Question

a(t)=(2,0,6t) V(0)=(0,1,0) r(0)=(0,1,-3) how to find unit tangent and unit normal vector when t=1

Answer 1

tangent and normal require knowledge of V9t) and\[v(t) = v(0) + a*t\]

Answer 2

i got T=(2t,1,3t^2)/(4t^2+1+9t^4)^(1/2) then i am stuck

Answer 3

I was wrong. here\[v(t)= v(0) + \int\limits_0^t a(t) dt\]

Answer 4

But you made a wrong computation

Answer 5

\[v(t) = (0,1,0) + (2t, 0, 3t^2)\]

Answer 6

v(t) =(2t,1,3t^2) am i right??

Answer 7

substitute t = 1

Answer 8

unit tangent = v(t)/Iv(t)I

Answer 9

so i got
(2t,1,3t^2)/(4t^2+1+9t^4)^(1/2)

Answer 10

It is clearly unneeded in general form.
1 Substitute
2 Compute || and divide by it

Answer 11

Do I need a equation to compute unit normal vector?

Answer 12

Just 3 numbers that you obtain by substitution of t=1 into the expression v(t)

Answer 13

i think i need a general form of unit tangent to compute unit normal vector
N=T'/IT'I

Answer 14

No just the |T| at one point.

Answer 15

thx i think u are right

Answer 16

close q.