Question

A circular ink blot grows at the rate of 2 cm^2/sec, find the rate at which its radius is increasing after 28/11 sec.

Answer 1

Area of circular ink blot , A = πr² , here r is the radius .
A = πr²
differentiate with respect to time ,
dA/dt = 2πr.dr/dt
Here given dA/dt = 2cm²/sec.
⇒2 = πr.dr/dt
⇒1 = πr.dr/dt ------------(1)

at t = 0, area = zero. According to question area increasing at the rate of 2cm²/sec . so, at t = 28/11 sec , area will be 2 × 28/11 = 56/11 cm²
e.g., 56/11 = πr² ⇒r = √{56/11π}
Now from equation (1) ,
1 = π√{56/11π}dr/dt
⇒1 = √{56π/11}.dr/dt
⇒dr/dt = 1/√{56π/11} = √{56 × 22/7 × 11} = √{1/16} = 1/4
Hence, dr/dt = 1/4 = 0.25 cm/sec

So, the increasing radius at the rate is 0.25 cm/sec