Is there a nice visualization of the length of a curve formula?

Question

We know that if there is a curve $$\Gamma=\{(x,y)\in\Bbb R^2\ :\ y=f(x), x\in[a,b]\}$$ then $$\text{length}(\Gamma)=\int\limits_a^b\sqrt{1+f'(t)^2}dt$$ and I get that this is because $$\text{length}(\Gamma)=\int\limits_\Gamma1=\int\limits_a^b1\cdot|\gamma'(t)|dt=\int\limits_a^b|(\gamma_1'(t),\gamma_2'(t))|dt=\int\limits_a^b\sqrt{\gamma_1'(t)^2+\gamma_2'(t)^2}dt$$ where $\gamma(t)=(\gamma_1(t),\gamma_2(t))=(t,f(t))$

But $\sqrt{1+f'(t)^2}$ is another function in itself and I would like to see what it corresponds to, is there a way of comparing the two functions in a more meaningful way then just the reasoning that I wrote above?

I am looking for an animation or a drawing or simply an explanation.

Answer 1

The formula arises from an analysis of the curve $(t,f(t))$. Suppose you move from a point $p_1=(t_1,f(t_1))$ to a point $p_2(t_2,f(t_2))$ on the curve, with $t_1<t_2$. If $t_1,t_2$ are sufficiently close to each other, the distance between these two points is close to the length of the piece of the curve between the points $p_1,p_2$. Now the distance is simply: $$d(p_1,p_2)=\sqrt{(t_2-t_1)^2+(f(t_2)-f(t_1))^2}$$ Assuming $f$ is differentiable, the mean value theorem guarantees that we can find a point $\xi\in(t_1,t_2)$ such that $$f(t_2)-f(t_1)=f'(\xi)(t_2-t_1)$$ so we can write $$d(p_1,p_2)=\sqrt{(t_1-t_2)^2+(f^{'}(\xi))^2(t_2-t_1)^2}=(t_2-t_1)\sqrt{1+(f'(\xi))^2}$$ Now if we partition the curve into points $(t_1,f(t_1)),(t_2,f(t_2)),\dots,(t_n,f(t_n))$ and add all these distances we just calculated, we obtain the sum $$\sum_{i=1}^n(t_{i+1}-t_i)\sqrt{(f'(\xi_i))^2+1}$$ which is a Riemman sum for the integral $\int_a^b\sqrt{1+(f'(x))^2}\,dx$.

Answer 2

If $\gamma(t)=(x(t),y(t))$ then $\gamma'(t)=(x'(t),y'(t))$ and $\gamma'(t)dt = (dx,dy)$, so the infinitesimal element of length is $\sqrt{dx^2+dy^2} = \|\gamma'(t)\|\,dt$ and $\|\gamma'(t)\|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$. If $\gamma$ is the graph of a function we have $x(t)=t$ and $y(t)=f(t)$, so $\|\gamma'(t)\|=\sqrt{1+f'(x)^2}$. Long story short, they are exactly the same formula.

Answer 3

I did not feel like drawing so I'm just going to write.

Imagine the graph of a function, and on it two points $f(x)$ and $f(x+\Delta x)$. Sketch lines through these points that are parallel to the axes - this will form two right triangles which are essentially the same.

On both of these triangles, one of the legs has length $\Delta x$, and the other $\vert f(x)-f(x+\Delta x) \vert.$ By Pythagoras, the hypothenuse therefore has length $$\sqrt{(\Delta x)^2 + (f(x)-f(x+\Delta x))^2}.$$ The hypothenuse is a good approximation of the length of the curve generated by $f$ between $x$ and $x+\Delta x$.

Now, $$\sqrt{(\Delta x)^2 + (f(x)-f(x+\Delta x))^2} = \Delta x \sqrt{1 + \left(\frac{f(x)-f(x+\Delta x)}{\Delta x}\right)^2},$$ which becomes $$ \sqrt{1+(f')^2}\;dx $$ when you let $\Delta x \to 0$. So you can think of this function as an infinitesimal length of the curve $f$ gives rise to.