Question

surface area of a cube is increasing at rate of 3.6 cm 2/s. find the rate at which its volume increases when length of side a is 10 cm​

Answer 1

ANSWER:

• The volume increases at a rate of 9 cm³/s.

GIVEN:

• Surface area of a cube is increasing at rate of 3.6 cm²/s.

• Length of side a is 10 cm.

TO FIND:

• The rate at which its volume increases.

EXPLANATION:

Surface area of cube = 6a²

$\sf \dfrac{d}{dt}6a^{2} = 12a\left(\dfrac{da}{dt}\right)$

$\sf \dfrac{dS}{dt} = 12a\left(\dfrac{da}{dt}\right)$

$\sf \dfrac{dS}{dt} = 3.6 \ cm^{2} /s.$

$\sf 3.6 = 12a\left(\dfrac{da}{dt}\right)$

$\sf 0.3 = a\left(\dfrac{da}{dt}\right)$

$\sf \dfrac{0.3}{a} = \left(\dfrac{da}{dt}\right)$

Volume of cube = a³

$\sf \dfrac{dV}{dt} =\dfrac{d}{dt} {a}^{3}$

$\sf \dfrac{d}{dt} {a}^{3} = 3 {a}^{2} \left(\dfrac{da}{dt}\right)$

$\sf We\ know\ that\ \left(\dfrac{da}{dt}\right)=\dfrac{0.3}{a}$

$\sf \dfrac{dV}{dt} = 3 {a}^{2} \left(\dfrac{0.3}{a} \right)$

$\sf \dfrac{dV}{dt} = 3 a (0.3)$

$\sf \dfrac{dV}{dt} = (0.9)a$

a = 10 cm

$\sf \dfrac{dV}{dt} = (0.9)10$

$\sf \dfrac{dV}{dt} = 9 \ cm^3/s$

Hence the volume increases at a rate of 9 cm³/s.