The volume of a cube is increasing at a rate of 9 cm^3/sec . How fast is the surface area increasing when the length of an edge is 12 cm?

Explanation

Then, `V=x^{3}` and `S=6 x^{2}` where `x` is a function of time `t`.


It is given that `frac{d V}{d t}=8 {cm}^{3} / {s}`.


Then, by using the chain rule, we have:


`therefore 8=frac{d V}{d t}=frac{d}{d t}(x^{3})=frac{d}{d x}(x^{3}) cdot frac{d x}{d t}=3 x^{2} cdot frac{d x}{d t}`


`Rightarrow frac{d x}{d t}=frac{8}{3 x^{2}}`


(1)


Now, `frac{d {S}}{d t}=frac{d}{d t}(6 x^{2})=frac{d}{d x}(6 x^{2}) cdot frac{d x}{d t}`


[By chain rule]


...