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 6 seconds can be described by the function#(6) f(theta)= 2cos(theta) + sqrt(3).Part 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 theta became 2theta what are the solutions in the interval [0, 211)? How do these compare to the original function? (5 points)Part C: A toddler is jumping on another pogo stick whose length of their spring can be represented by the function g(8) = 1-

Answer 1

$\begin{gathered} A)\theta=\frac{\pi}{2}+2\pi n,\:θ=\frac{3\pi}{2}+2\pi n \\ B)\theta=\frac{5\pi}{12},\frac{7\pi}{12},\frac{17\pi}{12},\frac{19\pi}{12} \end{gathered}$

A) Let's begin by writing out the equation for the given condition:

$\begin{gathered} f(\theta)=2\cos(\theta)+\sqrt{3} \\ 2\cos(\theta)+\sqrt{3}=\sqrt{3} \\ 2\cos(\theta)=0 \\ θ=\frac{\pi }{2}+2\pi n,\:θ=\frac{3\pi }{2}+2\pi n \end{gathered}$

Notice that in part A, we're determining all values in which the pogo stick is equal to its non-compressed length.

B) This part consists in solving the following equation:

$\begin{gathered} 2\cos \left(2θ\right)+\sqrt{3}=0,\:0\le \:θ<2\pi \\ 2\cos \left(2θ\right)+\sqrt{3}-\sqrt{3}=0-\sqrt{3} \\ 2\cos \left(2θ\right)=-\sqrt{3} \\ \frac{2\cos \left(2θ\right)}{2}=\frac{-\sqrt{3}}{2} \\ \cos \left(2θ\right)=-\frac{\sqrt{3}}{2} \\ 2θ=\frac{5\pi }{6}+2\pi n \\ \frac{2θ}{2}=\frac{\frac{5\pi }{6}}{2}+\frac{2\pi n}{2} \\ θ=\frac{5\pi }{12}+\pi n \\ \\ 2θ=\frac{7\pi }{6}+2\pi n \\ \frac{2θ}{2}=\frac{\frac{7\pi }{6}}{2}+\frac{2\pi n}{2} \\ θ=\frac{7\pi }{12}+\pi n \\ \theta=\frac{5\pi}{12},\:θ=\frac{7\pi}{12},\:θ=\frac{17\pi}{12},\:θ=\frac{19\pi}{12} \end{gathered}$

Note that since the pogo has a periodical movement and an interval was defined then the solutions above are defined for the interval.