Answer:
[tex]\displaystyle \frac{4}{3}\text{cm}^2/\text{min}[/tex]
Step-by-step explanation:
Given
[tex]\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}[/tex]
Solution
(1) Find the rate of the cube's edge length with respect to time at s=30:
[tex]\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}[/tex]
(2) Find the rate of the cube's surface area with respect to time at s=30:
[tex]\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}[/tex]
Therefore, the surface area increases when the length of an edge is 30 cm at a rate of [tex]\displaystyle \frac{4}{3}\text{cm}^2/\text{min}[/tex].
