Applications of differentiation qn

Two concentric circles have radii R and r, where R>r.

R increases at a constant rate of 1 cm per minute and the area between the two circles remains constant at 25cm square. At the instant when R = 5 cm, find the exact rate of increase of
(i) the area of the smaller circle,
(ii) r.

Answer to 5i is 10pi cm square/min and answer to 5 ii is sqrt (pi/(pi - 1) cm/min. Please explain clearly, thanks!

(i) The area of the annulus between the circles is
pi R^2 - pi r^2 = 25
Rearranging:
pi r^2 = pi R^2 - 25
Let the area of the smaller circle be A:
so, A = pi R^2 - 25
dA/dt = 2 pi R (dR/dt)
(R=5, dR/dt = 1):
dA/dt = 10 pi


(ii) if R=5, then pi (25-r^2)=25
25-r^2 = 25/pi
r^2 = 25 - 25/pi
r^2 = 25(1 - 1/pi)
r^2 = 25 (pi - 1) / pi
r = 5 sqrt [ (pi - 1) / pi ]

Since A = pi r^2
then dA/dt = 2 pi r dr/dt
10 pi = 2 pi times 5 sqrt [ (pi - 1) / pi ]
rearrangement gives your answer

A = pi * (R^2 - r^2)
dA/dt = pi * (2R * dR/dt - 2r * dr/dt)
Since the area doesn't change, we'll set that to 0

0 = pi * (2R * dR/dt - 2r * dr/dt)
0 = R * dR/dt - r * dr/dt
r * dr/dt = R * dR/dt

R = 5
dR/dt = 1

r * dr/dt = 5 * 1
r * dr/dt = 5

Let's solve for r by using our original formula:

A = pi * (R^2 - r^2)
25 = pi * (5^2 - r^2)
25 / pi = 25 - r^2
r^2 = 25 - 25/pi
r^2 = 25 * (pi - 1) / pi
r = 5 * sqrt((pi - 1) / pi)

r * dr/dt = 5
5 * sqrt((pi - 1) / pi) * dr/dt = 5
dr/dt = sqrt(pi / (pi - 1))


pi * r^2 => pi * 25 * (pi - 1) / pi = 25 * (pi - 1)

The rate of increase of the smaller circle is:

dA/dt = 2 * pi * r * dr/dt
dA/dt = 2 * pi * 5
dA/dt = 10pi

The rate of increase of the radius of the smaller circle is:
sqrt(pi / (pi - 1))