking the derivative with respect to time we get

$2x\frac{dx}{dt} +2y\frac{dy}{dt}=2z\frac{dz}{dt}$

So the speed of the 2nd boat is

$\frac{dy}{dt}=\frac{z\frac{dz}{dt}-x\frac{dx}{dt}}{y}$

From here you just need to plug in the rates and the points.

Note be careful with the signs
May 3rd 2009, 12:00 PM
gtgamer140@yahoo.com

wow you made the problem seem easy...thanks a bunch!(Talking)
May 3rd 2009, 12:53 PM
gtgamer140@yahoo.com

(the rate left out was 17mph) It makes sense how you got the final equation but what are the x and y values I plug in? It doesn't seem to be represented in the word problem.
May 3rd 2009, 04:25 PM
TheEmptySet

Quote:

Originally Posted by gtgamer140@yahoo.com

When equidistant from the marker the boats are 16 miles apart

This tells us we have an isosceles triangle(two sides have the same length) then by the pythagorean theorem we get

$x^2+x^2=16^2 \iff 2x^2=256 \iff x^2=128 \iff x=8\sqrt{2}$

So the distance is $8\sqrt{2}$