Solving Calculus Problems with Mechanics

[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?

 

[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,

 

[Chen]

You can break it down like this:

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

 

[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?

 

[Chen]

well I think I got that one but what about the normal vector? is it as easy as taking the derivative?

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...

is there a difference between the velocity vector and the tangent line?

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).

 

[Feynmanfan]

That was a great answer! Thanks a lot.