MHT CET2017MathematicsApplication of Derivatives
If the volume of spherical ball is increasing at the rate of 4 π c m 3 / sec then the rate of change of its surface area when the volume is 288 π c m 3 is
Options
- A4 3 π c m 2 / s e c
- B2 3 π c m 2 / s e c
- C4 π c m 2 / s e c
- D2 π c m 2 / s e c
Correct answer
A. 4 3 π c m 2 / s e c
Step-by-step solution
V = 4 3 π r 3 ⇒ d V d t = 4 π r 2 d r d t When V - 288 π 288 π = 4 3 π r 3 ⇒ r = 6 Now d V d t = 4 π ∴ 4 π r 2 d r d t = 4 π = d r d t = 1 r 2 A = surface area = 4 π r 2 ∴ d A d t = 8 π r d r d t = 8 π r × 1 r 2 = 8 π r = 8 π 6 = 4 π 3