I understand also that in order to find the solution, I need to use partial derivatives. However, the technique in my textbook functions for less complicated troubles -- I seem to be making a calculation error when I try to use the technique to this trouble.

You are watching: Scalar parametric equations for the line tangent to the graph of

Can anyone suggest just how to strategy this problem?

I discovered a *very* comparable difficulty and also solution below, yet the solution by the person who answered is difficult for me to follow. Unfortunately, I gain stuck at the line where he subtracts $fracpi6$ from $pi$ within the trigonometric functions.

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

Any sincere help would certainly be appreciated. Thank you.

calculus geometry multivariable-calculus trigonomeattempt parametric

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edited Feb 7 "16 at 13:05

Paolo

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asked Feb 21 "11 at 19:28

inter-base.net Studentinter-base.net Student

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$egingroup$ I have actually edited your title and also question. In basic attempt to save the title short and also capture the essence of the problem rather of having the whole problem statement as title. $endgroup$

–user17762

Feb 21 "11 at 21:48

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## 4 Answers 4

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So $ extbfr(t) = left$. Then $ extbfr"(t) = left$. So $t = -4 pi/6$. So $ extbfr"(-frac4 pi6) = left$. So the equation of the tangent line would be

$$x = cos(-frac4 pi6)+tleft(-sinleft( -frac4 pi6 ight) ight)$$ $$y = sin(-frac4 pi6)+tleft(cos left(-frac4 pi6 ight) ight)$$ and also $$z = -frac4 pi6+t$$

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edited Nov 9 "16 at 21:03

Vivek

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answered Feb 21 "11 at 20:07

PrimeNumberPrimeNumber

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I"m pretty certain that the answer to which you attached on Yahoo! Answers is just wrong. The parametric equations $x=cos t$, $y=sin t$, $z=t$ define a spiral on a cylinder of radius 1 coaxial via the $z$-axis. Because there is only 1 parameter, these parametric equations cannot describe a 2-dimensional surconfront. The approach provided in your second attach appears appropriate—the direction vector of the tangent line at any type of allude on $langle x(t),y(t),z(t) angle=langlecos t,sin t,t angle$ is $langle x'(t),y'(t),z'(t) angle=cdots$ (no partial derivatives needed) and also you know a allude on the line, so you can create a parametric equation for the tangent line.

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answered Feb 21 "11 at 20:08

IsaacIsaac

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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 approximated to 4 digits by plugging in (-4pi/6) into the equations for x,y,z

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answered Aug 31 "13 at 19:26

Nate ColleyNate Colley

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

this is because the suggest that lies on the parametric lines are characterized by inputting the $t$ worth right into the original parametric equation

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edited Sep 17 "13 at 3:42

dfeuer

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answered May 13 "13 at 23:36

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