Question

Find the parametric equations for the tangent line to the curve with the given parametric equations x=e^t, y=te^t and z=te^t^2 at the point (1,0,0).

Answer 1

Would you do the same for this as you would a normal or perpendicular vector?

If so is x(t) = 1+t(0), y(t) = 0+te^t and z(t) = 0+te^t^2 correct?

Answer 2

i believe the tangent line at point (a,b,c) will be in the form
\[x(t) = a+x'(a) t\]
and so forth for y and z

Answer 3

So you will need to use the vector to find the Tangential component (Vector T = vector V / |vector V|? Using that as the new vector, substitute it with the point for the parametric equation?

Answer 4

well you need the slope of the curve at the point....thus find dx/dt, dy/dt, dz/dt

Answer 5

isn't vector T as you defined it above just the unit vector though ?

Answer 6

Well we are studying Motion in Space:Velocity and acceleration at the moment. I assumed it was to easy to pull the vector from the original statement and substitute... But as for what I need to do with vector V <0,e^t, e^t^2> I am assuming that I would need to find the Tangential vector (which my class uses Vector T) but I am pretty lost with this topic...

Answer 7

lets do this
look at the attached file

Answer 8

parameric form will be

Answer 9

I will be back in 45 minutes, but wouldn't r(t) = 0i? because the original equation is x = e^t not x=e^t +te^t? and is r'(t) tangent to r(t)?
So confused...

Answer 10

yes r'(t) is tangent to r(t) .and r(t)=e^t .. not r(t)=0i