# Help with parametric curve question

• amb1989
In summary, the conversation discusses a parametric equation for a thrown object and a Ferris wheel, and the task of finding the speed and angle of inclination of the object when t=t0. The solution involves taking the derivative of the equation and substituting t=t0 to find the velocity vector and then using the dot product and inverse cosine to find the angle. The conversation also mentions finding a simulation online to plot the equations simultaneously.
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.

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.

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.

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

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.

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...

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.

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.

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!

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

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]

Last edited by a moderator:
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.

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?

## 1. What is a parametric curve?

A parametric curve is a mathematical curve that is defined by a set of parametric equations. These equations use one or more parameters to determine the coordinates of points on the curve.

## 2. How do I plot a parametric curve?

To plot a parametric curve, you will need to first determine the range of values for the parameter(s) and then evaluate the parametric equations for each value to get a set of coordinates. You can then plot these points on a graph to create the curve.

## 3. What is the purpose of using parametric curves?

Parametric curves are often used to describe motion or changes over time. They are also useful for representing complex curves that cannot be easily described by a single equation.

## 4. How do I differentiate a parametric curve?

To differentiate a parametric curve, you will need to use the chain rule and differentiate each parametric equation separately with respect to the parameter. The resulting derivatives will give you the slope of the curve at each point.

## 5. Can parametric curves be used in real-world applications?

Yes, parametric curves have many real-world applications such as in physics, engineering, and computer graphics. They are used to model and analyze various systems and phenomena, including motion, trajectories, and curves in 3D space.

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