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Integral area and arc length

  1. May 31, 2014 #1
    I learned in my calc 1 class that to calculate the arc length of a curve, we are to compute the integral of the function. For example, the integral of a function that describes the path of a thrown baseball would give the total distance traveled by the baseball (I hope i'm using the term arc length correctly to describe this).

    However, we also learned that finding the integral of a function gives you the area under the curve described by the function. I must have a fundamental misunderstanding of either one or both of these statements, because if they are accurate, then what happens when you take the integral of a semi-circle? Then multiply it by 2? Something doesn't work out because I know the length of the curve (circumference of a circle, C=2pi(r)) is different than the area under/in-between the curves (A=pi(r)^2).

    I would greatly appreciate if someone could please help me understand this apparent paradox.
     
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  3. May 31, 2014 #2

    micromass

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    Let's say you have a positive function ##y=f(x)## (for example the semi-circle ##y=\sqrt{1-x^2}##.

    To find the area below ##f## between the points ##a## and ##b##, you need to calculate

    [tex]\int_a^b f(x)dx[/tex]

    So to find the area below the semi-circle, that would be

    [tex]\int_{-1}^1 \sqrt{1-x^2}dx[/tex]

    which will yield ##\pi/2##.

    Now, to find the arclength of ##f## between the points ##a## and ##b##, you need to find

    [tex]\int_a^b \sqrt{1 + (f^\prime(x))^2}dx[/tex]

    Now, if ##f## is the semi-circle, then ##f^\prime(x) = -x/\sqrt{1-x^2}##, so you need to find

    [tex]\int_{-1}^1 \sqrt{1 + \frac{x^2}{1-x^2}}dx = \int_{-1}^1 \frac{1}{\sqrt{1-x^2}}dx[/tex]

    Calculating this will yield ##\pi##.
     
  4. May 31, 2014 #3
    Thank you for replying so quickly this late at night. Going by what you posted, calculating the length of a curve is a different computation than that of the area under the curve? Your computations make sense, but this directly contradicts what I have learned, for example my textbook gives a function describing the motion of a particle, then it says that to find the total distance traveled by the particle (arc length of the curve), you simply find the integral of the function. This is the exact same approach the book teaches to find the area under a curve. Are they in some way simplifying the procedure without telling us, or perhaps a circle/semi-circle a special case? If so, could you explain to me conceptually why finding the length of the curve of a circle using an integral is different than finding the area of a circle using an integral? And why this is different than finding them for a normal curve?
     
    Last edited: May 31, 2014
  5. May 31, 2014 #4

    micromass

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    Yes, computing the area under a curve should be a very different computation than to compute arclength.

    Is there any chance you could show us what your textbook says on this?
     
  6. May 31, 2014 #5
    Here is the picture of the relevant problem from my book, perhaps I am misusing the term arc-length to describe the length of the curve. However, I am still stumped regardless.
    XeRsv5e.jpg
     
  7. Jun 1, 2014 #6

    SteamKing

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    In the example, the form of the velocity-distance relationship used pertains to one particular form of motion: rectilinear motion, where the particle is traveling in a straight line. For particles traveling along more complex paths, you would need to solve the system of differential equations which describe the motion of the particle.

    The arc-length formula describes how one computes the length of a static curve knowing something about its geometry.
     
  8. Jun 1, 2014 #7

    HallsofIvy

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    Do you understand the difference between "displacement" and "distance traveled"? If you go 30 miles due East, then turn around and drive 20 miles back west, you total "distance traveled" is 30+ 20= 50 miles. But your net "displacement" is only 30- 20= 10 miles, your final distance from your starting point.
     
  9. Jun 1, 2014 #8

    micromass

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    What you're asked to do is not to find the length of the curve ##v(t)##. The formula

    [tex]\int_a^b \sqrt{1+ (v^\prime(t))^2}dt[/tex]

    would be the arclength of ##v(t)##. But that has no physical significance and is not what you need to find.

    You are given at any point ##t##, the velocity of a particle ##v(t)##. This velocity changes over time. You are asked to find the total distance that the particle traveled. This is not the same as the arc-length of ##v(t)##. It does happen to coincide with the area below the ##v## graph.
     
  10. Jun 1, 2014 #9
    Thank you for the help, I was under the impression that the given function was position-time, not realizing it was in fact velocity-time. This confused me because I knew summing the distances between the points of a position-time graph would give total distance traveled, and then I saw what I thought was the book finding that same result using integration. This then led me to believe that the length of the curve of the graph was equivalent to the area under the curve (which doesn't make sense for a number of reasons).
     
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