Parametric Equations Homework - Find Curve & Position Vector

In summary, the problem involves finding parametric equations for the curve of intersection between two cylinders and using them to write a position vector. The choice for t can be the angular coordinate in the x,y plane, which leads to x=2*cos(t) and y=2*sin(t). Z can then be found using these coordinates. Setting t=x^2 does not work as it still results in a square root for y. Finally, the correct position vector is <2cos(t), 2sin(t), 4-4cos^2(t)>.
  • #1
chrsr34
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0

Homework Statement


Consider the curve of intersection of the cylinders [x^2+y^2=4] and [z+x^2=4]. Find parametric equations for this curve and use them to write a position vector.

Homework Equations


Thats what I am looking for. What to set t equal to.


The Attempt at a Solution


I set t=x and got a square root for y. So if i set t=x^2, i get rid of the square root for y, but I am not sure if this is correct. i really don't know the rules for parametric equations. Any input is appreciated.

Chris
 
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  • #2
An easy choice for t is the angular coordinate in the x,y plane. So x=2*cos(t), y=2*sin(t). Given this, can you figure out what z is in terms of t? Unfortunately, I don't think there are any general rules for doing this. You just have to make a picture in your mind of what the curve looks like and then look for a choice for t.
 
  • #3
I see. Yes i could find z with those coordinates. But is there a reason you chose those? Are those cylindrical coordinates? (im a little rusty).
Is there anything wrong with setting t=x^2?
 
  • #4
The x^2+y^2=4 cylinder intersects the x,y plane in a circle. It's easy to parametrize that in polar coordinates. If you take t=x^2 then you still have a square root for y (contrary to what you said), so you'd have to define the curve as a union of pieces.
 
  • #5
All yes your right.
So my position vector should be: r = < 4cos^2(t), 4sin^2(t), 4-4cos^2(t) >
Is this correct?
 
  • #6
Noooo. x=2*cos(t), not 4cos^2(t)!
 
  • #7
yea i just realized i did that.
So, r = < 2cos(t), 2sin(t), 4-4cos^2(t) >

This look good?
 

FAQ: Parametric Equations Homework - Find Curve & Position Vector

1. What are parametric equations?

Parametric equations are a set of equations that express the coordinates of a point on a curve as functions of one or more independent variables, known as parameters. These equations are commonly used to describe the motion of objects in space.

2. How do I find the curve of a parametric equation?

To find the curve of a parametric equation, you can plot a set of points by substituting different values for the parameters and then connecting the points. You can also use a graphing calculator or computer software to generate a graph of the curve.

3. How do I find the position vector of a point on a curve from a parametric equation?

To find the position vector of a point on a curve from a parametric equation, you can substitute the values of the parameters into the equations to find the coordinates of the point. The position vector of the point is then the vector from the origin to that point.

4. Can parametric equations be used in three-dimensional space?

Yes, parametric equations can be used to describe curves and surfaces in three-dimensional space. In this case, the equations will have three parameters and will result in a three-dimensional graph.

5. How are parametric equations used in real-world applications?

Parametric equations have a wide range of applications in fields such as physics, engineering, and computer graphics. They are used to describe the motion of objects, the shape of curves and surfaces, and the behavior of systems over time. They are also used in creating computer animations and simulations.

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