Chapter 1.2, Problem 86E

i.

To determine

To calculate: The resistance $y$ (in ohms) of 1000 feet of solid copper wire at 68 degrees Fahrenheit is $y=\frac{10370}{x^2}$ where $x$ is the diameter of the wire in mils (0.001 inches). Complete the table.

  $\begin{array}{cccccccccccc}x& 5& 10& 20& 30& 40& 50& 60& 70& 80& 90& 100\\ y& & & & & & & & & & & \end{array}$

Answer to Problem 86E

  $\begin{array}{cccccccccccc}x& 5& 10& 20& 30& 40& 50& 60& 70& 80& 90& 100\\ y& 414.80& 103.7& 25.95& 11.52& 6.48& 4.15& 2.88& 2.12& 1.62& 1.28& 1.04\end{array}$

Explanation of Solution

Given information: $y=\frac{10370}{x^2}$

Calculation: Put all given values of $x$ in $y=\frac{10370}{x^2}$

  $\begin{array}{cccccccccccc}X& 5& 10& 20& 30& 40& 50& 60& 70& 80& 90& 100\\ Y& \begin{array}{l}\frac{10370}{5^2}\\ =\frac{10370}{25}\\ =414.80\end{array}& \begin{array}{l}\frac{10370}{{10}^2}\\ =\frac{10370}{100}\\ =103.7\end{array}& \begin{array}{l}\frac{10370}{{20}^2}\\ =\frac{10370}{400}\\ =25.95\end{array}& \begin{array}{l}\frac{10370}{{30}^2}\\ =\frac{10370}{900}\\ =11.52\end{array}& \begin{array}{l}\frac{10370}{{40}^2}\\ =\frac{10370}{1600}\\ =6.48\end{array}& \begin{array}{l}\frac{10370}{{50}^2}\\ =\frac{10370}{2500}\\ =4.15\end{array}& \begin{array}{l}\frac{10370}{{60}^2}\\ =\frac{10370}{3600}\\ =2.88\end{array}& \begin{array}{l}\frac{10370}{{70}^2}\\ =\frac{10370}{4900}\\ =2.12\end{array}& \begin{array}{l}\frac{10370}{{80}^2}\\ =\frac{10370}{6400}\\ =1.62\end{array}& \begin{array}{l}\frac{10370}{{90}^2}\\ =\frac{10370}{8100}\\ =1.28\end{array}& \begin{array}{l}\frac{10370}{{100}^2}\\ =\frac{10370}{10000}\\ =1.04\end{array}\end{array}$

ii.

To determine

To graph: The resistance $y$ (in ohms) of 1000 feet of solid copper wire at 68 degrees Fahrenheit is $y=\frac{10370}{x^2}$ where $x$ is the diameter of the wire in mils (0.001 inches). Use the table of value in part (a) to sketch a graph of the model. Then use your graph to estimate the resistance when $x=85.5$

Answer to Problem 86E

Resistance is $y\approx 1.419$ ohms when $x=85.5$

Explanation of Solution

Given information: all the values of $y$ are calculated in above table corresponding to the given values of $x$ . Using these values we plot the graph. Points are $\left(5,414.8\right),\left(10,103.5\right),\left(20,25.92\right),\left(30,11.52\right),\left(40,6.48\right),\left(50,4.15\right),\left(60,2.88\right),\left(70,2.12\right),\left(80,1.62\right),$

  $\left(90,1.28\right),\left(100,1.04\right)$

  

From the above graph we can find the value of $y$ when $x=85.5$

  

here $y\approx 1.419$ ohms.

iii.

To determine

To calculate: The resistance $y$ (in ohms) of 1000 feet of solid copper wire at 68 degrees Fahrenheit is $y=\frac{10370}{x^2}$ where $x$ is the diameter of the wire in mils (0.001 inches). Use the model to confirm algebraically the estimate you found in part (b).

Answer to Problem 86E

  $y\approx 1.419$ ohms

Explanation of Solution

Given information: The given function is $y=\frac{10370}{x^2}$

Calculation: Putting $x=85.5$ in above function, we get

  $\begin{array}{l}y=\frac{10370}{{85.5}^2}=\frac{10370}{7310.25}\\ y=1.418556\text{ohms}\end{array}$

Conclusion:

iv.

To determine

To conclude: The resistance $y$ (in ohms) of 1000 feet of solid copper wire at 68 degrees Fahrenheit is $y=\frac{10370}{x^2}$ where $x$ is the diameter of the wire in mils (0.001 inches). What can you conclude about the relationship between the diameter of the copper wire and the resistance?

Explanation of Solution

From the above table and graph we can observe that as the diameter $x$ of the wire increases, the resistance $y$ of the wire decreases.