Single Variable Calculus Questions and Answers

Start Your Free Trial

Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 12

Expert Answers info

eNotes eNotes educator | Certified Educator

calendarEducator since 2007

write13,548 answers

starTop subjects are Math, Literature, and Science

a.) Recall that the volume of the cube is $\text{Vol } = x^3$ where $x$ is the sides of the cube.

$ \begin{equation} \begin{aligned} \frac{dV}{dx} &= \frac{d}{dx} (x^3)\\ \\ \frac{dV}{dx} &= 3x^2 \end{aligned} \end{equation} $

when $x = 3$mm

$ \begin{equation} \begin{aligned} \frac{dV}{dx} &= 3(3)^2\\ \\ \frac{dV}{dx} &= 27 \frac{mm^3}{mm} \end{aligned} \end{equation} $

$\displaystyle \frac{dV}{dx} = 27$ means that the volume is increasing at a rate of $\displaystyle 27 \frac{mm^3}{mm}$ when the length of the cube is 3$mm$.

b.) Recall that the surface area of the cube is equal to $A = 6s^2$ and the rate of change of the volume of the cube is...

$ \begin{equation} \begin{aligned} V'(s) &= 3s^2 && \text{;where } s^2 = \frac{A}{6}\\ \\ V'(s) &= 3 \left( \frac{A}{6}\right)\\ \\ V'(s) &= \frac{A}{2} \end{aligned} \end{equation} $