Related rate?

Question

A circle is inside a square. 

The radius of the circle is increasing at a rate of 4 meters per minute and the sides of the square are decreasing at a rate of 4 meters per minute. 

When the radius is 5 meters, and the sides are 18 meters, then how fast is the AREA outside the circle but inside the square changing? 

Answer 1

The radius(r) of the circle is increasing at a rate of 4 meter per minute ⟹ dr/dt = 4 m/t.

The sides(x) of the square are decreasing at a rate of 4 meter per minute ⟹ - dx/dt = 4 m/t.

The radius r = 5 meters and each side = 18 meters.

The area outside the circle but inside the square A = (x² - πr²) meters².

The rate change of area dA/dt = (d/dt)(x² - πr²).

                                                      = (d/dt)(x²) - π(d/dt)(r²).

                                                      = 2x(dx/dt) - π2r(dr/dt).

                                                      = 2(18)(-4) - 2(22/7)(5)(4).

                                                      = -144 - 880/7.

                                                      = - (1008 + 880)/7.

                                                      = - 1888/7.

                                                      = - 269.7 m²/t.

dA/dt = - 269.7 m²/t.

The area between is decreasing at a rate of 269.66 sq. m./min.