differentiation question...?

Question

a liquid is poured on a flat surface. it forms a circular patch which grows at a steady rate of 5cm2/s. find, in terms of pi, the rate of change of the radius after 20s.

pls show the steps

Answer 1

Let r = radius (cm) of the patch
and A = its area in (cm)^2.

A = (pi) r^2
dA/dt = 5 = 2(pi)r. dr/dt ... ... ( 1 )

Area increasing at a steady rate of 5 (cm)^2/s will be 100 (cm)^2 at the end of 5 s.

If r' = radius after 20 s, then (pi)r'^2 = 100 gives r' = 10/sqrt(pi)

Putting this value of r' for r in eqn. ( 1 ),

5 = 2(pi)*[10/sqrt(pi)](dr/dt) which gives dr/dt = 1 / [4sqrt(pi)]

Answer 2

S = pi*r^2
dS/dt = 5 = dS/dr* dr/dt
dS/dr= 2pi*r
5=2pir*dr/dt so

dt =2/5*pi* r*dr
t =1/5 pi r^2 +C and supposing that at t = 0 r=0 C=0

20=1/5pi*r^2 so r = 10/sqrt(pi)
and dr/dt= 5/(20sqrt pi)= 1/(4sqrt(pi)