last ap calculus problem?

a. Let's name P The perimeter of the square

and a the square's side.We know that

a=r√2.So P=4a=4r√2 (1) and C=2pir (2)

From (1),(2)--> C=(pi√2/4)*P -->

C'(t)=(pi√2/4)*P'(t)-->

C'(t)=(pi√2/4)*P'(t)

C'(t)=(pi√2/4)*8=2pi√2 in/s

b. Let's call S the area of the square

Then S=a^2=2r^2 (3)

the area enclosed between the square

and the circle is A-S=(pi-2)r^2 (4)

From (a) above we know that

The perimeter of the square is increasing

at the rate of 8 inches per second

P'(t)=4r'(t)√2-->8=4r'(t)√2-->r'(t)=√2 in/s

Now (A-S)'(t)=2*(pi-2)*r(t)*r'(t)-->

(the moment that S=16 in^2-->a=4 in

-->r=4/√2=2√2 in.So:

(A-S)'(t)=2*(pi-2)*2√2*√2=8(pi-2) sq in/s

perimeter square = 4*side

therefore length of side = 8*t / 4 = 2t inches/sec (t=time)

therefore radius of circle = t* sqr2 (sqr = square root)

C = 2pi * radius

a) C = 2pi * t *sqr2 inches/sec

b) Acir = 2pi * (t *sqr2)^2

Asqare = 2t*2t

enclose A = Acir-Asquare