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

  • Context: MHB 
  • Thread starter Thread starter Amathproblem22
  • Start date Start date
  • Tags Tags
    Addition Sinusoids
Click For Summary
SUMMARY

The discussion focuses on determining the times when the combined heights of two Ferris wheels, modeled by the equations F1 and F2, reach 27.5 meters. The height equations are F1: $$h=-12\cos\frac{\pi}{10}t+12.5$$ and F2: $$h=-12\cos\frac{\pi}{30}t+15$$. By equating the sum of these height functions to 27.5, the equation $$\cos\left(\frac{\pi}{10}t\right)+\cos\left(\frac{\pi}{30}t\right)=0$$ is derived. The solution involves applying a triple-angle identity and factoring to find the values of t that satisfy the conditions.

PREREQUISITES
  • Understanding of trigonometric functions and identities
  • Familiarity with cosine equations and their properties
  • Basic knowledge of algebraic manipulation and factoring
  • Ability to solve equations involving angles and periodic functions
NEXT STEPS
  • Study the application of triple-angle identities in trigonometry
  • Learn how to solve trigonometric equations involving multiple angles
  • Explore the behavior of periodic functions and their graphical representations
  • Investigate the implications of cosine functions in real-world applications, such as modeling circular motion
USEFUL FOR

Mathematicians, physics students, engineers, and anyone interested in solving trigonometric equations related to periodic functions and real-world applications like amusement park rides.

Amathproblem22
Messages
12
Reaction score
0
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?
 
Physics news on Phys.org
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:

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

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

Similar threads

  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 9 ·
Replies
9
Views
13K
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
12K