# Arc Length

## Arc Length in Rectangular Coordinates

Let a curve $$C$$ be defined by the equation $$y = f\left( x \right),$$ where $$f$$ is continuous on an interval $$\left[ {a,b} \right].$$ We will assume that the derivative $$f^\prime\left( x \right)$$ is also continuous on $$\left[ {a,b} \right].$$

The length of the curve $$y = f\left( x \right)$$ from $$x = a$$ to $$x = b$$ is given by

$L = \int\limits_a^b {\sqrt {1 + {{\left[ {f^\prime\left( x \right)} \right]}^2}} dx} .$

If we use Leibniz notation for derivatives, the arc length is expressed by the formula

$L = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} .$

We can introduce a function that measures the arc length of a curve from a fixed point of the curve. Let $${P_0}\left( {a,f\left( a \right)} \right)$$ be the initial point on the curve $$y = f\left( x \right),$$ $$a \le x \le b.$$ Then the arc length of the curve from $${P_0}\left( {a,f\left( a \right)} \right)$$ to a point $$Q\left( {x,f\left( x \right)} \right)$$ is given by the integral

$S\left( x \right) = \int\limits_a^x {\sqrt {1 + {{\left[ {f'\left( t \right)} \right]}^2}} dt} ,$

where $$t$$ is an internal variable of the integral.

The function $$S\left( x \right)$$ is called the arc length function.

## Arc Length of a Parametric Curve

If a curve $$C$$ is given in parametric form by the equations

$x = x\left( t \right),\;\; y = y\left( t \right),$

where the parameter $$t$$ runs between $${t_1}$$ and $${t_2},$$ the arc length of the curve is

$L = \int\limits_{{t_1}}^{{t_2}} {\sqrt {{{\left[ {x'\left( t \right)} \right]}^2} + {{\left[ {y'\left( t \right)} \right]}^2}} dt} .$

## Arc Length in Polar Coordinates

The arc length of a polar curve given by the equation $$r = r\left( \theta \right),$$ with $$\theta$$ ranging over some interval $$\left[ {\alpha ,\beta } \right],$$ is expressed by the formula

$L = \int\limits_\alpha ^\beta {\sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the length of the line segment given by the equation $y = 7x + 2$ from $$x = 2$$ to $$x = 6.$$

### Example 2

Find the arc length of the semicubical parabola $y = {x^{\frac{3}{2}}}$ from $$x = 0$$ to $$x = 5.$$

### Example 3

Prove that the circumference of a circle of radius $$R$$ is $$2\pi R.$$

### Example 4

Calculate the arc length of the curve $y = \ln x$ from $$x = \sqrt{3}$$ to $$x = \sqrt{15}.$$

### Example 1.

Find the length of the line segment given by the equation $y = 7x + 2$ from $$x = 2$$ to $$x = 6.$$

Solution.

Let's solve the more general problem. Consider an arbitrary straight line that is defined by the equation $$y = mx+n.$$ What is the length of the line segment in the interval $$\left[ {a,b} \right]?$$

It's obvious that

$y^\prime = f^\prime\left( x \right) = \left( {mx + n} \right)^\prime = m.$

Using the arc length formula, we have

$L = \int\limits_a^b {\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx} = \int\limits_a^b {\sqrt {1 + {m^2}} dx} = \sqrt {1 + {m^2}} \left( {b - a} \right).$

So, for the line segment $$y = 7x + 2,$$ we obtain

$L = \sqrt {1 + {7^2}} \left( {6 - 2} \right) = 4\sqrt {50} = 20\sqrt 2 .$

### Example 2.

Find the arc length of the semicubical parabola $y = {x^{\frac{3}{2}}}$ from $$x = 0$$ to $$x = 5.$$

Solution.

We use the arc length formula

$L = \int\limits_a^b {\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx} .$

Here $$f\left( x \right) = {x^{\frac{3}{2}}},$$ $$f^\prime\left( x \right) = \left( {{x^{\frac{3}{2}}}} \right)^\prime = \frac{3}{2} {x^{\frac{1}{2}}},$$ $$a = 0,$$ $$b = 5.$$ Then

$L = \int\limits_0^5 {\sqrt {1 + {{\left( {\frac{3}{2}{x^{\frac{1}{2}}}} \right)}^2}} dx} = \int\limits_0^5 {\sqrt {1 + \frac{{9x}}{4}} dx} = \frac{1}{2}\int\limits_0^5 {\sqrt {4 + 9x} dx} .$

We make the substitution

${u^2} = 4 + 9x,\;\; \Rightarrow u = \sqrt {4 + 9x} ,\;\; \Rightarrow 2udu = 9dx,\;\; \Rightarrow dx = \frac{{2udu}}{9}.$

When $$x = 0,$$ $$u = 2,$$ and when $$x = 5,$$ $$u = 7.$$ So the integral becomes

$L = \frac{1}{2}\int\limits_2^7 {\left( {u \cdot \frac{2}{9}u} \right)du} = \frac{1}{9}\int\limits_2^7 {{u^2}du} = \frac{1}{9} \cdot \left. {\frac{{{u^3}}}{3}} \right|_2^7 = \frac{1}{{27}}\left( {{7^3} - {2^3}} \right) = \frac{{335}}{{27}}.$

### Example 3.

Prove that the circumference of a circle of radius $$R$$ is $$2\pi R.$$

Solution.

We calculate the circumference of the upper half of the circle and then multiply the answer by $$2.$$ The upper half of the circle is defined by the function

$y = \sqrt {{R^2} - {x^2}}.$

Take the derivative:

$y^\prime = f^\prime\left( x \right) = \left( {\sqrt {{R^2} - {x^2}} } \right)^\prime = - \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {{R^2} - {x^2}} }} = - \frac{x}{{\sqrt {{R^2} - {x^2}} }}.$

Then the circumference of the circle is given by

$L = 2\int\limits_{ - R}^R {\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx} = 2\int\limits_{ - R}^R {\sqrt {1 + {{\left( { - \frac{x}{{\sqrt {{R^2} - {x^2}} }}} \right)}^2}} dx} = 2\int\limits_{ - R}^R {\sqrt {1 + \frac{{{x^2}}}{{{R^2} - {x^2}}}} dx} = 2\int\limits_{ - R}^R {\sqrt {\frac{{{R^2}}}{{{R^2} - {x^2}}}} dx} = 2R\int\limits_{ - R}^R {\frac{{dx}}{{\sqrt {{R^2} - {x^2}} }}} = 2R\left. {\arcsin \frac{x}{R}} \right|_{ - R}^R = 2R\left[ {\arcsin 1 - \arcsin \left( { - 1} \right)} \right] = 2R\left[ {\frac{\pi }{2} - \left( { - \frac{\pi }{2}} \right)} \right] = 2\pi R.$

### Example 4.

Calculate the arc length of the curve $y = \ln x$ from $$x = \sqrt{3}$$ to $$x = \sqrt{15}.$$

Solution.

Since $$y^\prime = \left( {\ln x} \right)^\prime = \frac{1}{x},$$ we can write

$L = \int\limits_a^b {\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx} = \int\limits_{\sqrt 3 }^{\sqrt {15} } {\sqrt {1 + {{\left( {\frac{1}{x}} \right)}^2}} dx} = \int\limits_{\sqrt 3 }^{\sqrt {15} } {\sqrt {\frac{{{x^2} + 1}}{{{x^2}}}} dx} = \int\limits_{\sqrt 3 }^{\sqrt {15} } {\frac{{\sqrt {{x^2} + 1} }}{x}dx} .$

To evaluate the latter integral we rewrite it in the form

$I = \int {\frac{{\sqrt {{x^2} + 1} }}{x}dx} = \int {\frac{{x\sqrt {{x^2} + 1} }}{{{x^2}}}dx}$

and change the variable:

${x^2} + 1 = {u^2},\;\; \Rightarrow xdx = udu,\;\; \Rightarrow {x^2} = {u^2} - 1.$

This results in

$I = \int {\frac{{{u^2}}}{{{u^2} - 1}}du} = \int {\frac{{{u^2} - 1 + 1}}{{{u^2} - 1}}du} = \int {\left( {1 + \frac{1}{{{u^2} - 1}}} \right)du} = u + \frac{1}{2}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right| = \sqrt {{x^2} + 1} + \frac{1}{2}\ln \left| {\frac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 1} + 1}}} \right|.$

Returning back to the definite integral, we find the arc length $$L:$$

$L = \left. {\left[ {\sqrt {{x^2} + 1} + \frac{1}{2}\ln \left| {\frac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 1} + 1}}} \right|} \right]} \right|_{\sqrt 3 }^{\sqrt {15} } = \left( {4 + \frac{1}{2}\ln \frac{{4 - 1}}{{4 + 1}}} \right) - \left( {2 + \frac{1}{2}\ln \frac{{2 - 1}}{{2 + 1}}} \right) = 2 + \frac{1}{2}\left( {\ln \frac{3}{5} - \ln \frac{1}{3}} \right) = 2 + \frac{1}{2}\ln \frac{9}{5}.$