Ferris Wheel Problem: Find Donna's Position After 36s

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In summary, Donna is moving up in an inverted cosine curve as the Ferris wheel moves in a counter-clockwise direction.
  • #1
fluffertoes
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HELP!

Donna is riding a 100 foot diameter Ferris Wheel with a center located 55 feet above the ground. Assume the center of the ferris wheel is on the y-axis, and that the ferris wheel turns 1 revolution every 20 seconds in the clockwise direction.

a. Write parametric equations to model Donna's motion at any time if she is at the bottom of the wheel at time t=0.

So here, I found my two equations, please check to see if they are correct!
x(t)= 50Sin(pi*t)/10)
y(t)= 50Cos(pi*t)/10) + 55
b. What would Donna's position (x,y) be after 36 seconds?

I inputted 36 as t in the above equations and solved. I ended up with (-47.553, 70.451). Are both of my answers here (the equations and the coordinates at the time t=36) correct?
 
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  • #2
This is how I would answer the questions (with the parameter time $t$ measures in seconds and $x(t),\,y(t)$ measured in ft.):

a).

\(\displaystyle x(t)=50\cos\left(-\frac{2\pi}{20}t-\frac{\pi}{2}\right)=-50\sin\left(\frac{\pi}{10}t\right)\)

\(\displaystyle y(t)=50\sin\left(-\frac{2\pi}{20}t-\frac{\pi}{2}\right)+55=-50\cos\left(\frac{\pi}{10}t\right)+55\)

b.)

\(\displaystyle \left(x(36),y(36)\right)=\left(-50\sin\left(\frac{\pi}{5}18\right),-50\cos\left(\frac{\pi}{5}18\right)+55\right)=\left(25\sqrt{\frac{1}{2}(5+\sqrt{5})},55-\frac{25}{2}(\sqrt{5}-1)\right)\approx(47.55,39.55)\)
 
  • #3
Why is cos negative in your equations??
 
  • #4
fluffertoes said:
Why is cos negative in your equations??

Consider:

\(\displaystyle u(\theta)=\sin\left(-\theta-\frac{\pi}{2}\right)=\sin\left(-\left(\theta+\frac{\pi}{2}\right)\right)\)

Now, since sine is a odd function, this means $\sin(-x)=-\sin(x)$, and so we now have:

\(\displaystyle u(\theta)=-\sin\left(\theta+\frac{\pi}{2}\right)\)

Now, using the angle sum identity for sine:

\(\displaystyle \sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)\)

We have:

\(\displaystyle u(\theta)=-\left(\sin(\theta)\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\cos(\theta)\right)=-\left(\sin(\theta)\cdot0+1\cdot\cos(\theta)\right)=-\cos(\theta)\)

Another way to look at it is to again consider:

\(\displaystyle u(\theta)=\sin\left(-\theta-\frac{\pi}{2}\right)=\sin\left(-\left(\frac{\pi}{2}-(-\theta)\right)\right)\)

Using the fact that sine is odd, we have:

\(\displaystyle u(\theta)=-\sin\left(\frac{\pi}{2}-(-\theta)\right)\)

Using a co-function identity $\sin\left(\dfrac{\pi}{2}-x\right)=\cos(x)$, we obtain:

\(\displaystyle u(\theta)=-\cos(-\theta)\)

Using the fact that cosine is an even function (i.e. $\cos(-x)=\cos(x)$), we have:

\(\displaystyle u(\theta)=-\cos(\theta)\)
 
  • #5
So would it always be negative for other ferris wheel problems?
 
  • #6
fluffertoes said:
So would it always be negative for other ferris wheel problems?

Consider that the way we have our coordinate axes oriented, the rider begins on the $y$-axis. Now, as the wheel begins moving in a clockwise direction, the rider's height above the ground increases and after an angular displacement of \(\displaystyle \frac{\pi}{2}\) radians, the rider is at the same level as the hub of the wheel, and after a total angular displacement of $\pi$ radians, the rider is at the maximum height. After that, the rider's height begins to decrease. This is how a negative cosine function behaves...does that make sense?

If we think about the unit circle, the rider begins with an angular displacement of \(\displaystyle -\frac{\pi}{2}\) radians (the bottom of the circle), and when the wheel moves in the clockwise position, the angular motion is in the negative direction...that's why I used as the argument for the trig. functions:

\(\displaystyle \theta=-\frac{2\pi}{20}t-\frac{\pi}{2}\)
 
  • #7
Donna is riding a 100 foot diameter Ferris Wheel with a center located 55 feet above the ground. Assume the center of the ferris wheel is on the y-axis, and that the ferris wheel turns 1 revolution every 20 seconds in the clockwise direction.

a. Write parametric equations to model Donna's motion at any time if she is at the bottom of the wheel at time t=0.

often times it helps to first make a sketch of the individual position graphs by plotting a few points

for x as a function of time, x(0) = 0, x(5) = 50, x(10) = 0, x(15) = -50, and x(20) = 0

plot those points on a graph of position vs time and note the points define a sine curve with equation x = 50sin(pi/10*t)do the same for y as a function of time, y(0) = 5, y(5) = 55, y(10) = 105, y(15) = 55, and y(20) = 5

plot those points on a graph of position vs time and note the points define an "inverted" cosine curve with an upward shift , equation y = -50cos(pi/10*t) + 55
 

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  • #8
JorgeLuna said:
often times it helps to first make a sketch of the individual position graphs by plotting a few points

for x as a function of time, x(0) = 0, x(5) = 50, x(10) = 0, x(15) = -50, and x(20) = 0

plot those points on a graph of position vs time and note the points define a sine curve with equation x = 50sin(pi/10*t)

You have the Ferris wheel moving in a counter-clockwise direction. (Shake)
 
  • #9
MarkFL said:
You have the Ferris wheel moving in a counter-clockwise direction. (Shake)

Oops ... well, the y position curve is still good. Just reflect the x position curve over the horizontal axis making A = -50
 
  • #10
JorgeLuna said:
Oops ... well, the y position curve is still good. Just reflect the x position curve over the horizontal axis making A = -50

That doesn't take away from your good advice of plotting points to see what the sinusoidal parametric curves are going to look like. :D
 

Related to Ferris Wheel Problem: Find Donna's Position After 36s

1. What is the Ferris Wheel Problem?

The Ferris Wheel Problem is a mathematical problem that involves determining the position of a person on a Ferris wheel after a certain amount of time has passed.

2. How is the problem solved?

The problem is solved using trigonometric functions, specifically the sine and cosine functions, to calculate the vertical and horizontal components of the person's position on the wheel.

3. What information is needed to solve the problem?

To solve the Ferris Wheel Problem, you need to know the radius of the wheel, the initial position of the person, the angular velocity of the wheel, and the time at which the position is being calculated.

4. How do you account for the different speeds of the Ferris wheel?

The different speeds of the Ferris wheel are accounted for by using the angular velocity, which is measured in radians per second. This allows for a more accurate calculation of the person's position at any given time.

5. Are there any real-life applications of this problem?

Yes, the Ferris Wheel Problem has many real-life applications, such as predicting the position of satellites in orbit, determining the position of a person on a rotating platform, and calculating the trajectory of a moving object in a circular motion.

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