i found the unit tangent vector and its right T(t)= <1/(2t+1),(2t)/2t+1),(2t^(1/2))/(2t+1)> now i need to find the unit normal vector. i know the formula T'/|T'| but a truely can not come up with the answer that my proffessor gave me. the answer is N(t)=<(-2t^(1/2))/(2t+1),(2t^(1/2))/(2t+1),(1-2t)/(1+2t)> please help redoing this algebra over in over again is killing me.
unit normal vector problem
- Thread starter blakeadk
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Please reply showing your work so far, so that the volunteers can try to find where the errors, if any, are arising. Thank you! :wink:
HallsofIvy
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You found the tangent vector to what? I guess you found the tangent vector to some curve, given by parametric or vector functions of t in an xyz coordinate system- but you should have said that.
You say "T'=1/(2t+1)<0,2,t^(-1/2)>-(2)(2t+1)<1,2t,2t^(1/2)>"
That second term is wrong. You are using the product rule, of course, but the derivative of 1/(2t+1) is -2/(2t+1)^2. Perhaps that was a typo and you just forgot to add the "^2}. To combine those, you need a "common denominator so you need to multiply numerator and denominator of the first vector by "2t+ 1". That gives
T'= (1/(2t+1)^2)<0, 4t+ 2, 2t^{1/2}+ t^{-1/2}>- (1/(2t+1)^2)<2, 4t, 4t^{1/2}>
= (1/(2t+1)^2)<2, 8t^2+ 4t, -2t^{1/2}+ t^{-1/2}>
To find the length of that, you will need to calcuate \(\displaystyle \sqrt{4+ (8t^2+ 4t)^2+ (-2t^{1/2}+ t^{-1/2})^2}\). That doesn't look very easy! Of course, if the problem asked you to find the normal vector at a specific point on the curve or value of t, that would be much easier.
You say "T'=1/(2t+1)<0,2,t^(-1/2)>-(2)(2t+1)<1,2t,2t^(1/2)>"
That second term is wrong. You are using the product rule, of course, but the derivative of 1/(2t+1) is -2/(2t+1)^2. Perhaps that was a typo and you just forgot to add the "^2}. To combine those, you need a "common denominator so you need to multiply numerator and denominator of the first vector by "2t+ 1". That gives
T'= (1/(2t+1)^2)<0, 4t+ 2, 2t^{1/2}+ t^{-1/2}>- (1/(2t+1)^2)<2, 4t, 4t^{1/2}>
= (1/(2t+1)^2)<2, 8t^2+ 4t, -2t^{1/2}+ t^{-1/2}>
To find the length of that, you will need to calcuate \(\displaystyle \sqrt{4+ (8t^2+ 4t)^2+ (-2t^{1/2}+ t^{-1/2})^2}\). That doesn't look very easy! Of course, if the problem asked you to find the normal vector at a specific point on the curve or value of t, that would be much easier.