Comprimento entre dois pontos

Now that we've tried estimating the length of a curve, we can also find its exact value, this time using calculus:

$\textbf{Theorem:}$ Suppose $f(x)$ is a continuous function on the closed interval $[a,b]$. Then the arc length from $a$ to $b$ is equal to $$\int_a^b \sqrt{1+[f'(x)]^2}\, dx.$$

Let's try the very simple example $f(x)=2$ in the interval $[a,b]$:

Clearly, $f'(x)=0$, and so by the formula described above, the arc length is equal to $$\int_a^b \sqrt{1+[f'(x)]^2}\, dx=\int_a^b \sqrt{1+0^2}\, dx = \int_a^b dx=x\bigg|_a^b=b-a,$$ which is what we expect.

Suppose $f(x)$ is a continuous function on the closed interval $[a,b]$. First, we divide the interval into $n$ sections of equal length $\Delta x$, as shown below.

Observe that the arc length can be estimated by the sum $$L\approx \sum_{i=1}^n |P_{i-1}P_i|.$$ As $n$ increases, our estimate becomes closer to the actual value; so, we can say that $$L=\lim_{n\to\infty} \sum_{i=1}^n |P_{i-1}P_i|.$$ Now, let's try to find the value of $|P_{i-1}P_i|$. By the Distance Formula, $$|P_{i-1}P_i|=\sqrt{(x_i-x_{i-1})^2+(y_i-y_{i-1})^2}=\sqrt{(\Delta x)^2+(f(x_i)-f(x_{i-1}))^2}.$$ By the Mean Value Theorem for Integrals, we know that there exists $x_i'$ between $x_{i-1}$ and $x_i$ such that $$f'(x_i')=\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}\Longrightarrow f(x_i)-f(x_{i-1})\\=f'(x_i')(x_i-x_{i-1})=(\Delta x)f'(x_i').$$ Therefore, our equation becomes $$|P_{i-1}P_i|=\sqrt{(\Delta x)^2+([\Delta x]f'(x_i'))^2}\\=\sqrt{(\Delta x)^2+(\Delta x)^2\cdot [f'(x_i')]^2}=(\Delta x)\sqrt{1+[f'(x_i')]^2}.$$ So, it follows that $$L=\lim_{n\to\infty} \sum_{i=1}^n |P_{i-1}P_i|=\lim_{n\to\infty} \sum_{i=1}^n \sqrt{1+[f'(x_i')]^2}(\Delta x).$$ By the definition of definite integrals, we conclude that $$L=\int_{x_0}^{x_n} \sqrt{1+[f'(x)]^2}\, dx=\int_a^b \sqrt{1+[f'(x)]^2}\, dx.$$

Let's say we want to find the arc length of a curve $y=f(x)$ between the points $x=a$ and $x=b$, given that $f(x)$ is continuous on $[a,b]$. Let's start by finding the length of a very small segment of $f(x)$, one with width $dx$. Because of the property of local linearity in continuous functions, we an infinitely small segment of the curve can be drawn as a line segment:

It is easy to see that at the point $(x,y)$, the segment of the curve has length $\sqrt{1 + (f'(x))^2}dx$. Thus, the length of the whole curve from $x=a$ to $x=b$ is

$\int_a ^b \sqrt{1 + (f'(x))^2}dx$.

The formula for finding the arc length between two points is $\int \sqrt{1+(dy/dx)^2}\ dx$ If we want to find the arc length of $\sin(x)$ from 0 to 1, then we just plug those values into the function: $\int_0^1 \sqrt{1+(\cos(x)^2)} dx$ Note: I put in $\cos(x)$ instead of $\sin(x)$ because $\cos(x)$ is the derivative of $\sin(x)$.