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Unit tangent, unit normal, unit binormal, curvature 
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#1
Feb2312, 04:43 PM

P: 559

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! 


#2
Feb2312, 05:09 PM

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P: 21,215

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). 


#3
Feb2312, 05:17 PM

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P: 25,244

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.



#4
Feb2312, 05:58 PM

P: 559

Unit tangent, unit normal, unit binormal, curvature
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.) 


#5
Feb2312, 06:15 PM

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P: 25,244




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