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 period 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?

Answer 1

Part A.

The length of the non-compressed spring can be determined by setting

$\theta=0.$

Now, evaluating the given function at θ=0, we get:

$f(0)=2\cos (0)+\sqrt[]{3}.$

Recall that:

$\cos 0=1.$

Therefore:

$f(0)=2\cdot1+\sqrt[]{3}=2+\sqrt[]{3.}$

Now, to determine all the possible times at which the length will be equal to the non-compressed length, we set the following equation:

$2\cos \theta+\sqrt[]{3}=2+\sqrt[]{3}.$

Solving the above equation for θ, we get:

$\begin{gathered} 2\cos \theta=2, \\ \cos \theta=1. \end{gathered}$

The solutions to the above equation are:

$\theta=2n\pi,\text{ where n is a natural number or zero.}$

Answer part A:

$\theta=2n\pi,\text{ where n is a natural number or zero.}$

Part B: If we set θ to the double of itself, then, the solutions to the following equation

$f(2\theta)=2+\sqrt[]{3},$

are:

$2\theta=2n\pi,$

therefore:

$\theta=n\pi,\text{ where n is a whole number.}$

If

$\theta\in\lbrack0,2\pi),$

then

$\theta=0\text{ or }\theta=\pi.$

Answer part B: The difference is that the period is half of the original period.

$\theta=0\text{ or }\theta=\pi.$