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...

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.