Question #305330

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.


1
Expert's answer
2022-03-05T08:08:43-0500

(a)


V=x3V=x^3

Differentiate both sides with respect to tt


dVdt=3x2dxdt\dfrac{dV}{dt}=3x^2\dfrac{dx}{dt}

Then


dxdt=13x2dVdt\dfrac{dx}{dt}=\dfrac{1}{3x^2}\dfrac{dV}{dt}

Given dVdt=0.216 cm3s1,x=6 cm\dfrac{dV}{dt}=0.216\ cm^3\cdot s^{-1}, x=6\ cm


dxdt=13(6 cm)2(0.216 cm3s1)=0.002 cms1\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 cms1.0.002\ cm\cdot s^{-1}.


(b)


A=x2A=x^2

Differentiate both sides with respect to tt


dAdt=2xdxdt\dfrac{dA}{dt}=2x\dfrac{dx}{dt}

Given dxdt=0.002 cms1,x=6 cm\dfrac{dx}{dt}=0.002\ cm\cdot s^{-1}, x=6\ cm


dAdt=2(6 cm)(0.002 cms1)=0.024 cm2s1\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 cm2s1.0.024\ cm^2\cdot s^{-1}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS