Calculus/Arc length

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Arc length

Suppose that we are given a function f that is continuous on an interval [a,b] and we want to calculate the length of the curve drawn out by the graph of f(x) from x=a to x=b. If the graph were a straight line this would be easy — the formula for the length of the line is given by Pythagoras' theorem. And if the graph were a piecewise linear function we can calculate the length by adding up the length of each piece.

The problem is that most graphs are not linear. Nevertheless we can estimate the length of the curve by approximating it with straight lines. Suppose the curve C is given by the formula y=f(x) for a\le x\le b. We divide the interval [a,b] into n subintervals with equal width \Delta x and endpoints x_0,x_1,\ldots,x_n. Now let y_i=f(x_i) so P_i=(x_i,y_i) is the point on the curve above x_i. The length of the straight line between P_i and P_{i+1} is

 |P_iP_{i+1}| = \sqrt{ (y_{i+1} - y_{i})^2 + (x_{i+1}-{x_i})^2}

So an estimate of the length of the curve C is the sum

 \sum_{i=0}^{n-1} |P_iP_{i+1}|

As we divide the interval [a,b] into more pieces this gives a better estimate for the length of C. In fact we make that a definition.

Length of a Curve

The length of the curve y=f(x) for a\le x\le b is defined to be

L = \lim_{n\to \infty} \sum_{i=0}^{n-1} |P_{i+1}P_i|

The Arclength Formula[edit]

Suppose that f' is continuous on [a,b]. Then the length of the curve given by y=f(x) between a and b is given by

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

And in Leibniz notation

 L = \int_a^b \sqrt{ 1+ \left(\frac{dy}{dx}\right)^2 } dx

Proof: Consider y_{i+1} - y_i= f(x_{i+1}) -f(x_i). By the Mean Value Theorem there is a point  z_i in (x_{i+1},x_i) such that

 y_{i+1} - y_i= f(x_{i+1}) -f(x_i)= f'(z_i) (x_{i+1}-x_i)\,

So

\begin{align}|P_iP_{i+1}|&=\sqrt{ (y_{i+1} - y_{i})^2 + (x_{i+1}-x_i)^2}\\
&=\sqrt{ (f'(z_i))^2 (x_{i+1}-x_i)^2 + (x_{i+1}-x_i)^2 }\\
&=\sqrt{ (1+(f'(z_i))^2) (x_{i+1}-x_i)^2 }\\
&=\sqrt{ (1+(f'(z_i))^2)} \Delta x\end{align}

Putting this into the definition of the length of C gives

L=\lim_{n\to \infty} \sum_{i=0}^{n-1} \sqrt{ (1+(f'(z_i))^2)} \Delta x

Now this is the definition of the integral of the function g(x) = \sqrt{1+(f'(x))^2} between a and b (notice that g is continuous because we are assuming that f' is continuous). Hence

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

as claimed.

Example: Length of the curve y=2x from x=0 to x=1

As a sanity check of our formula, let's calculate the length of the "curve" y=2x from x=0 to x=1. First let's find the answer using the Pythagorean Theorem.

P_0=(0,0)

and

P_1=(1,2)

so the length of the curve, s, is

s=\sqrt{2^2+1^2}=\sqrt{5}

Now let's use the formula

s=\int_0^1 \sqrt{1+\left(\frac{d(2x)}{dx}\right)^2}=\int_0^1 \sqrt{1+2^2}=\sqrt{5}x\bigr|_0^1=\sqrt{5}


Exercises[edit]

1. Find the length of the curve y=x^{3/2} from x=0 to x=1.
\frac{13^{3/2}-8}{27}
2. Find the length of the curve y=\frac{e^x+e^{-x}}{2} from x=0 to x=1.
\frac{e-\frac{1}{e}}{2}

Solutions

Arclength of a parametric curve[edit]

For a parametric curve, that is, a curve defined by x=f(t) and y=g(t), the formula is slightly different:

L = \int_a^b \sqrt{ (f'(t))^2 + (g'(t))^2 }\,dt

Proof: The proof is analogous to the previous one: Consider y_{i+1} - y_i= g(t_{i+1}) -g(t_i) and x_{i+1} - x_i= f(t_{i+1}) -f(t_i). By the Mean Value Theorem there are points c_i and d_i in (t_{i+1},t_i) such that

 y_{i+1} - y_i= g(t_{i+1}) -g(t_i)= g'(c_i) (t_{i+1}-t_i)\,

and

  x_{i+1} - x_i= f(t_{i+1}) -f(t_i)= f'(d_i) (t_{i+1}-t_i)\,

So

\begin{align}|P_iP_{i+1}|&=\sqrt{ (y_{i+1} - y_{i})^2 + (x_{i+1}-x_i)^2}\\
&=\sqrt{ (g'(c_i))^2 (t_{i+1}-t_i)^2 + (f'(d_i))^2 (t_{i+1}-t_i)^2 }\\
&=\sqrt{ (f'(d_i))^2)+(g'(c_i))^2) (t_{i+1}-t_i)^2 }\\
&=\sqrt{ (f'(d_i))^2 + (g'(c_i))^2} \Delta t\end{align}

Putting this into the definition of the length of the curve gives

L=\lim_{n\to \infty} \sum_{i=0}^{n-1} \sqrt{ (f'(d_i))^2 + (g'(c_i))^2 } \Delta t

This is equivalent to:

L=\int_a^b \sqrt{ (f'(t))^2 + (g'(t))^2}\,dt

Exercises[edit]

3. Find the circumference of the circle given by the parametric equations x(t)=R\cos(t), y(t)=R\sin(t), with t running from 0 to 2\pi.
2\pi R
4. Find the length of one arch of the cycloid given by the parametric equations x(t)=R(t-\sin(t)), y(t)=R(1-\cos(t)), with t running from 0 to 2\pi.
8R

Solutions

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Arc length