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

[fluffertoes]

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) + 55b. 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?

 

[MarkFL]

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.)

\(\displaystyle \left(x(36),y(36)\right)=\left(-50\sin\left(\frac{\pi}{5}18\right),-50\cos\left(\frac{\pi}{5}18\right)+55\right)=\left(25\sqrt{\frac{1}{2}(5+\sqrt{5})},55-\frac{25}{2}(\sqrt{5}-1)\right)\approx(47.55,39.55)\)

 

[fluffertoes]

Why is cos negative in your equations??

 

[MarkFL]

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:

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

 

[fluffertoes]

So would it always be negative for other ferris wheel problems?

 

[MarkFL]

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:

\(\displaystyle \theta=-\frac{2\pi}{20}t-\frac{\pi}{2}\)

 

[JorgeLuna]

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

 

[MarkFL]

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)

 

[JorgeLuna]

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

 

[MarkFL]

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