Calculus problem particle motion in a plane

So I have a problem with the position of a particle in a plane:

X(t) = tI + ln(cos(t))J .. @ t = (-.6pi)/2

I am supposed to find the unit tangent vector, the unit normal vector, the acceleration vector and the curvature.

I know that the unit tangent vector is (1/sqrt(1+(-tan(-.6pi/2))^2))I + (-tan(-.6pi/2))/sqrt(1+(-tan(-.6pi/2))^2…

and that the acceleration vector is 0I + -sec^2(-0.6pi/2)J...

But I can't seem to get the right answer for the unit normal vector nor the curvature.

Whichever way I look at it, I'm getting zeros for the I/J values.. and these are not being accepted by the system. The curvature is having a similar problem.

One step a time...

unit tangent vector is

T(t) = x'/|x'|

x =
x' = <1, -tan(t)>
|x'| = sec²t

T(t) =

Note that -0.6π/2 = -3π/10

T(-3π/10) ≈ <0.588, -0.809>

The unit normal vector is

N(t) = T'/|T'|

T =
T' = <-sin(t),-cos(t)>
|T'| = 1

N(t) = <-sin(t),-cos(t)>

N(-3π/10) ≈ <0.809, 0.588>

The acceleration vector is

x'' = <0,-sec²t>

x''(-3π/10) ≈ <0,2.89>

The curvature is

κ = |T'|/|x'| = 1/sec²t = cos²t

κ(-3π/10) = cos²(-3π/10) ≈ 0.345

Yin

math should grow up and solve its on problems

3032#@R!R#@3 x 2





that is what that looks like to me