question_answer

An edge of a variable cube is increasing at the rate of 10 cm/s. How fast the volume of the cube is increasing when the edge is 5 cm long?

Answer:

Let x be the length of an edge of cube and V be the volume of the cube. Then, \[V={{x}^{3}}\] \[\therefore \]Rate of change of volume w.r.t. time, we get \[\frac{dx}{dt}=\frac{d}{dt}({{x}^{3}})=3{{x}^{2}}\frac{dx}{dt}\] It is given that edge of the cube is increasing at the rate of 10 cm/s. So,       \[\frac{dx}{dt}=10\,cm/s\] \[\therefore \]      \[\frac{dV}{dt}=3{{x}^{2}}(10)=30{{x}^{2}}c{{m}^{3}}/s\] Thus, when\[x=5\,cm\], then \[\frac{dV}{dt}=30\,{{(5)}^{2}}=750\,c{{m}^{3}}/s\] Hence, the volume of the cube is increasing at the rate of \[750\,c{{m}^{3}}/s\]when the edge is 5 cm long.