Let $f(x)$ be a twice differentiable function over $[a,b]$ with arc length $L$. Show that there exists a value $c \in [a,b]$ such that the angle $\theta$ between a horizontal line and the tangent line at $x = c$ satisfies $|\cos\theta| = \frac{b-a}{L}.$
Is my proof correct? Please give specific feedback. Thanks!
Using the trigonometric identites, we rearrange $|\cos\theta|$ into $\frac1{\sqrt{1+\tan^2\theta}}.$ We can think of $\theta$ as a function $\theta(x)$ with $\tan(\theta(x))=f'(x),$ so $$L=\int_a^b\sqrt{1+(f'(x))^2}\,dx=\int_a^b\sqrt{1+\tan^2\theta(x)}\,dx=\int_a^b\frac1{|\cos\theta(x)|}\,dx.$$ Using the Fundemental Therorem of Calculus we have $$f(b) = f(a) + \int_a^b f'(t)\, dt.$$ Rearranging this and dividing by $b-a,$ when $b > a$ we have $$\frac{f(b) - f(a)}{b-a} = \frac{1}{b-a}\int_a^b f'(t)\, dt$$ Therefore, by the Mean Value Theorem for Integrals, there exists $c\in[a,b]$ such that $$\int_a^b\frac{1}{|\cos\theta(x)|}\,dx=(b-a)\,\frac1{|\cos\theta(c)|}.$$ So, we have $$\frac{b-a}{\int_a^b\frac{1}{|\cos\theta(x)|}\,dx} = |\cos\theta(c)|.$$ Subbing in $L$ for our integral, we get $$|\cos\theta(c)|=\frac{b-a}L.$$ This shows that we have $x=c$ that satisfies $$|\cos\theta|=\frac{b-a}L.$$