Question

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 an edge is 12 cm?

Answer 1

$volume \: of \: cube = {x}^{3} \\ \frac{dv}{dt} = 8 {cm}^{3} \\ \\ = 3 {x}^{2} \frac{dx}{dt} = 8 \\ \frac{dx}{dt} = \frac{8}{3 {x}^{2} }$
now have to find rate of increment of surface area.
$surface \: area = 6{x}^{2} \\ \frac{ds}{dt} = 6 \times 2x \frac{dx}{dt} \\ = 12x \times \frac{8}{3 {x}^{2} } \\ \frac{ds}{dt} = \frac{32}{x}$
now put x=12
$\: \frac{ds}{dt} = \frac{32}{12} = \frac{8}{3}$

Answer 2

Given,
volume of cube is increasing at the rate of 8cm³/s. e.g., $\bf{\frac{dV}{dt}=8cm^3/s}$
edge length , a = 12cm

we know volume of cube , V = a³
now differentiate with respect to time,
$\bf{\frac{dV}{dt}=3a^2\frac{da}{dt}}$
put a = 12cm and dV/dt = 8cm³/s
now, 8 = 3(12)² da/dt
8 = 3 × 144 × da/dt
=> da/dt = 1/54 cm/s -----(1)

we also know surface area of cube , A = 6a²
differentiate A with respect to time,
$\bf{\frac{dA}{dt}=12a\frac{da}{dt}}$
put a = 12cm and from equation (1),
so, dA/dt = 12 × 12 × 1/54 = 144/54 = 8/3

hence, rate of change of surface area of cube is $\bf{\frac{dA}{dt}=\frac{8}{3}cm^2/s}$