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KCET2011MathematicsApplication of Derivatives

A sphere increases its volume at the rate of cc / s . The rate at which its surface area increases when the radius is 1 ~cm is

Options

  1. A2 sq cm / s
  2. Bsq cm / s
  3. C3 2 sq cm / s
  4. D2 sq cm / s

Correct answer

A. 2 sq cm / s

Step-by-step solution

Given rate of increase volume of sphere is d V d t = We know that, Volume of sphere; V= 4 3 r³ d V d t = 4 3 3 r² d r d t =4 r² d r d t =4 r² d r d t ; [from Eq. (i)] d r d t = 1 4 r² 5 Also, we know that, surface of sphere, aligned &S=4 r² & d S d t =8 r d r d t =8 r 1 4 r² [ from Eq. (ii)] & [ given r=1] & d S d t =2 aligned So, rate of increase in surface =2 sq cm / s

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