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Calculus/Calculus: related rates

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Expert: - 1/16/2011


Question
A rectangle is inscribed in a circle of radius 5 inches. If the length of the rectangle is decreasing at the rate of 2 inches per second, how fast is the area changing when the length is 6 inches?

I am stuck at finding w and w'
A = lw
A'= l'w + w'l

Answer
If the rectangle is inscribed in a circle, that means the corners touch the edge of the circle.
If the length is 6", then that means 3" are on either side of the center.
That is one leg of a right triangle and the hypotenuse is the radius.

Since the radius is 5", that means the length of the other leg is
sqrt(5^2 - 3^2) = sqrt(25 - 9) = sqrt(16).  Since it is a length, it is positive sqrt(16),
which is 4.  This is half the length of the other leg of the rectangle.

Now one legs of the triangle is w/2 and the other leg of the triangle is l/2,
It is known that (w/2)^2 + (l/2)^2 = r^2.  Mutiplying by 4/4 { 2^2/2^2 } gives
w^2 + l^2 = 4r^2.

Since r is 5, we can put that in and get w^2 + l^2 = 4*5^2, and 4*5^2 is 4*25, which is 100.
So if we have w^2 + l^2 = 100, that is really w^(t) + l^(t) = 100, since w and l
change over time.

The derivative of the equation (w/2)^2 + (l/2)^2 = 100 is 2w(t)w'(t) + 2l(t)l'(t) = 0.
It is known that w(t) = 2*4" = 8", l(t) = 2*3" = 6", and l'(t) = 6,
so from this, w'(t) can be found since I believe it works out to be w'(t) = -l(t)l'(t)/w(t).

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