How do you get the vector equation for the tangent line to the graph of r at t=1 given: r(1)= r'(1)= ||r'(1)||=6 r'''(1)=?

Question

How do you get the vector equation for the tangent line to the graph of r at t=1 given: r(1)= <-3,0,2> r'(1)=<5,2,0> ||r'(1)||=6 r'''(1)=<0,-8,3>?

anonymous · Answer

is it ||r''(1)|| = 6 or ||r'(1)||?   r' is <5,2,0>

anonymous · Answer

It's ||r'(1)|| and r'(1) is <5,2,0>

amistre64 · Answer

r' is the direction vector of the line when t=1
r is the point that it is attached to when t=1

amistre64 · Answer

L = r + tr'

amistre64 · Answer

L = r(1) + tr'(1)  that is :)

anonymous · Answer

Thanks. But how would the vector equation be written? Like this: <-3+5t, 2t, 2>?

amistre64 · Answer

that doesnt seem like a vector equation to me.

i believe the notation for that form is:  given a point P and a vector V, the line L is formed as:

L = P + tV

amistre64 · Answer

it can be parametricised as:  P=(px,py,pz), V=(vx,vy,vz)

L=:
x = px + t vx
y = py + t vy
z = pz + t vz

anonymous · Answer

So what would be the final answer to the problem?

amistre64 · Answer

L = r(1) + t r'(1)   ... as i stated before.

anonymous · Answer

Oh right! Thanks a bunch! :)

amistre64 · Answer

good luck ;)