Question

The edge of a cube was found to be 30 cm with a possible error in measurement of 0.4 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (Round your answers to four decimal places.) (a) the volume of the cube maximum possible error cm3 relative error percentage error % (b) the surface area of the cube maximum possible error cm2 relative error percentage error %

Answer 1

Answer:

Maximum possible error volume

$1080cm^3$

Maximum possible error surface

$144cm^2$

Relative error volume

0.04

Relative error surface

0.0266

% error volume

4%

% error surface

2.66%

Step-by-step explanation:

Let's call x the length of the edge of the cube, V its volume and S its surface.

We have x= 30 cm with a possible error of  0.4 cm

On the other hand,

$V=x^3$

and  

$S=6x^2$

Taking differentials

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

and

$\frac{dS}{dx}=12x$

So, the error when estimating the volume and surface, dV, dS are

$dV=3x^2dx$

and

$dS = 12xdx$

Where x = 30cm and dx=0.4cm

Replacing

$dV=3(30)^20.4=3(900)0.4=1080cm^3$

$dS=12(30)0.4=144cm^2$

The relatives errors are

$\frac{dV}{V}=\frac{3x^2dx}{x^3}=\frac{3dx}{x}=\frac{3(0.4)}{30}=\frac{1.2}{30}=0.04$

$\frac{dS}{S}=\frac{12xdx}{6x^2}=\frac{2dx}{x}=\frac{2(0.4)}{30}=\frac{0.8}{30}=0.0266$

The percentage errors are relative errors times 100%

$\frac{dV}{V}100\%=(0.04)100\%=4\%$

$\frac{dS}{S}100\%=(0.0266)100\%=2.66\%$