Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises - Page 234: 12

Answer

(a) When the side length is $3~mm$, the instantaneous rate of change of the volume with respect to the side length is $27~mm^2$ (b) The rate of change of the volume with respect to the side length is half the surface area.

Work Step by Step

(a) $V(x) = x^3$ $\frac{dV}{dx} = 3x^2$ When $x = 3~mm$: $\frac{dV}{dx} = 3(3)^2 = 27$ When the side length is $3~mm$, the instantaneous rate of change of the volume with respect to the side length is $27~mm^2$ (b) The area of each face of a cube is $x^2$ The surface area of a cube is $6x^2$ Since $\frac{dV}{dx} = 3x^2$, the rate of change of the volume with respect to the side length is half the surface area. Suppose the original volume of a cube is $x^2$ After we increase the sides by $\Delta x$, the new volume is $(x+\Delta x)^3$ We can find the change in volume: $\Delta V = (x+\Delta x)^3-x^3$ $\Delta V = x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3-x^3$ $\Delta V = 3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3$ $\Delta V \approx 3x^2\Delta x$ Note that we can say this because $3x(\Delta x)^2+(\Delta x)^3 \approx 0$ Then: $\Delta V \approx 3x^2\Delta x$ $\frac{\Delta V}{\Delta x} \approx 3x^2$
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