## University Calculus: Early Transcendentals (3rd Edition)

The given % error change in side length: $\frac{dx}{x}$=0.5% % error change=$\frac{change\ in\ dimension}{original\ dimension}$ $A=6x^2$ $dA=12x dx$ $V=x^3$ $dV=3x^2 dx$ % error in Area=$\frac{dA}{A}=\frac{12x dx}{6x^2}=2\times0.005=0.01$ % error in volume=$\frac{dV}{V}=\frac{3x^2dx}{x^3}=0.015$