Answer to Question #286042 in Calculus for zeyn

Question

The edge of a cube was found to be 30 cm with a possible error in measurement of .1 cm. Use differentials to estimate the percentage error (to the nearest hundredth) in computing (a) the volume of the cube and (b) the surface area of the cube.

Answer 1

(a) The Volume of the Cube: 

$V=x^3$

$\dfrac{dV}{dx}=3x^2$

$dV=3x^2 dx$

$\Delta V=3x^2\Delta x$

$\Delta V=3(30 cm)^2(0.1 cm)=270 \ cm^3$

Relative Error:

$\dfrac{\Delta V}{V}=\dfrac{270\ cm^3}{(30\ cm)^3}=0.01$

Percentage Error:

$0.01\cdot 100\%=1\%$

(b)  The Surface Area of the Cube: 

$A=6x^2$

$\dfrac{dA}{dx}=12x$

$dA=12xdx$

$\Delta A=12x\Delta x$

$\Delta A=12(30 cm)(0.1 cm)=36 \ cm^2$

Relative Error:

$\dfrac{\Delta A}{A}=\dfrac{36\ cm^2}{6(30\ cm)^2}=\dfrac{1}{150}$

Percentage Error:

$\dfrac{1}{150}\cdot 100\%=\dfrac{2}{3}\%\approx0.67$