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Prove that a given curve is planar

  1. Jul 26, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that the given parametric curve decribed by the following notations:

    x=cos(t), y=sin(t), z=2+2cos(t)

    lies in a single plane ,find the normal vector to this plane

    2. Relevant equations---

    r(t)=cos(t) i + sin(t) j + 2+2cos(t) k

    3. The attempt at a solution

    My attempt was finding the osculating plane to this parametric curve using the follwing equations

    Tangent(t)=(r'(t)/|r'(t)|) ===> Normal(t)=(T'(t)/|T'(t)|)
    inomial=T X B

    I tried this method , yet it takes a much longer time than I guess required for solving this question---and actually I was told there is a simple way to solve this problem ......

    Another attempt was trying to isolate t and substituing it in 'z' and 'y' .....Yet it doesn't help me any way

    my general idea is actually proving the normal vector points the same direction in every point on the curve

    I would like to mention ,someone who solved a simillar problem on another forum used the formula

    Normal vector to the plane = (r' x r'')/ |r' x r''|.
    Yet I couldn't find this formula or any proof of it in any of Calculus textbooks(Stewart,Thomas,Anton....) -and in class we have never encountered this formula -so we are not allowed to use it.....(BTW ,in which textbook can I find this formula and its proof?)

    Thank you!
  2. jcsd
  3. Jul 26, 2015 #2


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    If you could find a [itex]\mathbf{c}[/itex] and a [itex]\mathbf{d}[/itex] such that [tex]
    (\mathbf{r}(t) - \mathbf{d}) \cdot \mathbf{c} = 0
    [/tex] for every [itex]t[/itex] then you would know that your curve lies in a plane normal to [itex]\mathbf{c}[/itex]. Can you do that? Perhaps if you express [itex]\mathbf{r}(t)[/itex] in the form [itex]\mathbf{r}_0 + \mathbf{a}\cos(t) + \mathbf{b}\sin(t)[/itex] ...
  4. Jul 26, 2015 #3


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    You are given that x= cos(t), y= sin(t), z= 2+ 2 cos(t).

    Isn't it obvious that z- x= 2?
  5. Jul 26, 2015 #4
    Yes it is(that's what I did) , yet ,the answer key tells me the normal vector is different from (-1,0,1).....
  6. Jul 26, 2015 #5

    Ray Vickson

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    So, what did the answer key say was the normal vector?
  7. Jul 26, 2015 #6
    The complete solution is attached in this picture below.....(without that ,you will not be able to understand the answer)
    I do understnad the way it was solved yet , I still wanted to solve it by my own in the ways presented above.....

    Attached Files:

    • jkl.png
      File size:
      19.3 KB
  8. Jul 26, 2015 #7
    So again , is there any alternative way to solve this ?
  9. Jul 26, 2015 #8


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    Did you try my earlier suggestion?
  10. Jul 26, 2015 #9
    Oh sorry Pasmith, first of all thank you for your reply!
    I have never seen this formula before....it makes sense of course
    However, I need another way to solve it (as I mentioned, I guess I can't use it ,since I wasn't taught that....)

    BTW:what is the source of this formua ,is it taken from a textbook?
  11. Jul 26, 2015 #10

    Ray Vickson

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    We don't need to see the picture; we just need to see the final result, which gives the normal ##\vec{n}## as
    [tex] \vec{n} = \text{scalar} \cdot ( \vec{e}_z - 2 \vec{e}_x ) [/tex]
    Isn't that what the last line says? Isn't that consistent with the fact that on the curve we have ##z = 2 + 2x##?
  12. Jul 26, 2015 #11


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    It's obvious in comparing what HallsofIvy said with the published solution that HallsofIvy has a typo in his post.

    z = 2 + 2cos(t) gives z = 2 + 2x which results in -2x + z = 2 .

    This gives the normal as (-2, 0, 1) as desired.
  13. Jul 27, 2015 #12
    yes you are totally right , but take a look at the solution my lecturer uploaded - why didn't he just use that method ?
  14. Jul 27, 2015 #13


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    He may not have noticed there was a simpler solution, or he may have simply wanted to illustrate some concepts he wants you to learn.
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