Parametric Equations for Tangent Line at (cos 0pi/6, sin 0pi/6, 0pi/6)

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Homework Help Overview

The discussion revolves around finding the parametric equations for the tangent line at a specific point on a space curve defined by the equations x = cos(t), y = sin(t), z = t. The point of interest is (cos(0pi/6), sin(0pi/6), 0pi/6), which raises questions about the correctness of the input values.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to find derivatives of the parametric equations and how to apply them to the given point. There is uncertainty about the notation used for the angle, specifically whether it should be 0pi/6 or pi/6. Some participants attempt to clarify the representation of the tangent line and its components.

Discussion Status

Multiple interpretations of the tangent line equations are being explored, with some participants suggesting different forms for x(t), y(t), and z(t). There is a mix of agreement and confusion regarding the correct representation, particularly for x(t). Guidance has been offered, but discrepancies remain in the participants' understanding of the correct equations.

Contextual Notes

There is a noted concern about the accuracy of the input values and the potential for calculation errors. Participants express frustration with the problem, particularly in relation to the complexity of vector representation in calculus.

weckod
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need parametric equations to the tangent line at the point
(cos 0pi/6, sin 0pi/6, 0pi/6) on the curve x = cost, y = sint, z = t

x(t) = ?
y(t)=?
z(t)=?

now from my understanding, i have to find the derivatives of x, y, and z right? and i did this... now alll i should do is plug in the x, y, z pts? and get the answers? i don't know if the 0pi/6 is correct because it was printed in w/ the problem... it could be pi/6 only and not 0pi/6... could someone help maybe i did some kind of calculation error... if possible explain thanks
 
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anyone know how to help here?
 
Think of your function as a basic space curve:
[tex]\vec r\left( t \right) = \left\langle {\cos t,\sin t,t} \right\rangle[/tex]

where
[tex]\vec r\,'\left( t \right) = \left\langle { - \sin t,\cos t,1} \right\rangle[/tex]

Your point, [tex]\left( {1,0,0} \right)[/tex], can be represented by the positional vector [tex]\vec r \left( 0 \right)[/tex].

As you remember from earlier calculus, the tangent line will thus be parallel to
[tex]\vec r \, ' \left( 0 \right) = \left\langle {0,1,1} \right\rangle[/tex]

Now that you have an equation, you can represent the tangent line as :smile: :
[tex]\vec L\left( t \right) = \left\langle {1,0,0} \right\rangle + t\left\langle {0,1,1} \right\rangle \Rightarrow \vec L\left( t \right) = \left\langle {1,t,t} \right\rangle[/tex]

Or parametrically without vector notation:
[tex]L\left( t \right) = \left\{ \begin{gathered}<br /> x = 1 \hfill \\<br /> y = z = t \hfill \\ <br /> \end{gathered} \right\}[/tex]

:biggrin: Which basically says:
[tex]x = 1 , \,y = z[/tex]

**Hope this helps :smile:
 
Last edited:
so what is x(t)=? y(t)=? z(t)=? because i see what u did but the computer say its wrong... so i dontk now where it went wrong... i know what u did i did the same..
 
weckod said:
so what is x(t)=? y(t)=? z(t)=? because i see what u did but the computer say its wrong... so i dontk now where it went wrong... i know what u did i did the same..
[tex]\left\{ \begin{gathered}<br /> x\left( t \right) = 1 \hfill \\<br /> y\left( t \right) = t \hfill \\<br /> z\left( t \right) = t \hfill \\ <br /> \end{gathered} \right\}[/tex]

What's the problem?
 
Last edited:
well x(t) is not 1+t and same w/the others.. the computer say its wrong... that's what trippin me out
 
wow now the y(t) and z(t) is right but the x(t) is still wrong...
 
yay its just 1...
 
thanks a lot dude! u helped a lot i hate cal 3 its just hard for me for some reasons... its the vectors... i can't picture them...
 
  • #10
No problem :smile: Welcome to PF
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