# Tangent line

1. Jun 1, 2004

### Feynmanfan

Hello everybody!

I'm having trouble with this calculus problem, where I don't know if I can apply what I've learned in MECHANICS.

"given a path s(t)=(t+1,E^t) calculate it's tangent line and the normal line at
this point s(0)"

and this is another version of the problem in R3

"given a path s(t)=(2t,t^2,Lnt) calculate the velocity vector and the tangent line at (2,1,0)"

How do I solve this? Is it just the derivative and that's all?

2. Jun 1, 2004

### matt grime

the derivative gives a tangent vector at that point. you need to then finsd the equation of a line passing through that point in the direction of the tangent vector,

3. Jun 1, 2004

### Chen

You can break it down like this:

If s(t) gives the path of the object, then its coordinates at all times are:
sx(t) = 2t
sy(t) = t2
sz(t) = ln(t)
Then the velocity in every direction (axis) is:
|vx(t)| = sx'(t) = 2
|vy(t)| = sy'(t) = 2t
|vz(t)| = sz'(t) = 1/t
For t = 1, when the object is at (2, 1, 0), you have |vx| = 2, |vy| = 4 and |vz| = 1. In other words, the velocity vector is 2i + 4j + k or (2, 4, 1).

4. Jun 1, 2004

### Feynmanfan

thanks.
well I think I got that one but what about the normal vector? is it as easy as taking the derivative? is there a difference between the velocity vector and the tangent line?

5. Jun 1, 2004

### Chen

The first problem you're not supposed to do with vectors, I don't think. Once you find the tangent line y = ax + b, the normal line is y = -x/a + c. You just need to find c...

The velocity vector determines the direction of the speed as well as its magnitude. Because of this you can't manipulate it however you want, because while the direction wouldn't change, the speed might. The tangent vector, on the other hand, only represents direction, which means its magnitude doesn't matter. Therefore you can multiply or divide it by any scalar k, i.e above we found the velocity vector to be (2, 4, 1) and that's it, but the tangent line can also be (4, 8, 2) or (1/2, 1, 1/4).

Last edited: Jun 1, 2004
6. Jun 1, 2004

### Feynmanfan

That was a great answer! Thanks a lot.