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

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 their spring can be represented by the function g of theta equals 1 minus sine squared theta plus radical 3 period At what times are the springs from the original pogo stick and the toddler's pogo stick lengths equal? (5 points)

Answer 1

Hope this helps ^^

Explanation:

Part A: To determine the values where the pogo stick's spring will be equal to its non-compressed length, we set the function equal to zero:

f(θ) = 2cos(θ) + √3

To find the values of θ that satisfy this equation, we solve:

2cos(θ) + √3 = 0

Subtracting √3 from both sides:

2cos(θ) = -√3

Dividing by 2:

cos(θ) = -√3/2

Using the unit circle, we can find the angles where cosine is equal to -√3/2. These angles are π/6 and 11π/6.

Therefore, the values where the pogo stick's spring will be equal to its non-compressed length are θ = π/6 and 11π/6.

Part B: If the angle θ is doubled, that is, θ becomes 2θ, we substitute 2θ into the original function:

f(2θ) = 2cos(2θ) + √3

Using the double angle formula for cosine:

f(2θ) = 2(2cos²(θ) - 1) + √3

Expanding and simplifying:

f(2θ) = 4cos²(θ) - 2 + √3

Comparing this to the original function f(θ), we see that the new function has the same form but with different coefficients. The solutions for f(2θ) in the interval [0, 2π) will be the same as the solutions for f(θ), but they will occur at double the angles. In other words, if θ is a solution for f(θ), then 2θ will be a solution for f(2θ).

Part C: To find the times when the lengths of the springs from the original pogo stick and the toddler's pogo stick are equal, we set the two functions equal to each other:

f(θ) = g(θ)

2cos(θ) + √3 = 1 - sin²(θ) + √3

Rearranging the equation:

sin²(θ) + 2cos(θ) = 0

Using the Pythagorean identity sin²(θ) = 1 - cos²(θ), we substitute:

1 - cos²(θ) + 2cos(θ) = 0

Rearranging and simplifying:

cos²(θ) - 2cos(θ) + 1 = 0

Factoring:

(cos(θ) - 1)² = 0

Taking the square root:

cos(θ) - 1 = 0

cos(θ) = 1

This occurs when θ is a multiple of 2π.

Therefore, the lengths of the springs from the original pogo stick and the toddler's pogo stick are equal at θ = 2πn, where n is an integer.