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  • The volume of a sphere is increasing at the rate of 8 cm3/s. Find the rate at which its surface area is increasing when the radius of the sphere is 12 cm.      </

The volume of a sphere is increasing at the rate of 8 cm3/s. Find the rate at which its surface area is increasing when the radius of the sphere is 12 cm.

 

 

 

 
 
 
 
 

Answers (1)

we have, volume of sphere \left ( v \right )= \frac{4}{3}\pi r^{3}
where r is the radius of spere
Now, differentiating v w.r.t t, we get
\frac{dv}{dt}= \frac{4}{3}\pi 3r^{2}\frac{dr}{dt}= 8 cm^{3}/s
\Rightarrow 4\pi r^{2}\frac{dr}{dt}= 8cm^{3}/s
\Rightarrow \frac{dr}{dt}= \frac{2}{\pi r^{2}}\, cm^{3}/s---(i)
and surface area of sphere \left ( s \right )= 4\pi r^{2}
Then, differentiating s w.r.t t, we get
\frac{ds}{dt}= 4\pi \cdot 2r\frac{dr}{dt}
\Rightarrow \frac{ds}{dt}= 8\pi r \cdot \frac{2}{\pi r^{2}}\left [ using(i) \right ]
\Rightarrow \frac{ds}{dt}=\frac{16}{r}\, cm^{2}/s
when r= 12cm \; \left [ \frac{ds}{dt} \right ]_{r= 12}= \frac{16}{12}= \frac{4}{3}\, cm^{2}/s
hence the surface area is increasing at the rate of \frac{4}{3}\, cm^{2}/s when radius of sphere is 12 cm

Posted by

Ravindra Pindel

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