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?

The difference in length of a spring on a pogo stick from its noncompressed length when a teenager is jumping on it after θ seconds can be described by the func class=

Respuesta :

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

[tex]f( \theta) = 2 \cdot cos( \theta) + \sqrt{3} [/tex]

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

[tex]f( \theta) = 2 \cdot cos( \theta) + \sqrt{3} = 0[/tex]

[tex] 2 \cdot cos( \theta) = - \sqrt{3} [/tex]

[tex]\theta= arccos \left( \frac{ - \sqrt{3} }{2} \right) [/tex]

Which gives;

[tex] \theta = \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]

[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]

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

[tex]f( \theta) = 2 \cdot cos(2 \cdot \theta) + \sqrt{3} = 0[/tex]

Therefore;

[tex] 2 \cdot cos(2 \cdot \theta) = - \sqrt{3} [/tex]

Which gives;

[tex] \theta = \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]

[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]

Between 0 and 2•π, we have;

[tex] \theta = \frac{5\cdot \pi}{12} [/tex]

[tex] \theta = \frac{ 17\cdot \pi}{12} [/tex]

Part C;

The toddler's pogo stick is presented as follows;

[tex]g( \theta) = 1- son^2( \theta) + \sqrt{3} [/tex]

Integrating the original function between 0 and theta gives;

[tex]2 \cdot sin( \theta) + \sqrt{3}\cdot \theta [/tex]

The original length =

Therefore, when the lengths are equal, we have;

[tex]1- son^2( \theta) + \sqrt{3} = 2 \cdot sin( \theta) + \sqrt{3}\cdot \theta [/tex]