Chapter 8.8, Problem 30E

(a)

To determine

To Find: The expression for the linear and quadratic approximations.

Answer to Problem 30E

The quadratic approximation is ${\rho }_2\left(t\right)={\rho }_{20}+\alpha {\rho }_{20}\left(t-20\right)+\frac{1}{2}{\alpha }^2\rho \left(20\right){\left(t-20\right)}^2$ and the linear approximation is ${\rho }_1\left(t\right)={\rho }_{20}+\alpha {\rho }_{20}\left(t-20\right)$ .

Explanation of Solution

Given:

The given expression for the resistivity is $\rho \left(t\right)={\rho }_{20}{e}^{\alpha \left(t-20\right)}$ .

The temperature coefficient is $\alpha$ and $t$ is the temperature in degree Celsius.

Calculation:

Consider the Taylor series of $f\left(x\right)$ at $x=a$ .

  $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(a\right)}{n!}}{\left(x-a\right)}^n$

Consider the expression for the resistivity is $\rho \left(t\right)={\rho }_{20}{e}^{\alpha \left(t-20\right)}$ .

  ${\rho }^{\prime }\left(t\right)=\alpha {\rho }_{20}{e}^{\alpha \left(t-20\right)}$

Then, the linear approximation is $T_1\left(1\right)={\rho }_1\left(t\right)$ that is,

  $\begin{array}{c}{\rho }_1\left(t\right)={\rho }_{20}{e}^{\alpha \left(20-20\right)}+\left(t-20\right){\rho }^{\prime }\left(20\right)\\ ={\rho }_{20}+\left(t-20\right)\left[\alpha {\rho }_{20}{e}^{\alpha \left(20-20\right)}\right]\\ ={\rho }_{20}+\alpha {\rho }_{20}\left(t-20\right)\end{array}$

Consider the quadratic approximation is $T_2\left(t\right)={\rho }_2\left(t\right)$ that is,

  $\begin{array}{c}{\rho }_2\left(t\right)={\rho }_{20}{e}^{\alpha \left(20-20\right)}+\left(t-20\right){\rho }^{\prime }\left(20\right)+{\left(t-20\right)}^2\frac{1}{2}{\rho }^{″}\left(20\right)\\ ={\rho }_{20}+\left(t-20\right)\left[\alpha {\rho }_{20}{e}^{\alpha \left(20-20\right)}\right]+\frac{1}{2}{\left(t-20\right)}^2\left[{\alpha }^2{{\rho }^{\prime }}_{20}{e}^{\alpha \left(20-20\right)}\right]\\ ={\rho }_{20}+\alpha {\rho }_{20}\left(t-20\right)+\frac{1}{2}{\alpha }^2\rho \left(20\right){\left(t-20\right)}^2\end{array}$

The quadratic approximation is ${\rho }_2\left(t\right)={\rho }_{20}+\alpha {\rho }_{20}\left(t-20\right)+\frac{1}{2}{\alpha }^2\rho \left(20\right){\left(t-20\right)}^2$ and the linear approximation is ${\rho }_1\left(t\right)={\rho }_{20}+\alpha {\rho }_{20}\left(t-20\right)$ .

(b)

To determine

To Find: The graph for the resistivity of the copper.

Answer to Problem 30E

The required graph is shown in Figure 2

Explanation of Solution

Given:

The thermal coefficient is $\alpha =\text{0}\text{.039}/\text{°C}$ and the resistivity is ${\rho }_{20}=1.7\times {10}^{-8}\Omega \cdot \text{m}$ .

Calculation:

Consider the linear approximation is,

  ${\rho }_1\left(t\right)={\rho }_{20}+\alpha {\rho }_{20}\left(t-20\right)$

Then,

  $\begin{array}{c}{\rho }_1\left(t\right)=1.7\times {10}^{-8}+0.0039\left(1.7\times {10}^{-8}\right)\left(t-20\right)\\ =1.7\times {10}^{-8}+6.63\times {10}^{-11}\left(t-20\right)\end{array}$

Consider the quadratic approximation is,

  ${\rho }_2\left(t\right)={\rho }_{20}+\alpha {\rho }_{20}\left(t-20\right)+\frac{1}{2}{\alpha }^2\rho \left(20\right){\left(t-20\right)}^2$

Then,

  $\begin{array}{c}{\rho }_2\left(t\right)=1.7\times {10}^{-8}+6.63\times {10}^{-11}\left(t-20\right)+\frac{1}{2}{\left(0.0039\right)}^2\left(1.7\times {10}^{-8}\right){\left(t-20\right)}^2\\ =1.7\times {10}^{-8}+6.63\times {10}^{-11}\left(t-20\right)+1.29285\times {10}^{-13}{\left(t-20\right)}^2\end{array}$

Enter the 3 functions after pressing the Y= button as shown in Figure 1

  

Figure 1

Then, the graph is shown in Figure 2

  

Figure 2

(c)

To determine

To Find: The values of $t$ that the linear approximation agree with the exponential expression to within one percent.

Answer to Problem 30E

The temperature range is $19.974°\text{C}\le t\le 22.564°\text{C}$ .

Explanation of Solution

Calculation:

The linear approximation is larger by 1 percent then,

  $\begin{array}{l}1.7\times {10}^{-8}+6.63\times {10}^{-11}\left(t-20\right)=\left(1.01\right)1.7\times {10}^{-8}\\ 6.63\times {10}^{-11}\left(t-20\right)=1.7\times {10}^{-10}\\ t=22.564°\text{C}\end{array}$

The case when the linear approximation is less than one percent then,

  $\begin{array}{l}1.7\times {10}^{-8}+6.63\times {10}^{-11}\left(t-20\right)=\left(0.99\right)1.7\times {10}^{-8}\\ 6.63\times {10}^{-11}\left(t-20\right)=\left(-0.01\right)1.7\times {10}^{-10}\\ t=19.974°\text{C}\end{array}$

Thus, the temperature range is $19.974°\text{C}\le t\le 22.564°\text{C}$ .