When do the combined heights of two Ferris wheels reach 27.5m?

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Discussion Overview

The discussion revolves around determining the times at which the combined heights of two Ferris wheels reach 27.5 meters. Participants explore the mathematical modeling of the heights of the Ferris wheels using trigonometric equations and seek to solve for the time variable.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the equations for the heights of two Ferris wheels, F1 and F2, and asks how to find when their combined height equals 27.5 meters.
  • Another participant suggests equating the sum of the height functions to 27.5, leading to a cosine equation that involves a substitution for simplification.
  • Further elaboration on the cosine equation is provided, including the application of a triple-angle identity and the factorization of the resulting expression.
  • Subsequent posts detail the solutions for the angles where the cosine function equals zero and where it equals ±√(1/2), along with the corresponding time calculations.

Areas of Agreement / Disagreement

Participants appear to agree on the mathematical approach to solving the problem, but there is no consensus on the final interpretation or implications of the solutions derived.

Contextual Notes

The discussion includes various mathematical steps and identities, but some assumptions and dependencies on definitions are not fully articulated, leaving certain aspects unresolved.

Amathproblem22
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So I have figured out two equations model a Ferris wheel ride(two different ferris wheels).
F1 = Ferris wheel one
F2= Ferris wheel two

F1, $$h=-12\cos\frac{\pi}{10}t+12.5$$

F2, $$h=-12\cos\frac{\pi}{30}t+15$$

Now from these two equations, I want to know when F2+F1= 27.5m i.e I want to find the times when the combined height of each Ferris wheels seat adds up to 27.5m.
How would I go about doing this?
 
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If we equate sum of the two height functions to 27.5, we ultimately obtain:

$$\cos\left(\frac{\pi}{10}t\right)+\cos\left(\frac{\pi}{30}t\right)=0$$

Let's let:

$$\theta=\frac{\pi}{30}t$$

And we may write:

$$\cos(3\theta)+\cos(\theta)=0$$

Applying a triple-angle identity for cosine, we ultimately get:

$$2\cos^3(\theta)-\cos(\theta)=0$$

Factor:

$$\cos(\theta)(2\cos^2(\theta)-1)=0$$

Can you proceed?
 
MarkFL said:
If we equate sum of the two height functions to 27.5, we ultimately obtain:

$$\cos\left(\frac{\pi}{10}t\right)+\cos\left(\frac{\pi}{30}t\right)=0$$

Let's let:

$$\theta=\frac{\pi}{30}t$$

And we may write:

$$\cos(3\theta)+\cos(\theta)=0$$

Applying a triple-angle identity for cosine, we ultimately get:

$$2\cos^3(\theta)-\cos(\theta)=0$$

Factor:

$$\cos(\theta)(2\cos^2(\theta)-1)=0$$

Can you proceed?
\[ \cos \left(θ\right)=0\quad \mathrm{or}\quad \:2\cos ^2\left(θ\right)-1=0 \]

\[ \cos \left(θ\right)=0: θ=\frac{\pi }{2}+2\pi n,\:θ=\frac{3\pi }{2}+2\pi n \]

\[ 2\cos ^2\left(θ\right)-1=0: θ=\arccos \left(\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=2\pi -\arccos \left(\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=\arccos \left(-\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=-\arccos \left(-\sqrt{\frac{1}{2}}\right)+2\pi n \] ?
 
Yes, for:

$$\cos(\theta)=0$$

This implies (where \(k\in\mathbb{Z}\)):

$$\theta=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2k+1)$$

$$t=\frac{30}{\pi}\theta=15(2k+1)$$

And for:

$$\cos(\theta)=\pm\frac{1}{\sqrt{2}}$$

This implies:

$$\theta=\frac{\pi}{4}+\frac{\pi}{2}k=\frac{\pi}{4}(2k+1)$$

$$t=\frac{30}{\pi}\theta=\frac{15}{2}(2k+1)$$
 

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