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Tangent to a curve

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the tangent vector to a curve $$r=R(1-\varepsilon \sin ^2 \varphi)$$ as a function of ##\varphi##

    2. Relevant equations


    3. The attempt at a solution
    Ok, I tried like this:

    I defined a vector $$(f(\varphi),\varphi)=(R(1-\varepsilon \sin ^2 \varphi),\varphi)$$ and than I calculated the derivative $$\frac{\partial }{\partial \varphi}$$ therefore my result is $$(-\varepsilon R \sin(2\varphi),1).$$

    But I have a friend claiming that the result is $$(-\varepsilon R \sin(2\varphi),R(1-\varepsilon \sin ^2 \varphi)).$$ So, my question to you is... which is right? :/
     
  2. jcsd
  3. May 24, 2015 #2

    hunt_mat

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    Looks like you've computed the normal. You need to find a vector such that your vector dotted with it give zero. I think both of you are wrong.
     
  4. May 24, 2015 #3

    Orodruin

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    First of all: What basis are you using to write down your vector? Unless you specify a basis, your two numbers mean nothing. What basis is your friend using?

    Edit: Second, how do you define the tangent vector of a curve?
     
  5. May 24, 2015 #4
    It's polar basis ##r## being the distance from the origin and ##\varphi ## the polar angle.

    The function $$R(1-\varepsilon \sin ^2\varphi)$$ turns out to be an ellipse. ##R## and ##\varepsilon ## are two constants, where ##\varepsilon \ll 1##.
     
  6. May 24, 2015 #5

    Orodruin

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    And what is the definition of the tangent vector of a curve?
     
  7. May 25, 2015 #6
    Yeah that could be the reason why my result is wrong.

    If I am not mistaken, the vector is ##(f(\varphi),\varphi)## and normalized tangent vector should be $$t=\frac{(\frac{\partial f(\varphi)}{\partial \varphi},\frac{\partial }{\partial \varphi}\varphi)}{|(f(\varphi),\varphi)|}$$
     
  8. May 25, 2015 #7

    Orodruin

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    Your big problem is that you are trying to only deal with the components of a vector, and even the components you quote are wrong. For any curve ##\vec r(t)##, the tangent vector is given by ##\vec t = d\vec r/dt##. In polar coordinates, the position vector ##\vec r## is given by ##\vec r = r \vec e_r##, where ##\vec e_r## is the radial unit vector. There is no component in the angular direction.

    However, you do need to note that ##\vec e_r## is not a constant vector and in order to differentiate this with respect to time, you will also need to consider the time derivative of ##\vec e_r##. This is a big disadvantage of trying to write everything as a list of numbers only. The first step is to stop referring to vectors as a list of numbers without any reference as to what basis is being used.
     
  9. May 25, 2015 #8
    Ok, lets write $$\vec r=R(1-\varepsilon \sin ^2 \varphi)\vec e_r=s(\varphi)\vec e_r$$ than tangent vector ##\vec t## is $$\vec t=\frac{\partial }{\partial \varphi }s(\varphi) \vec e_r+s\frac{\partial }{\partial \varphi }\vec e_r.$$ The second term is $$\frac{\partial }{\partial \varphi }e_r=\frac{\partial }{\partial \varphi }(\cos\varphi, \sin \varphi)=(-\sin\varphi,\cos\varphi) $$ but IF (and only if) I am not mistaken, than by definition $$\vec e_\varphi =s(-\sin\varphi,\cos\varphi).$$ Meaning the tangent vector is $$t=\frac{\partial s(\varphi)}{\partial \varphi }\vec e_r + \vec e_\varphi .$$
     
  10. May 25, 2015 #9

    Orodruin

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    There is an ##s## too much in front here. Remember that ##\vec e_\varphi## is a unit vector.

    And one too few in front of the ##\vec e_\varphi## here. Otherwise you are doing well now.
     
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