- #1
Kumar8434
- 121
- 5
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?
$$\alpha=\frac{\int_a^b\tan^{-1}{(f'(x))}}{b-a}$$
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: