Chapter 6.2, Problem 56E

a)

To determine

To find: The relationship between $d\&\theta$ .

Answer to Problem 56E

The required equation is $d^2=149-140\cos \theta$

Explanation of Solution

Given information: Three rollers of radii $3inches,4inches\&6inches$ and centre of the rollers are d meter apart.

Calculation: Substitute the measures of the triangle given in the diagram into the law of cosines.

  $\begin{array}{l}{d}^2={\left(3+4\right)}^2+{\left(6+4\right)}^2-2\left(3+4\right)\left(6+4\right)\cos \theta \\ d^2=49+100-140\cos \theta \\ d^2=149-140\cos \theta \end{array}$

Therefore, the required equation is $d^2=149-140\cos \theta$ .

b)

To determine

To write: $\theta$ as a function of d.

Answer to Problem 56E

The required function is $\theta =\arccos \left(\frac{-d^2+149}{140}\right)$ .

Explanation of Solution

Given information: The obtained equation of between $d\&\theta$ is $d^2=149-140\cos \theta$

Calculation: Solve for theta,

  $\begin{array}{l}{d}^2=149-140\cos \theta \\ \cos \theta =\frac{-d^2+149}{140}\\ \theta =\arccos \left(\frac{-d^2+149}{140}\right)\end{array}$

Therefore, the required function is $\theta =\arccos \left(\frac{-d^2+149}{140}\right)$ .

c)

To determine

To write: s as a function of $\theta$ .

Answer to Problem 56E

The required function is $s=\frac{\pi \left(360-\theta \right)}{45}$ .

Explanation of Solution

Given information: $\theta =\arccos \left(\frac{-d^2+149}{140}\right)$

Calculation: Solve for theta,

  $\begin{array}{l}\theta =\arccos \left(\frac{-d^2+149}{140}\right)\\ s=\frac{\left(360-\theta \right)}{360}\left(2\pi 4\right)\\ s=\frac{\pi \left(360-\theta \right)}{45}\end{array}$

Therefore, the required function is $s=\frac{\pi \left(360-\theta \right)}{45}$ .

d)

To determine

To fill: To complete the given table.

Answer to Problem 56E

  $\begin{array}{cccccccc}d\left(\text{inches}\right)& 9& 10& 12& 13& 14& 15& 16\\ \theta \left(\text{degrees}\right)& 60.94& 69.51& 87.95& 98.21& 109.21& 122.88& 139.84\\ s\left(\text{inches}\right)& 20.88& 20.28& 18.99& 18.28& 17.48& 16.55& 15.37\end{array}$

Explanation of Solution

Given information: $\theta =\arccos \left(\frac{-d^2+149}{140}\right)$

Get the value of $\theta$ by substituting the value d in the function $\theta =\arccos \left(\frac{-d^2+149}{140}\right)$ .

Then plug the values of $\theta$ into function $s=\frac{\pi \left(360-\theta \right)}{45}$ to get s.

The required table has been filled with the help above method.

  $\begin{array}{cccccccc}d\left(\text{inches}\right)& 9& 10& 12& 13& 14& 15& 16\\ \theta \left(\text{degrees}\right)& 60.94& 69.51& 87.95& 98.21& 109.21& 122.88& 139.84\\ s\left(\text{inches}\right)& 20.88& 20.28& 18.99& 18.28& 17.48& 16.55& 15.37\end{array}$