Unit tangent, unit normal, unit binormal, curvature

s3a

1. The problem statement, all variables and given/known data
Question:
"Find the unit tangent, normal and binormal vectors T, N, B, and the curvature of the curve
x = 4t, y = -3t^2, z = -4t^3 at t = 1."

Answer:
T = 0.285714285714286 i - 0.428571428571429 j - 0.857142857142857 k
N = -0.75644794981871 i + 0.448265451744421 - 0.476282042478447 k
B = 0.588348405414552 i + 0.784464540552736 j - 0.196116135138184
ϰ = 0.0445978383113072


2. Relevant equations
N = dT/dt / |dT/dt|


3. The attempt at a solution
I tried to use the equation from the "Relevant equations" part above. I know there are alternative ways but I want to figure out what I am doing wrong for this method.

I (successfully) get the unit tangent vector to be:
T = (4 i - 6t j - 12t^2 k)/sqrt(4^2 + 6^2 * t^2 + 12^2 * t^4)
T = 2/7 i - 3/7 * t j - 6/7 * t^2 k

I (unsuccessfully) get the unit normal vector to be:
N = (3/7 i - 12/7*t k)/sqrt( (3/7)^2 + (12/7)^2 * t^2)

What am I doing wrong?

Any input would be greatly appreciated!
Thanks in advance!

Mark44

The above is really T(t), the tangent vector for an arbitrary value of the parameter t.

The problem asks for the unit tangent vector and unit normal vector at t = 1. IOW, it's looking for T(1), N(1), and B(1).See above. I didn't check your work, so if you have errors, I wasn't looking for them. Again, you want T(1), N(1), and B(1) - the unit vectors at a particular value of t.

Dick

When you went from T = (4 i - 6t j - 12t^2 k)/sqrt(4^2 + 6^2 * t^2 + 12^2 * t^4) to T = 2/7 i - 3/7 * t j - 6/7 * t^2 k you put t=1 in the denominator. That's means you can't use the second expression to find dT/dt. You eliminated some of the t dependence. You need to use the first and use the quotient rule. BTW this is quite a messy problem.

s3a

Sorry, I was doing a lot of stuff in my head so the t = 1 went on and off.

I get |dT/dt| to be sqrt(144t^4 + 36t^2 + 16) and it doesn't seem that I can get rid of the square root.

So, I am assuming it's hard to do it this way and that I shouldn't do it this way assuming it is possible. Is it possible though? (I am not asking for it computed that way but I would just like to know if it is possible to get past that step without using a computer or something of the sort.)

Dick

I get something a LOT messier for |dT/dt|. I'm using a computer for this one and would feel sorry for someone who wasn't. It's pretty bad.