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B Average angle made by a curve with the ##x-axis##

  1. Feb 24, 2017 #1
    The average angle made by a curve ##f(x)## between ##x=a## and ##x=b## is:
    $$\alpha=\frac{\int_a^b\tan^{-1}{(f'(x))}}{b-a}$$
    I don't think there should be any questions on that. Since ##f'(x)## is the value of ##\tan{\theta}## at every point, so ##tan^{-1}{(f'(x))}##, should be the angle made by the curve at that point.

    Now, I expected this to hold:
    $$\tan^{-1}\left({\frac{f{(b})-f{(a)}}{b-a}}\right)=\alpha=\frac{\int_a^b\tan^{-1}{(f'(x))}}{b-a}$$
    because, ##\tan^{-1}\left({\frac{f{(b})-f{(a)}}{b-a}}\right)## is also the 'average angle' made by the curve between ##x=a## and ##x=b##.
    It was true only approximately for ##f(x)=\log{|\sec{x}|}## when I checked for ##a=0## and ##b=\frac{\pi}{4}##. It obviously holds for linear functions and I checked that it only approximately holds for quadratic functions. I don't know anything beyond high-school calculus, so couldn't check it for polynomials of degree greater than ##2##.

    I also tried root-mean-square instead of average angle but that expression too didn't hold accurately.

    I took one step further and replaced ##\tan^{-1}{x}## with any function ##g(x)## and expected this to hold:
    $$g\left({\frac{f{(b})-f{(a)}}{b-a}}\right)=k=\frac{\int_a^bg{(f'(x))}}{b-a}$$
    But this one too only holds approximately for some ##g(x)## that I checked.
    So, why don't these expressions hold as expected?
     
    Last edited: Feb 24, 2017
  2. jcsd
  3. Feb 24, 2017 #2

    jedishrfu

    Staff: Mentor

    arctan depends on x so you can't just pull it out of the integral over x like that.

    In general, you can't pull any function dependent on x out of an integral over x.

    If you changed the integral to a summation and ask yourself about this simpler example:

    atan( f(x1) + f(x2) + f(x3) )

    - vs -

    atan( f(x1) ) + atan( f(x2) ) + atan( f(x3) )
     
  4. Feb 24, 2017 #3

    Dale

    Staff: Mentor

    Because your expectation was wrong.
     
  5. Feb 24, 2017 #4

    jedishrfu

    Staff: Mentor

    Have you tried using the chain rule to evaluate the righthand side to see if it matches the lefthand side:

    $$\tan^{-1}(f(b)-f(a)) = \int_a^b\tan^{-1}(f'(x)) dx$$

    to see if you can derive your conclusion. It may also explain why the results are different from what you expected.

    https://en.wikipedia.org/wiki/Chain_rule

    In particular heck out the first proof which uses limits to prove the rule.
     
    Last edited: Feb 24, 2017
  6. Feb 24, 2017 #5

    Mark44

    Staff: Mentor

    Thread closed. As @Dale pointed out in post #3, your expectation is wrong. In future posts, please make more of an attempt on proving your conjectures instead of asking why something doesn't work.
     
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