Question

The volume of a cube is increasing at the rate of 9 cm³ /s. How fast is its surface area increasing when the length of an edge is 10 cm?

Answer 1

Let V and S be the volume and surface area of a cube of side x cm respectively.

Given,

\(\frac{dV}{dt}\) = 9 cm³/sec

We require \(\frac{dS}{dt}]_{x=10cm}\)

Now,

V = x³

⇒ \(\frac{dV}{dt}\) = 3x².\(\frac{dx}{dt}\)

 ⇒ \(9\) = 3x².\(\frac{dx}{dt}\)

 ⇒ \(\frac{dV}{dt}\) = \(\frac{9}{3x^2}\)= \(\frac{3}{x^2}\)

Again,

∵ S = 6x²

[By formula for surface area of cube]