Applications Of Derivatives Question 216
The volume of a spherical balloon is increasing at the rate of 40 cubic centimetre per minute. The rate of change of the surface area 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 }=5 $
BETA
