Unit Tangent Vector at a Point

In summary, to find the tangent vector r'(t) and the corresponding unit tangent vector u(t) at point P:(.5, 3.5,0), you can first plug the components of point P into both r'(t) and u(t) to get the corresponding values of t. Then, evaluate r'(t) and u(t) at that value of t to get the desired vectors. Remember to use radians instead of degrees in your calculations.
  • #1
Wildcat04
34
0

Homework Statement


r(t) = costi + 2 sint j
Find the tangent vector r'(t) and the corresponding unit tangent vector u(t) at point P:(.5, 3.5,0)


Homework Equations


r'(t) = r(t)dt
u(t) = r'(t) / |r'(t)|


The Attempt at a Solution



r'(t) = -sinti + 2costj

|r'(t)| = [sin2t + 4cos2t].5
= [1-3cos2t].5

u(t) = {-sinti + 2costj} / {[1-3cos2t].5}

I think I am right so far, however I don't know what I am supposed to due with Point P to find the unit tangent vector at that point.

Thanks in advance for the help.
 
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  • #2
The point P corresponds to a value of the parameter t.
r'(t) is the tangent vector at the point r(t)
 
  • #3
So all that is required is to plug the i and j components of point P into both r'(t) and the u(t) equation to "evaluate" them at that point?
 
  • #4
How would you plug in the components of one vector into another vector?

No, the idea is that you plug some t, which corresponds to the point P, into both r'(t) and u(t). You can consider r(t) as describing the position of a particle at time t, and r'(t) its velocity at that time. You can reformulate the question as: "Give the velocity of the particle when it is at P" or, equivalently: "Give the velocity of the particle at that time, at which its position vector is P".
 
  • #5
Ahh...so you get =>

P = r(t)
<.5, 3.5,0> = <cos t, 2sin t, 0>

=> t = 60

From there evaluate r'(t) and u(t) at t=60.

Is this correct?
 
  • #6
yes. Although you should be thinking "[itex]\pi/3[/itex]" rather than "60" at this point.
 
  • #7
I know...degrees have always been hard to get out of my head. I need to start thinking in radians.

Thank you very much for the help!
 

What is a unit tangent vector at a point?

A unit tangent vector at a point is a vector that is tangent to a curve or surface at that particular point and has a magnitude of 1. It represents the direction of the curve or surface at that point.

How is a unit tangent vector calculated?

A unit tangent vector is calculated by taking the derivative of the curve or surface at the given point and then normalizing it to have a magnitude of 1.

What is the significance of the unit tangent vector?

The unit tangent vector provides information about the direction and rate of change of a curve or surface at a specific point. It is also used in various applications such as finding the curvature of a curve or the normal vector of a surface.

How is the unit tangent vector used in physics?

In physics, the unit tangent vector is used to determine the direction of motion of an object along a curved path. It is also used in calculating the velocity and acceleration of the object.

Can the unit tangent vector change at different points on a curve or surface?

Yes, the unit tangent vector can change at different points on a curve or surface. This is because the direction and rate of change of the curve or surface can vary at different points.

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