The volume of a cube is increasing at a rate of 10 cm^3/min. How fast is the surface area increasing when the length of an edge 90 cm?

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The volume of a cube is increasing at a rate of 10 cm^3/min. How fast is the surface area increasing when the length of an edge 90 cm?

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Answer 1 · 2018-05-24T23:53:13

$\frac{12}{27}$ $12/27$ cms squared per minute.

Explanation:

Let the length of a side of the cube = $x$ $x$

Then volume of cube $v$ $v$ = $x^{3}$ $x^3$ ......... $[1]$ $[1]$
Surface area of cube $s$ $s$ will = $6 x^{2}$ $6x^2$ ...... $[2]$ $[2]$

Differentiating $[1]$ $[1]$ implicitly w.r.t. $t$ $t$ [ time], $\frac{d v}{d t} = 3 x^{2} \frac{d x}{d t}$ $[dv]/dt=3x^2dx/dt$ , but we know from the question that $\frac{d v}{d t} = 10$ $[dv]/dt=10$ , therefore $\frac{d x}{d t} = \frac{10}{3 x^{2}}$ $dx/dt=[10]/[3x^2$ ...... $[3]$ $[3]$

Differentiating....... $[2]$ $[2]$ w.r.t. $t$ $t$ , $\frac{d s}{d t} = 12 x \frac{d x}{d t}$ $[ds]/dt=12xdx/dt$ and substituting for $\frac{d x}{d t}$ $dx/dt$ from ..... $[3]$ $[3]$ we obtain,

$\frac{d s}{d t} = [12 x \frac{10}{3 x^{2}}]$ $[ds]/dt=[12x[10]/[3x^2]]$ = $\frac{120}{3 x}$ $[120]/[3x]$ , and so we have the rate of change of the side length of the cube with respect to time, and when $x = 90$ $x=90$ , $\frac{d s}{d t} = \frac{120}{3 [90]}$ $[ds]/dt=120/[3[90]]$ = $\frac{12}{27}$ $12/27$ .

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The volume of a cube is increasing at a rate of 10 cm^3/min. How fast is the surface area increasing when the length of an edge 90 cm?

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