Vector valued functions: finding tangent line

bfusco
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Homework Statement


Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P

r(t)= <2sin(t), 2cos(t), 4sin2(t)>, P(1, √3, 1)

The Attempt at a Solution


I found T(t) using the formula T(t)= r'(t)/||r'(t)||

r'(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>

∴ T(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>/ 2+8sin(t)cos(t)
-note: i could do more to the denominator as far as trig functions and algebra but i don't know if it is useful.

What i don't understand is how the point is used to find the parametric equation.
 
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bfusco said:

Homework Statement


Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P

r(t)= <2sin(t), 2cos(t), 4sin2(t)>, P(1, √3, 1)

The Attempt at a Solution


I found T(t) using the formula T(t)= r'(t)/||r'(t)||

r'(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>

∴ T(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>/ sqrt([STRIKE]2[/STRIKE] 4+8sin(t)cos(t))
-note: i could do more to the denominator as far as trig functions and algebra but i don't know if it is useful.

What i don't understand is how the point is used to find the parametric equation.
You have left out the square root for ||r' ||, as well as there is another error as noted above.

Do you know the general form of a parametric equation of a line in vector form?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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