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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 $x = \cos(t), y = \sin(t), z=t$

I understand that in order to find the solution, I need to use partial derivatives. However, the method in my textbook works for simpler problems -- I seem to be making a calculation error when I try to apply the method to this problem.

Can anyone suggest how to approach this problem?

I found a very similar problem and solution here, but the solution by the person who answered is hard for me to follow. Unfortunately, I get stuck at the line where he subtracts $\frac{\pi}{6}$ from $\pi$ within the trigonometric functions.

Here is the "simple" method that I was originally using.

Any sincere help would be appreciated. Thank you.

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I have edited your title and question. In general try to keep the title short and capture the essence of the problem instead of having the entire problem statement as title. – user17762 Feb 21 '11 at 21:48

So $\textbf{r}(t) = \left< \cos t, \sin t, t \right>$. Then $\textbf{r}'(t) = \left<-\sin t, \cos t, 1 \right>$. So $t = -4 \pi/6$. So $\textbf{r'}(-\frac{4 \pi}{6}) = \left<-\sin( -\frac{4 \pi}{6}), \cos \left( -\frac{4 \pi}{6} \right), 1 \right>$. So the equation of the tangent line would be

$$x = -\frac{4 \pi}{6}+t\left(-\sin\left( -\frac{4 \pi}{6}\right)\right)$$ $$y = -\frac{4 \pi}{6}+t\left(\cos \left(-\frac{4 \pi}{6} \right) \right)$$ and $$z = -\frac{4 \pi}{6}+t$$

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Thank you for the great response. Actually, the first two answers (x and y) are what I have been getting--for some reason, they are not considered correct by the program I am using to check the answers, although I am using the same input. Z was considered correct. Any ideas? – Math Student Feb 23 '11 at 4:17
I'm pretty sure your $x$ and $y$ equations are wrong—the given point matches your $z$ equation when $t=0$, but not the $x$ and $y$ equations when $t=0$. – Isaac Feb 23 '11 at 18:25

I'm pretty sure that the answer to which you linked on Yahoo! Answers is just wrong. The parametric equations $x=\cos t$, $y=\sin t$, $z=t$ describe a spiral on a cylinder of radius 1 coaxial with the $z$-axis. Since there is only 1 parameter, these parametric equations cannot describe a 2-dimensional surface. The method used in your second link seems appropriate—the direction vector of the tangent line at any point on $\langle x(t),y(t),z(t)\rangle=\langle\cos t,\sin t,t\rangle$ is $\langle x'(t),y'(t),z'(t)\rangle=\cdots$ (no partial derivatives needed) and you know a point on the line, so you can write a parametric equation for the tangent line.

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X= cos(-4pi/6) + t(-sin(-4pi/6)) y= sin(-4pi/6) + t(cos(-4pi/6)) z= -4pi/6 + t

x= .9993 + t(0.0365) y= -0.0365 + t(0.9993) z= -2.094 + t

All of these values were estimated to 4 digits by plugging in (-4pi/6) into the equations for x,y,z

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Welcome to MSE! It really helps readability to format answers using MathJax (see FAQ). Also, when can get exact representations of these equations. Regards – Amzoti Aug 31 '13 at 19:48

the correct answer for $x$ and $y$ are: $$ \cos(-4\pi/6) + t(-\sin(-4\pi/6))\\ \sin(-4\pi/6) + t(\cos(-4\pi/6)) $$

this is because the point that lies on the parametric lines are defined by inputting the $t$ value into the original parametric equation

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Are you able to show the details, for the benefit of the person who asked the question? – Henry T. Horton May 13 '13 at 23:59

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