Question

The difference in length of a spring on a pogo stick from its non-compressed length when a teenager is jumping on it after θ seconds can be described by the function f of theta equals 2 times cosine theta plus radical 3 periodPart A: Determine all values where the pogo stick's spring will be equal to its non-compressed length. (5 points)Part B: If the angle was doubled, that is θ became 2θ, what are the solutions in the interval [0, 2π)? How do these compare to the original function? (5 points)

Answer 1

Part A)

Since the function f represents the difference in length of the spring from its non-compressed position, then, when the spring is not compressed, the value of f(θ) is equal to 0:

`f(\theta)=0`

Replace the expression for f(θ) and solve for θ:

`\begin{gathered} 2\cos (\theta)+\sqrt[]{3}=0 \\ \Rightarrow2\cos (\theta)=-\sqrt[]{3} \\ \Rightarrow\cos (\theta)=-\frac{\sqrt[]{3}}{2} \\ \Rightarrow\theta=\arccos (-\frac{\sqrt[]{3}}{2}) \\ \\ \therefore\theta=\pm(5)/(6)\pi+2\pi k;k\in\Z \end{gathered}`

Part B)

Replace θ=2θ and find the solutions in the given interval:

`\begin{gathered} \Rightarrow2\theta=\pm(5)/(6)\pi+2\pi k \\ \Rightarrow\theta=\pm(5)/(12)\pi+\pi k \end{gathered}`

For k=0, the positive solution lies in the interval [0,2π):

`\theta_1=(5)/(12)\pi`

For k=1, both solutions lie in the interval [0,2π):

`\begin{gathered} \theta_2=-(5)/(12)\pi+\pi=(7)/(12)\pi \\ \theta_3=(5)/(12)\pi+\pi=(17)/(12)\pi \end{gathered}`

For f=2, the negative solution lies in the interval [0,2π):

`\theta_4=-(5)/(12)\pi+2\pi=(19)/(12)\pi`

Then, the solutions in the interval [0,2π) are:

`\mleft\lbrace(5)/(12)\pi,(7)/(12)\pi,(17)/(12)\pi,(19)/(12)\pi\mright\rbrace`