Find parametric equations for the tangent line at the point

[undrcvrbro]

Homework Statement

Find parametric equations for the tangent line at the point

(\cos(\frac{4 \pi}{6}) ,\sin(\frac{4 \pi}{6}) ,\frac{4 \pi}{6}) )

on the curve


x=\cos t,\ y=\sin t, \ z=t

Homework Equations

The Attempt at a Solution

Took the derivative
r'(t)=( -sint, cost, 1)

Here's where I start to fall apart...I'm not sure what to do...but..is this close...

cos(4pi/6)= -1/2

r'(-1/2) = (-sin(-1/2), cos(-1/2), 1)

so,

x= (-1/2) + (-sin(-1/2))t
y= (sin(4pi/6)) + (cos(-1/2)t
z= 4pi/6 + t

So where did I go wrong? I hope I'm not that far off. If anything needs clarified, please ask.

 

[lanedance]

Homework Statement

Find parametric equations for the tangent line at the point

(\cos(\frac{4 \pi}{6}) ,\sin(\frac{4 \pi}{6}) ,\frac{4 \pi}{6}) )

on the curve


x=\cos t,\ y=\sin t, \ z=t

Homework Equations

The Attempt at a Solution

Took the derivative
r'(t)=( -sint, cost, 1)

Here's where I start to fall apart...I'm not sure what to do...but..is this close...

cos(4pi/6)= -1/2

r'(-1/2) = (-sin(-1/2), cos(-1/2), 1)

based on what you have said why choose the point t = -1/2? i would think t = 4.pi/6 is a better choice ;) and what the question asks for

so,

x= (-1/2) + (-sin(-1/2))t
y= (sin(4pi/6)) + (cos(-1/2)t
z= 4pi/6 + t

So where did I go wrong? I hope I'm not that far off. If anything needs clarified, please ask.

then once you have the tangent vector ^r', and the point of the curve say p, the equation o the tangent line will be

f(s) = s.r'+p

 

[undrcvrbro]

Thank you! I have the right answer, but I don't understand why you chose t= 4pi/6..can you explain?

 

[lanedance]

compare you original parametric equations with the point you are asked to evaluate it at, this gives you the correct t value

 

[undrcvrbro]

compare you original parametric equations with the point you are asked to evaluate it at, this gives you the correct t value

Ohhh. Yikes, I can't believe I didn't notice that! Thanks again!