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So we are given a twice differentiable function $f(x)$ and its arc length from $a$ to $b$ is $L$. The angle between the tangent line and the horizontal line at $x=c$ at some point on $(a,b)$ is given by $L\cos(r)=b-a$ where $r$ is the angle.

So what i did is i set $f'(x)$ equal to $\tan(r)$. then i used algebra and trigonometry to get $\cos(r)=\frac{1}{\sqrt{f'(x) + 1}}$. Then i integrated the top and the bottom (I don't know why, but it caused me to get the needed form, no justification) from $a$ to $b$, and then i got $\cos(r) = \frac{b-a}{L}$ so $L\cos(r)=b-a$.

I apologize because my english, grammar is not good and I don't know how to make the fractions and numbers appear right.

I was wondering how to actually prove that there exist a $c$, because i did not do real proof i just did algebra, it does not guarantee that there is a $c$, and i dont think that integration step is even valid.

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    Did you use the Mean Value Theorem? – 2017-01-28
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    What is the Mean Value Theorem? – 2017-01-28
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    Given a continuous function on an interval, there must be a value such that the derivative of the function is equal to the average rate of change. mean value theorem for integrals says that the average value of the function exists and is equal to 1/(b-a)$\int_{a}^{b} f(x)$ – 2017-01-28

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