Q 126. - Precalculus Enhanced with Graphing Utilities

Question

Piston Engines In a certain piston engine, the distance x (in cm) from the center of the drive shaft to the head of the piston is given by the function.

$x\left(\theta \right)=\cos \theta +\sqrt{16+0·5 \cos \left(2\theta \right)}$

where $\theta$ is the angle between the crank and the path of the piston head. See the figure. Find $x$ when $\theta =30°$ and $\theta =45°$.

Step-by-Step Solution

Answer

The value of $x$ when $\theta =30°$ is $\left(\sqrt{\frac{3}{2} +}\frac{\sqrt{65}}{2}\right)$ and value of x when $\theta =45°$ is $2\sqrt{2}$ .

Step 1. Given information .

Consider the given value of  $\theta =30°$and $\theta =45°$.

Step 2 .Finding x θ when θ = 30 ° .

$x\left(\theta \right)=\cos \theta +\sqrt{16+0·5 \cos \left(2\theta \right)}$

Substitute the value in given formula .

$$\begin{gathered} x\left(30°\right)=\cos 30°+\sqrt{16+0·5 \cos \left(2\times 30°\right)} \\ x\left(30°\right)=\frac{\sqrt{3}}{2}+\sqrt{16+\frac{1}{4}} \\ =\frac{\sqrt{3}}{2}+\frac{\sqrt{65}}{2} \end{gathered}$$

Step 3. Finding the value of x θ when θ = 45 ° .

Substitute the value in given formula $\theta =45°$

$$\begin{gathered} x\left(\theta \right)=\cos \theta +\sqrt{16+0·5 \cos \left(2\theta \right)} \\ x\left(45°\right)=\cos 45°+\sqrt{16+0·5 \cos 90°} \\ =\frac{1}{\sqrt{2}}+4 \\ =2\sqrt{2} \end{gathered}$$