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.

Part A: Determine all values where the pogo stick's spring will be equal to its non-compressed length.

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?

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?

Answer 1

The given function for the difference in length is presented as follows;

`f( \theta) = 2 \cdot cos( \theta) + √(3)`

When the pogo stick will be equal to its non compressed length, the difference is zero, therefore;

`f( \theta) = 2 \cdot cos( \theta) + √(3) = 0`

`2 \cdot cos( \theta) = - √(3)`

`\theta= arccos \left( ( - √(3) )/(2) \right)`

Which gives;

`\theta = (12 \cdot \pi \cdot n1 + 5\cdot \pi)/(6)`

`\theta = -(12 \cdot \pi \cdot n1 + 5\cdot \pi)/(6)`

Part B; If the angle was doubled, we have;

`f( \theta) = 2 \cdot cos(2 \cdot \theta) + √(3) = 0`

Therefore;

`2 \cdot cos(2 \cdot \theta) = - √(3)`

Which gives;

`\theta = (12 \cdot \pi \cdot n1 + 5\cdot \pi)/(12)`

`\theta = -(12 \cdot \pi \cdot n1 + 5\cdot \pi)/(12)`

Between 0 and 2•π, we have;

`\theta = (5\cdot \pi)/(12)`

`\theta = ( 17\cdot \pi)/(12)`

Part C;

The toddler's pogo stick is presented as follows;

`g( \theta) = 1- son^2( \theta) + √(3)`

Integrating the original function between 0 and theta gives;

`2 \cdot sin( \theta) + √(3)\cdot \theta`

The original length =

Therefore, when the lengths are equal, we have;

`1- son^2( \theta) + √(3) = 2 \cdot sin( \theta) + √(3)\cdot \theta`