Vector Calc - Tangent unit vector

[xxjeepdude44xx]

I need to find the tangent unit vector to r(t). where Vector r(t) = <3t^2+6cos(t), 4sin(t), ln(t-1)>

I know that vector T(t) = r'(t)/|r'(t)|... when i go to try and figure out |r'(t)| i get a crazy mess.

sqrt(20(t^2-2t+1)(sin(t))^2-72t(t^2-2t+1)sin(t)+36t^4-72t^3+52t^2-32t+17) all that miltiplied by |1/t-1|

my instructor say "Just differentiate each piece individually. Then reassemble and divide by the length." but i don't understand what he means..

 

[pka]

I need to find the tangent unit vector to r(t). where Vector r(t) = <3t^2+6cos(t), 4sin(t), ln(t-1)>

\(\displaystyle r'(t) = \left\langle {6t - 6\sin (t),~4\cos (t),~\dfrac{1}{{t - 1}}} \right\rangle \)

What is \(\displaystyle \|r'(t)\| = ~?\)

 

[xxjeepdude44xx]

Should i just leave it as <6t-6sin(t), 4cos(t), 1/(t-1)> / sqrt ((6t-6sin(t)^2)+(4cos(t)^2)+(1/(t-1)^2))? and then just evaluate it for the values of t?

 

[xxjeepdude44xx]

\(\displaystyle r'(t) = \left\langle {6t - 6\sin (t),~4\cos (t),~\dfrac{1}{{t - 1}}} \right\rangle \)

What is \(\displaystyle \|r'(t)\| = ~?\)

|r'(t)| = sqrt(X^2 + Y^2 + Z^2)
which if I try and actually add them ends up in the big mess..

 

[pka]

|r'(t)| = sqrt(X^2 + Y^2 + Z^2)
which if I try and actually add them ends up in the big mess..

You will learn that by not trying to "simplify things", just leave as is, in vector analysis your life will be a lot easier.

 

[xxjeepdude44xx]

Okay so I left it and evaluated at t=4 for (for the required problem) then got T'(t) and then found N'(t) by T'(t)/|T'(t)|

Thanks for your help.