Question

77. the volume of a cube is increasing at a rate of %5Cmathrm%7Bmin%7D" id="TexFormula1" title="10 \mathrm{~cm}^{3} / \mathrm{min}" alt="10 \mathrm{~cm}^{3} / \mathrm{min}" align="absmiddle" class="latex-formula"> . how fast is the surface area increasing when the length of an edge is $30 \mathrm{~cm} ?$

Answer 1

Answer:

$\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$

Step-by-step explanation:

Given

$\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}$

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

$\displaystyle V=s^3\\\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\ \\10=3(30)^2\frac{ds}{dt}\\ \\10=3(900)\frac{ds}{dt}\\\\10=2700\frac{ds}{dt}\\\\\frac{10}{2700}=\frac{ds}{dt}\\\\\frac{ds}{dt}=\frac{1}{270}\text{cm}/\text{min}$

(2) Find the rate of the cube's surface area with respect to time at s=30:

$\displaystyle SA=6s^2\\\\\frac{d(SA)}{dt}=12s\frac{ds}{dt}\\ \\\frac{d(SA)}{dt}=12(30)\biggr(\frac{1}{270}\biggr)\\\\\frac{d(SA)}{dt}=\frac{360}{270}\biggr\\\\\frac{d(SA)}{dt}=\frac{4}{3}\text{cm}^2/\text{min}$

Therefore, the surface area increases when the length of an edge is 30 cm at a rate of $\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$.