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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 sine theta plus radical 2 period

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 θ became 2θ, what are the solutions in the interval [0, 2π)? How do these compare to the original function? (5 points)

Part C: A toddler is jumping on another pogo stick whose length of its spring can be represented by the function g of theta equals 1 minus cosine squared theta plus radical 2 period At what times are the springs from the original pogo stick and the toddler's pogo stick lengths equal? (5 points)

The function f(θ) = 2sin(θ) + √2 describes the difference in length of the spring from its non-compressed length. This means f(θ) gives the amount by which the spring is either compressed or extended compared to its natural (non-compressed) length.

The spring is at its non-compressed length when this difference is zero, which corresponds to f(θ) = 0. Therefore, we need to solve this equation:

[tex]f(\theta) = 0\\\\\\2\sin(\theta) + \sqrt{2} = 0 \\\\\\2\sin(\theta) = -\sqrt{2}\\\\\\\sin(\theta) = -\dfrac{\sqrt{2}}{2}[/tex]

According to the unit circle, sin(θ) = -√2 / 2 when the values of θ are:

[tex]\theta = \dfrac{5\pi}{4}, \dfrac{7\pi}{4}[/tex]

So, all values where the pogo stick's spring will be equal to its non-compressed length are:

[tex]\theta = \dfrac{5\pi}{4}+2\pi n\; \textsf{ and }\; \theta= \dfrac{7\pi}{4}+2\pi n[/tex]

where n is any integer.

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Part B

If the angle is doubled, the function becomes f(2θ) = 2sin(2θ) + √2, so we need to find the values of θ where f(2θ) = 0:

[tex]f(2\theta) = 0\\\\\\2\sin(2\theta) + \sqrt{2} = 0 \\\\\\2\sin(2\theta) = -\sqrt{2}\\\\\\\sin(2\theta) = -\dfrac{\sqrt{2}}{2}[/tex]

The values of 2θ when sin(2θ) = -√2 / 2 are:

[tex]2\theta = \dfrac{5\pi}{4}+2\pi n\; \textsf{ and }\; 2\theta= \dfrac{7\pi}{4}+2\pi n[/tex]

Dividing each by 2, we get the values of θ:

[tex]\theta = \dfrac{5\pi}{8}+\pi n,\;\theta= \dfrac{7\pi}{8}+\pi n[/tex]

So, when the angle is doubled, the solutions in the interval [0, 2π) are:

[tex]\theta = \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8}[/tex]

In practical terms for the pogo stick scenario, f(θ) represents the spring's length when the teenager jumps at a certain frequency, while f(2θ) represents the spring's length when the teenager jumps at twice that frequency. Since the period in the second function is halved, this means there will be twice as many solutions in any given interval compared to those of the first function. Additionally, the difference in length of the spring in the second function is half that in the first function.

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Part C

The function for the toddler's pogo stick is g(θ) = 1 - cos²θ + √2.

To determine when the lengths of the springs on both pogo sticks are equal, set the two functions equal and solve for θ.

[tex]f(\theta) = g(\theta)\\\\\\2\sin(\theta) + \sqrt{2}= 1 - \cos^2(\theta) + \sqrt{2} \\\\\\ 2\sin(\theta)=1 - \cos^2(\theta)[/tex]

Using the Pythagorean identity, sin²θ + cos²θ = 1, we can substitute sin²θ = 1 - cos²θ:

[tex]2\sin (\theta)=\sin^2(\theta)[/tex]

Solve for θ:

[tex]\sin^2(\theta)-2\sin (\theta) = 0 \\\\\\ \sin(\theta)(\sin(\theta)-2)=0[/tex]

Therefore, we have 2 equations to solve.

Equation 1:

[tex]\sin (\theta)=0 \\\\\\\theta = 2\pi n, \theta = \pi + 2\pi n[/tex]

Equation 2:

[tex]\sin(\theta)-2=0 \\\\\\ \sin(\theta)=2 \\\\\\ \theta\textsf{ is unde\:\!fined as $-1\leq \sin(\theta)\leq 1$.}[/tex]

Therefore, the values of θ in the interval [0, 2π) are:

[tex]\theta = 0, \pi, 2\pi[/tex]

So, the springs from the original pogo stick and the toddler's pogo stick are of equal length at:

[tex]\theta = 0, \pi, 2\pi[/tex]