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Unit tangent vector and equation of tangent line to curve

GreenGoblin

Member
Feb 22, 2012
68
"find a unit tangent vector and the equation of the tangent line to the curve r(t) = (t, t^2, cost), t>=0 at the point r(pi/2)." NOW, what I don't get is, how is that a curve? This is not like the example I have studied and I don't really get the question. So I don't know where to start. Once I find a starting point I can probably do it but I don't recognise the form of the problem which is a major issue. Any hinters? Gracias.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,904
"find a unit tangent vector and the equation of the tangent line to the curve r(t) = (t, t^2, cost), t>=0 at the point r(pi/2)." NOW, what I don't get is, how is that a curve? This is not like the example I have studied and I don't really get the question. So I don't know where to start. Once I find a starting point I can probably do it but I don't recognise the form of the problem which is a major issue. Any hinters? Gracias.
Hi GreenGoblin! :)

If you fill in successive values for t, you'll get a number of dots with coordinates (x,y,z).
Connecting those dots gets you a curve.
You might also write it as:

x(t)=t
y(t)=t^2
z(t)=cos(t)

To find the tangent, you're supposed to take the derivative with respect to t, which will give you a tangent vector.
 

GreenGoblin

Member
Feb 22, 2012
68
Hi, thanks for your reply. So how can I express this? do i solve the derivatives at pi/2? I get (1, pi, -1). What can I do with this? I am really bad at this topic and I don't know where to go with it. What part of the question is this the solution for? I don't even know. I need a 'unit tangent vector', and a 'tangent line to the curve'. Are these both associated to the point pi/2, or the curve in general? I don't know. can i convert these parametric coords to cartesian. i try to do it but with three variables i cant figure.
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,904
Hi, thanks for your reply. So how can I express this? do i solve the derivatives at pi/2? I get (1, pi, -1). What can I do with this? I am really bad at this topic and I don't know where to go with it. What part of the question is this the solution for? I don't even know. I need a 'unit tangent vector', and a 'tangent line to the curve'. Are these both associated to the point pi/2, or the curve in general? I don't know
You have just found a tangent vector to the curve at $t=\pi/2$.
To make it a unit tangent vector you need to divide each of its components by the length of the vector.
Let's call this vector $\mathbf d$.

If you fill in pi/2 in the curve specification, you get the point on the curve that you need.
Let's call this point $\mathbf{p} = (p_x, p_y, p_z) = (x(\pi/2), y(\pi/2), z(\pi/2))$.

The line has to be a line that goes through this point in the direction of that tangent vector.
The parameter representation of the line is $\mathbf{r}(t) = (p_x + d_x t, ~ p_y + d_y t, ~ p_z + d_z t)$.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
As a quick and sloppy way of thinking about it, a curve is something with one degree of freedom. In this case, we have something like $r: I \subset \mathbb{R} \to \mathbb{R}^3$. If we had a surface, we would have something such as $\varphi: U \subset \mathbb{R}^2 \to \mathbb{R}^3$.

More generally, given a map $\gamma: V \subset \mathbb{R}^d \to \mathbb{R}^n$, with $d \leq n$, you can have a maximum of $d$ degrees of freedom. This results in a $d$-dimensional plane in your $n$-dimensional space.

I hope this helps you identify and understand curves and other geometric objects defined in similar ways. (Yes)

- Fantini