Answer to Question #305939 in Calculus for Nurul

Question

The volume, V cm³, of a metallic cube of side length x cm, is increasing at the constant rate of 0.216 cm3 s^-1 .

(a) Determine the rate at which the side of the cube is increasing when the side length reaches 6 cm.

(b) Find the rate at which the surface area of the cube, A cm², is increasing when the side length reaches 6 cm.

Answer 1

(a)

$$V=x^3$$

Differentiate both sides with respect to $t$

$$\dfrac{dV}{dt}=3x^2\dfrac{dx}{dt}$$

Then

$$\dfrac{dx}{dt}=\dfrac{1}{3x^2}\dfrac{dV}{dt}$$

Given $\dfrac{dV}{dt}=0.216\ cm^3\cdot s^{-1}, x=6\ cm$

$$\dfrac{dx}{dt}=\dfrac{1}{3(6\ cm)^2}(0.216\ cm^3\cdot s^{-1})=0.002\ cm\cdot s^{-1}$$

The side of the cube is increasing at the of $0.002\ cm\cdot s^{-1}.$

(b)

$$A=x^2$$

Differentiate both sides with respect to $t$

$$\dfrac{dA}{dt}=2x\dfrac{dx}{dt}$$

Given $\dfrac{dx}{dt}=0.002\ cm\cdot s^{-1}, x=6\ cm$

$$\dfrac{dA}{dt}=2(6\ cm)(0.002\ cm\cdot s^{-1})=0.024\ cm^2\cdot s^{-1}$$

The surface area of the cube is increasing at the of $0.024\ cm^2\cdot s^{-1}.$