Help with parametric curve question

[amb1989]

Homework Statement

I am working on my calc 3 project where she has given two parametric equations, one models the motion of a Ferris wheel and the other models an object being thrown to a person on the Ferris wheel.

OK so I have this parametric equation that models an object being thrown and I'll try to type this as clearly as possible.
r1(t)= <22-8.03(t-t0), 1+11.47(t-t0)-4.9(t-t0)^2>

The second equation is for the Ferris wheel
r2(t)= <15sin((pi*t)/10),16-15cos((pi*t)/10)

the t0 are t subscript 0 just for clarification. So I have that and it is asking me to find the speed and angle of inclination of the object when t=t0

I also want to make it clear that this question has to do with the first equation I only brought the second one in for completeness and in case I'm wrong and I actually need that one too.

Homework Equations

just gave them

The Attempt at a Solution

I've tried to plug in t for t0 and obviously that turns to 0 and then take an integral to get the equation that models velocity and then I would have to get the magnitude of that. However when I do that I get t showing up and I have no value for t or anything.

I've also tried to take the integral of the equation before plugging in for t0 but I am not sure how I would go about doing that since I have two variables in there.

So all in all I'm stuck and would appreciate some help.

 

[Inferior89]

When you find the velocity you do a derivative not an integral.
To find the speed you need to find \mathbf{v}(t_0) = \dot{\mathbf{r}}(t_0) =( \dot{r_1}(t_0), \dot{r_2}(t_0))
The dot symbolize taking the time derivative.

Edit: r_2 (t_0) is not the motion for the Ferris wheel but the second component of the motion of the object being thrown.

The speed is then |\mathbf{v}(t_0)|.

To find the angle you know that A \cdot B = |A||B|\cos \theta
Here you can take A as your velocity vector and B as the x-axis.

 

[amb1989]

But how do I take the deriviative of this if t = t₀ then I'm left with all constants

 

[Inferior89]

Take the derivative before you put in t = t_0...

If I told you to find the derivative of x^3 at x = 2 would you put in x = 2 to get 8 and the take the derivative of 8 which is 0..
No, you would first differentiate to get 3x^2 then put in x = 2 to get 12.

Same here.

 

[amb1989]

Right I get what you're saying but this t0 is confusing me because I'm not sure what to do with it when I take the derivative. I feel like I should treat it like a constant so I go through and take the derivative of the whole thing and simplifying and I get

-8.03, 11.47-4.9t-9.8t0

plugging in t

-8.03, 11.47-4.9t-9.8t => -8.03, 11.47-14.7t

then to find the magnitude

[(8.03)^2+11.47^2+14.7^2)^1/2]

14+14.7t

and now I am stuck with t because I'm not given anything for t. Do I need to solve for t from the second equation, or maybe I could set it equal to 0 since it's t0...

 

[Inferior89]

You are supposed to turn every t to t_0 after you have differentiated.

Your derivative of the first component is correct but the derivative of
4.9(t-t0)^2 is 9.8(t-t0) just like the derivative of (x-c)^2 = 2(x-c).

If you feel like it take t_0 = 0 and substitute t = 0 after differentiating instead. Same thing.

 

[amb1989]

Oh right that would be chain rule duh. Sorry I feel like I'm still in an integration mind set and I foiled everything out then tried to take the derivative of that.

 

[amb1989]

Alright I think I got it now

I got -8.03, 11.47-9.8(t-t0) and plugging in t=t0 I get -8.03, 11.47

the magnitude of that is 14 so woo got an answer.

For the angle I did -8.03i+11.47j (dot) i+0j and that is -8.03.

Dividing by the magnitude of the vectors which is just 1*14 I get -0.5735 and the inverse cos of that is 125 degrees. Subtracting 125 from 180 and I get 55 degrees and I think that makes sense!

 

[amb1989]

Also if you know of anywhere online that I can go to to plug those equations into and get a simulation?

 

[Inferior89]

Looks good.

Here you can plugin the parametric equation. I did the throwing formula for you.
The from 0 to 2 means between what time interval you want to plot between.

http://www.wolframalpha.com/input/?i=parametric+plot+(22-8.03t,+1%2B11.47t-4.9t^2)+from+0+to+2 [/URL]

 

[Inferior89]

By the way. Just fyi if you are interested in finding the acceleration you get (0,-9.8) which is just what would be expected since the force downward is gravity which gives you an acceleration of about -9.8 m/s^2.

 

[amb1989]

Hey thank you I really appreciate all the help you have given me. Is there anyway I can plot both of those simultaneously so I can see when the apple hits the Ferris wheel?