Chapter 3.9, Problem 27E

(a)

To determine

To estimate: The maximum possible error, relative error and percentage error when computing the volume of the cube.

Answer to Problem 27E

The maximum possible error is $dV=270{\text{ cm}}^3$, the relative error is 0.01 and the percentage error is $1\%$.

Explanation of Solution

Given:

The edge of the cube is 30 cm and possible error in measurement is 0.1 cm.

Calculation:

Let x be the length of the side of the cube.

The volume of the cube is $V=x^3$.

Consider the function $f\left(x\right)=x^3$

The differential is $dV=f^{\prime }\left(x\right)dx$ (1)

Derivative of the function $f\left(x\right)=x^3$ is $f^{\prime }\left(x\right)=3x^2$.

Substitute $f^{\prime }\left(x\right)=3x^2$ in equation (1),

$dV=3x^{2}dx$

Substitute $x=30\text{ and }dx=0.1$,

$\begin{array}{c}dV=3{\left(30\right)}^2\left(0.1\right)\\ =3\left(900\right)\left(0.1\right)\\ =2700\left(0.1\right)\\ =270\end{array}$

Thus, the maximum possible error in computing the volume of the cube is $dV=270\text{ c}{\text{.m}}^3$

Note that, the relative error is $\frac{\Delta V}{V}$ and $\Delta V$ is approximately equal to $dV$.

$\begin{array}{c}\frac{\Delta V}{V}\approx \frac{dV}{V}\\ =\frac{3x^{2}dx}{x^3}\\ =\frac{3dx}{x}\end{array}$

Substitute $x=30\text{ and }dx=0.1$,

$\begin{array}{c}\frac{\Delta V}{V}\approx \frac{3\left(0.1\right)}{30}\\ =0.01\end{array}$

Therefore, the relative error is 0.01.

Note that, the percentage error is the product of relative error and 100%.

The percentage error is computed as follows,

$\begin{array}{c}\text{Percentage error}=0.01\times 100\%\\ =1\%\end{array}$

Therefore, the percentage error is $1\%$.

(b)

To determine

To estimate: The maximum possible error, relative error and percentage error when computing the surface area of the cube.

Answer to Problem 27E

The maximum possible error is $dS=36{\text{ cm}}^2$, the relative error is $0.00\overline{6}$.and the percentage error is $0.\overline{6}\%$.

Explanation of Solution

Given:

The edge of the cube is 30 cm and possible error in measurement of 0.1 cm.

Calculation:

The surface of the cube is $S=6x^2$.

Consider the function $f\left(x\right)=6x^2$

Derivative of the function $f\left(x\right)=6x^2$ is $f^{\prime }\left(x\right)=12x$.

Substitute $f^{\prime }\left(x\right)=12x$ in the differential $dS=f^{\prime }\left(x\right)dx$,

$\begin{array}{c}dS=12xdx\\ =12\left(30\right)\left(0.1\right)\left(x=30\right)\\ =360\left(0.1\right)\\ =36\end{array}$

Therefore, the maximum possible error in computing the surface area of the cube is $dS=36{\text{ cm}}^2$

The relative error is computed by dividing the change of the surface area of the cube $\left(\Delta S\right)$ by the surface area of the cube S.

That is, the relative error is $\frac{\Delta S}{S}$.

Note that, the value $\Delta S$ is approximately equal to $dS$.

$\begin{array}{c}\frac{\Delta S}{S}\approx \frac{dS}{S}\\ =\frac{12xdx}{6x^2}\\ =\frac{2dx}{x}\end{array}$

Substitute $x=30\text{ and }dx=0.1$,

$\begin{array}{c}\frac{\Delta S}{S}\approx \frac{2\left(0.1\right)}{30}\\ =\frac{0.2}{30}\\ =0.00\overline{6}\end{array}$

Therefore, the relative error is $0.00\overline{6}$.

Note that, the percentage error is the product of relative error and 100%.

The percentage error is computed as follows,

$\begin{array}{c}\text{Percentage error}=0.00\overline{6}\times 100\%\\ =0.\overline{6}\%\end{array}$

Therefore, the percentage error is $0.\overline{6}\%$.