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Distance physics question

  1. Aug 6, 2006 #1

    I don't know if I'm in the the good section to ask my question, I don't really know which type of mathematics my question is about.

    If I'm in a car, according to a parabolic trajectory, which would be the distance traversed by the car between locations A and B?

    Thanks a lot


    NB : Sorry for the mistakes
  2. jcsd
  3. Aug 6, 2006 #2
    Your question is not very clear. A car is following a parabolic trajectory with you in it but what(or where) are points A and B? If this question is from some textbook please provide it as it is in the book.
  4. Aug 6, 2006 #3
    Excuse me, I have some (many) problems to express myself in English.

    No, it is not a question from a textbook, but from my mind. In fact, this example of a car is just another way to say : what is the equation to know the distance between any points of a parabola? Not the equation d² = (x'-x)² + (y'-y)², which is for a line, but the "curved distance".

  5. Aug 6, 2006 #4


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    You posted this in the "general math" section but the answer requires calculus so this may not help you. This is a definitely "non-trivial" problem!

    In general, if y= f(x), the "curved distance" (arclength) on the graph of y= f(x), from (a, f(a)) to (b, f(b)) is given by
    [tex]\int_a^b\sqrt{1+ \frac{dy}{dx}}dx[/itex]

    In particular, for a "parabolic trajectory", we can take y= ax2 so that [itex]\frac{dy}{dx}= 2ax[/itex] and the integral can be written as
    [tex]\int_a^b\sqrt{1+ 4a^2x^2}dx[/itex]

    That can be done by a trigonometric substitution: let [itex]2ax= tan\theta[/itex]. Then [itex]2adx= sec^2\theta d\theta[/itex] and
    [tex]\sqrt{1+ 4a^2x^2}= \sqrt{1+ tan^2\theta}= \sqrt{sec^2\theta}= sec\theta[/tex]

    The integral becomes
    [tex]\int_{arctan a}^{arctan b} sec^3\theta d\theta= \int_{arctan a}^{arctan b} \frac{d\theta}{cos^3\theta}[/tex]
    Since cosine is to an odd power, multiply both numerator and denominator by [itex]cos\theta[/itex]
    [tex]\int_{arctan a}^{arctan b}\frac{cos\theta d\theta}{cos^4\theta}=\int_{arctan a}^{arctan b}\frac{cos\theta d\theta}{(1- sin^2\theta)^2}[/tex]

    Now make the substitution [itex]u= sin\theta[/itex] so that [itex]du= cos\theta d\theta[/itex]. The integral becomes
    [tex]\int_{sin(arctan a)}^{sin(arctan b)}\frac{du}{(1- u^2)^2}[/tex]
    which can be done by partial fractions.
    (To find [itex]sin(arctan a)[/itex], draw a right triangle with opposite side of length a, near side of length b, so that the angle is [/itex]arctan a[/itex]. By the Pythagorean theorem, the hypotenus has length [itex]\sqrt{a^1+ 1}[/itex]. [itex]sin(arctan a)= \frac{a}{\sqrt{a^2+1|}}[/itex]. Do the same for [itex]sin(arctan b)[/itex].)
  6. Aug 7, 2006 #5
    Thanks HallsofIvy for the explanation.

    But, in the first equation, you wrote dy/dx. Then, you wrote that because y = ax², dy/dx = 2ax and you substitute dy/dx in the first equation for 4a²x² in the second. My question is : in the first equation, isn't (dy/dx)² ?

    Thanks! :smile:
  7. Aug 7, 2006 #6
    Yes it should be.
  8. Aug 7, 2006 #7

    If y has more parameters, like y = a(x-h)²+k, what would be dy/dx? Or if y = ax²+bx+c, would dy/dx equal 2ax+b ?

  9. Aug 7, 2006 #8
    That would be correct. And in the first case where you have y = a(x-h)²+k then dy/dx would be 2a(x-h) which shouldn't be hard to see.
  10. Aug 7, 2006 #9
    Thanks for the answer.
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