The volume of a cube is increasing at the rate of 8 cm^{3}/S. How fast is the
surface area increasing when the length of its edge is 12 cm?

 

 

 

 
 
 
 
 

Answers (1)

Let x be the strength of side, v be the volume & s be the surface area of cube
Then v= x^{3}\: and\: s= 6x^{2},where x is a function of
line t
Now \frac{dv}{dt}= 8\frac{cm^{3}}{s} \left [ given \right ]
\therefore 8= \frac{dv}{dt}= \frac{d}{dx}\left ( x^{3} \right )= 3x^{2}\frac{dx}{dt}\left ( chain \: rule \right )
8= 3x^{2}\frac{dx}{dt}
\frac{8}{3x^{2}}= \frac{dx}{dt}-(i)
Now, \frac{ds}{dt}= 12x\frac{8}{3x^{2}}\left [ using(i) \right ]\Rightarrow \frac{32}{x}
Hence when x= 12cm
then, \frac{ds}{dt}= \frac{32}{12}= \frac{8}{3}\: cm^{2}/S
 

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