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Parametric Equation for Tangent of Logarithmic Spiral

  1. Sep 20, 2012 #1
    [a]Give a parametric equation for the line tangent to this curve at [itex]t = \frac{pi}{4}.[/itex]
    [tex] \vec{r(t)} <e^tcost, e^tsint> [/tex]

    Give the equation for this same tangent line in the form [tex]ax + by = c[/tex]

    My attempt
    [tex]\vec{r(\frac{pi}{4})} = <e^\frac{pi}{4}cos\frac{pi}{4}, e^\frac{pi}{4}sin\frac{pi}{4}[/tex]
    [tex] = e^\frac{pi}{4}<\frac{1}{2}, \frac{1}{2}>[/tex]
    [tex] \vec {r'(t)} = e^t<cost - sint, cost + sint>[/tex]
    [tex] \vec {r'(\frac{pi}{4})} = e^\frac{pi}{4}<cos\frac{pi}{4} - sin\frac{pi}{4}, cos\frac{pi}{4} + sin\frac{pi}{4}>[/tex]
    [tex] \vec {r'(\frac{pi}{4})} = e^\frac{pi}{4}<0,1>[/tex]

    [tex] x = \frac{e^\frac{pi}{4}}{2} [/tex]
    [tex] y = \frac{e^\frac{pi}{4}}{2} + e^\frac{pi}{4}t [/tex]


    My answers aren't right. I suck.
    Couldn't even solve for
     
    Last edited: Sep 20, 2012
  2. jcsd
  3. Sep 20, 2012 #2

    LCKurtz

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    ##\cos\frac \pi 4 = \frac 1 {\sqrt 2}## and ##\sin\frac \pi 4 = \frac 1 {\sqrt 2}##. See if that helps.
     
  4. Sep 20, 2012 #3
    Well, this is a question from a test, and he didn't count off for the 1/2. The fact that I knew sin(pi/4) and cos(pi/4) were equivalent was enough. The problem is I don't know how an equation derived from a vector dependent on a variable [itex] \vec {r(t)} [/itex]
     
    Last edited: Sep 20, 2012
  5. Sep 20, 2012 #4

    LCKurtz

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    When ##t = \frac \pi 4##, you know ##r(\frac \pi 4)## is a point on the line and ##r'(\frac \pi 4)## is a direction vector for the line. And you know how to write the parametric equation of a line given a point on the line and a direction vector, right? And I would suggest using some letter other than ##t## for your parameter for the tangent line.
     
  6. Sep 20, 2012 #5
    Actually, saying my answers are wrong in misleading. I got this answer wrong on the test, but this is my redo. We get to re-work the problem and turn it in for half credit (basically having us earn our curve).

    So I'm just curious as to what you think of the answer I got on this page. I still think it's wrong, but I just don't know what else to do.

    Unfortunately I have to use t as my parameter.
     
  7. Sep 21, 2012 #6

    HallsofIvy

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    You may be asked to use the letter "t" as parameter but that has nothing to do with the "t" in the original function. Just use whatever parameter is simplest and call it "t".
     
  8. Sep 21, 2012 #7

    LCKurtz

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    All you have to do to fix your answer is make the correction I suggested. You can use any letter for your parameter for the tangent line. The only reason I suggested you not use ##t## is that makes it clear they aren't the same parameters.
     
  9. Sep 21, 2012 #8
    Yeah you're right. I figured that I had actually done everything correctly, and that was my only mistake.
    I just hadn't actually understood that [itex]x[/itex] is constant in my parametric, which meant that the vector tangent to the curve was actually a vertical line, tangent to the point [itex]t = \frac{\pi}{4} [/itex]

    FcOeI.png

    The problem I was having with this was the fact that to solve for the standard equation, using a parametric equation, you have to solve one parametric for [itex]t[/itex], and then plug it into your other parametric. The whole idea was to cancel out the [itex]t[/itex] so that would would just have the [itex]ax + by = c[/itex] form.
    But since I didn't have a [itex]t[/itex] in my [itex]x[/itex] parametric, I got confused.

    Today I learned how to do it. I had to make use of a vector normal to the tangent at [itex]t = \frac{\pi}{4}[/itex].

    Thank you for your help and sorry about being so hard headed about your correction. I just hadn't understood the geometry of what was going on.
     
    Last edited: Sep 21, 2012
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