# 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?

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Let one side of the rectangle inscribed be x, then the other side of the rectangle is = 2*times the distance of the side x from the centre = 2*sqrt(r^2- (x/2)^2) .

Therefore area of the rectangle A(x) = product of the sides = x* sqrt{r^2-x^2/4.

Therefore the rate of change in area A(x) is A'(x) when x is incresing.

A'(x) = {x*sqrt(x^2-/4)}'.

A'x) = {x*sqrt(r^2-x^2/4)}'.

A'(x) = x*{sqrt(r^2-x^2/4)}'+ x' *{sqrt(r^2-x^2/4).

A'(x) = x*(1/2)*(r^2-x^2/4)^(1/2-1) * (-2x/4) +sqrt(r^2-x^2/4}

A'(6) = 6*(1/2)*(5^2-6^2/4)^(-1/2)*(-2*6/4) +sqrt(5^2-6^2/4)

A'(6) = 3*(25-9)^(-1/2) * (-3) +sqrt(25-9)

A'(6) = (3/4)(-3) + 4

A'(6) = -2.25+4 = 0.5.

Therefore the rate increase in area = 1.75 sq units when x = 6.