- Thread starter
- #1

#### GreenGoblin

##### Member

- Feb 22, 2012

- 68

- Thread starter GreenGoblin
- Start date

- Thread starter
- #1

- Feb 22, 2012

- 68

- Admin
- #2

- Mar 5, 2012

- 9,598

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.

- Thread starter
- #3

- 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:

- Admin
- #4

- Mar 5, 2012

- 9,598

You have just found a tangent vector to the curve at $t=\pi/2$.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

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 $\mathbf{r}(t) = (p_x + d_x t, ~ p_y + d_y t, ~ p_z + d_z t)$.

- Feb 29, 2012

- 342

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.

- Fantini