Applications Of Derivatives Question 216
Question: The volume of a spherical balloon is increasing at the rate of 40 cubic centrimetre per minute. The rate of change of the surface of the balloon at the instant when its radius is 8 centimetre, is
[Roorkee 1983]
Options:
A) $ \frac{5}{2} $ sq cm/min
B) 5 sq cm/min
C) 10 sq cm/min
D) 20 sq cm/min
Show Answer
Answer:
Correct Answer: C
Solution:
Here $ f’(x)=2x\log x+x $ and $ S=4\pi r^{2} $
therefore $ \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt} $
therefore $ f’’(1)=3+2{\log_{e}}1 $
$ \therefore \frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi \times 8\times \frac{5}{32\pi }=10 $ .
BETA
