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Expert: - 9/14/2010


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

Answer
If the rectangle is inscribed in a circle of radius 5 and has one side that measures 6, draw this on graph paper.  Make the circle has center at (0,0).  Make the line of length 6 inches hit the circle on top at x=-3 and x=3.  Since x^2 + y^2 = 5^2, this means that
9 + y^2 = 25, or y^2 = 16, so y = 4 and -4.  The area is known to be xy, and y = sqrt(25-x^2).  This means at that point y = -4 and 4.

This means the area is xy = x*sqrt(25-x^2), so now area is
A(x) = x*sqrt(25-x^2).  Now noting that we really have
A(t) = x(t)*sqrt(25-x^2(t)), we know that

dA/dt =x(t)(-2x(t)x’(t)/sqrt(25-x^2) + sqrt(25-x^2)x’(t).
This reduces to
dA/dt = x’(-2x*sqrt(25-x^2)/(25-x^2) + sqrt(25-x^2)(25-x^2)/(25-x^2)
where x is really x(t) and x' is x'(t).

It is known that dx/dt = -2 and x = 3, so put in these values to solve for dA/dt.

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