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%

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

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

Answer 2

The maximum possible error, relative error, and percentage error are;

A) Maximum error = 1080 cm³

Relative error = 0.04

Percentage error = 4%

B) Maximum error = 1440 cm³

Relative error = 0.0267

Percentage error = 2.67%

We are told that;

Edge of cube; x = 30 cm

Possible error; dx = 0.4cm

All sides of a cube are equal. Thus, volume of cube is;

V = x³

Meanwhile, surface area is;

S = 6x²

1) For volume;

We will start by differentiating the volume equation.

dV/dx = 3x²

Thus; dV = 3x²dx

Plugging in relevant values, we have;

dV = 3(30²) × 0.4

Maximum error; dV = 3 × 900 × 0.4

Maximum error = 1080 cm³

Relative error is gotten by dividing the change in volume by the volume likewise change in length.

Thus; Relative error = dV/V = dx/x

Relative error = (3x²dx)/x³

Relative error = (3/x)dx

Relative error = (3/30) × 0.4

Relative error = 0.04

Percentage error is just converting the relative error to percentage. Thus;

%error = 0.04 × 100% = 4%

2) We will start by differentiating the surface area equation.

dS/dx = 12x

Thus; dS = 12x dx

Plugging in relevant values, we have;

dS = 12(30) × 0.4

Maximum error; dS = 1440 cm²

Relative error is gotten by dividing the change in volume by the volume likewise change in length.

Thus; Relative error = dS/S = dx/x

Relative error = (12x dx)/6x²

Relative error = (2/x)dx

Relative error = (2/30) × 0.4

Relative error = 0.0267

Percentage error is just converting the relative error to percentage. Thus;

%error = 0.0267 × 100% = 2.67%

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