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If the volume of a cube is increasing at the rate of $6\text{cm}^3\text{s}^{-1}$ when the side length is $2\text{cm}$, find the rate at which the total surface area is increasing at that time.

I've tried relating the rate of change of surface Area with the volume but I'm not getting it.

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  • $\begingroup$ Yes the length (side) $\endgroup$ – Orestes Dante Nov 21 '17 at 8:55
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Let $V$ be the volume, $s$ the side length and $A$ the surface area. Then $$ V = s^3\implies s = \sqrt[3]{V}\\ A = s^2 \implies A = \sqrt[3]{V}^2 $$ Now you can use the chain rule to differentiate $A$ with respect to time and evaluate, knowing that $V' = 6$ and $V = 2^3 = 8$ at the relevant moment.

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$V(x(t)) =x^3(t).$

$\dfrac{d}{dt} V (x(t)) = 3x^2\dfrac{dx}{dt}$; i.e.

$6= 3×2^2( \dfrac{dx}{dt})_{x=2}.$

$(\dfrac{dx}{dt})_{x=2} = 1/2.$

Area: $A(x) = 6x^2.$

$\dfrac{d}{dt}A(x(t)) = 12 x (\dfrac{dx}{dt})_{x=2}= $

$12 ×2×1/2 = 12 [cm^2/s].$

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Hint. Let $x(t)$ be the side length of the cube at time $t$, then volume and the area of the cube are given by $V(t)=x(t)^3$, and $S(t)=6x(t)^2$, respectively. Now $$x(t)=2,\quad V'(t)=3x(t)^2x'(t)=6\implies x'(t)=\frac{6}{3x(t)^2}=\frac{6}{3\cdot 2^2}=\frac{1}{2}.$$ Then what is the value of $S'(t)=6\cdot2x(t)x'(t)$?

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