# 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

${x = x\left( t \right),\;\;}\kern0pt{y = y\left( t \right),}$

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 5

Find the arc length of the curve $$y = \ln \sec x$$ from $$x = 0$$ to $$x = \large{\frac{\pi }{3}}\normalsize.$$

### Example 6

Find the arc length of the curve $$y = {e^x}$$ from $$x = 0$$ to $$x = 1.$$

### Example 7

Find the length of one arc of the cycloid given in parametric form by the equations $$x\left( t \right) = t – \sin t,$$ $$y\left( t \right) = 1 – \cos t.$$

### Example 8

Find the arc length of the astroid $$x\left( t \right) = {\cos ^3}t,$$ $$y\left( t \right) = {\sin^3}t.$$

### Example 9

Find the length of the cardioid $$r = 1 + \cos \theta .$$

### Example 10

Find the length of the first turn of Archimedean spiral $$r\left( \theta \right) = \theta.$$

### 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 $$f\left( x \right) = {x^{\frac{3}{2}}},$$ $$f^\prime\left( x \right) = \left( {{x^{\frac{3}{2}}}} \right)^\prime = \large{\frac{3}{2}}\normalsize{x^{\frac{1}{2}}},$$ $$a = 0,$$ $$b = 5.$$ 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:

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

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 $$y^\prime = \left( {\ln x} \right)^\prime = \large{\frac{1}{x}}\normalsize,$$ 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 $$L:$$

${L \text { = }}\kern0pt{\left. {\left[ {\sqrt {{x^2} + 1} + \frac{1}{2}\ln \left| {\frac{{\sqrt {{x^2} + 1} – 1}}{{\sqrt {{x^2} + 1} + 1}}} \right|} \right]} \right|_{\sqrt 3 }^{\sqrt {15} } }={ \left( {4 + \frac{1}{2}\ln \frac{{4 – 1}}{{4 + 1}}} \right) – \left( {2 + \frac{1}{2}\ln \frac{{2 – 1}}{{2 + 1}}} \right) }={ 2 + \frac{1}{2}\left( {\ln \frac{3}{5} }-{ \ln \frac{1}{3}} \right) }={ 2 + \frac{1}{2}\ln \frac{9}{5}.}$

### Example 5.

Find the arc length of the curve $$y = \ln \sec x$$ from $$x = 0$$ to $$x = \large{\frac{\pi }{3}}\normalsize.$$

Solution.

The derivative of the given function is

${y^\prime = f^\prime\left( x \right) }={ \left( {\ln \sec x} \right)^\prime }={ \frac{1}{{\sec x}}\left( {\sec x} \right)^\prime }={ \frac{{\left( { – {{\sec }^2}x} \right)\left( { – \sin x} \right)}}{{\sec x}} }={ \sec x\sin x }={ \tan x.}$

The arc length is determined by the integral

${L = \int\limits_a^b {\sqrt {1 + {{\left[ {f’\left( x \right)} \right]}^2}} dx} }={ \int\limits_0^{\frac{\pi }{3}} {\sqrt {1 + {{\tan }^2}x} dx} }={ \int\limits_0^{\frac{\pi }{3}} {\sqrt {{{\sec }^2}x} dx} }={ \int\limits_0^{\frac{\pi }{3}} {\sec xdx} .}$

Since

$\int {\sec xdx} = \ln \left| {\sec x + \tan x} \right| + C,$

${L \text{ = }}\kern0pt{\left. {\left[ {\ln \left| {\sec x + \tan x} \right|} \right]} \right|_0^{\frac{\pi }{3}} }={ \ln \left| {\sec \frac{\pi }{3} + \tan \frac{\pi }{3}} \right| }-{ \ln \left| {\sec 0 + \tan 0} \right| }={ \ln \left( {2 + \sqrt 3 } \right) – \underbrace {\ln \left( {1 + 0} \right)}_0 }={ \ln \left( {2 + \sqrt 3 } \right).}$

### Example 6.

Find the arc length of the exponential curve $$y = {e^x}$$ from $$x = 0$$ to $$x = 1.$$

Solution.

We compute the arc length using by integration:

${L = \int\limits_0^1 {\sqrt {1 + {{\left[ {y^\prime\left( x \right)} \right]}^2}} dx} }={ \int\limits_0^1 {\sqrt {1 + {e^{2x}}} dx} .}$

To evaluate the integral, we make the following substitution:

${1 + {e^{2x}} = {u^2},}\;\; \Rightarrow {{e^{2x}}dx = udu,}\;\; \Rightarrow {dx = \frac{{udu}}{{{e^{2x}}}} = \frac{{udu}}{{{u^2} – 1}}.}$

The indefinite integral is given by

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

Then the definite integral is written as

${L \text{ = }}\kern0pt{\left. {\left[ {\sqrt {1 + {e^{2x}}} + \frac{1}{2}\ln \left| {\frac{{\sqrt {1 + {e^{2x}}} – 1}}{{\sqrt {1 + {e^{2x}}} + 1}}} \right|} \right]} \right|_0^1 }={ \left( {\sqrt {1 + {e^2}} + \frac{1}{2}\ln \frac{{\sqrt {1 + {e^2}} – 1}}{{\sqrt {1 + {e^2}} + 1}}} \right) }-{ \left( {\sqrt 2 + \frac{1}{2}\ln \frac{{\sqrt 2 – 1}}{{\sqrt 2 + 1}}} \right).}$

We simplify the fractions under the logarithm sign:

${\frac{{\sqrt {1 + {e^2}} – 1}}{{\sqrt {1 + {e^2}} + 1}} }={ \frac{{{{\left( {\sqrt {1 + {e^2}} – 1} \right)}^2}}}{{{{\left( {\sqrt {1 + {e^2}} } \right)}^2} – {1^2}}} }={ \frac{{{{\left( {\sqrt {1 + {e^2}} – 1} \right)}^2}}}{{{e^2}}} }={ {\left( {\frac{{\sqrt {1 + {e^2}} – 1}}{e}} \right)^2},}$

${\frac{{\sqrt 2 – 1}}{{\sqrt 2 + 1}} }={ \frac{{{{\left( {\sqrt 2 } \right)}^2} – {1^2}}}{{{{\left( {\sqrt 2 + 1} \right)}^2}}} }={ {\left( {\frac{1}{{\sqrt 2 + 1}}} \right)^2}.}$

Hence, the arc length is equal to

${L = \sqrt {1 + {e^2}} }-{ \sqrt 2 }+{ \frac{1}{2}\ln {\left( {\frac{{\sqrt {1 + {e^2}} – 1}}{e}} \right)^2} }-{ \frac{1}{2}\ln {\left( {\frac{1}{{\sqrt 2 + 1}}} \right)^2} }={ \sqrt {1 + {e^2}} }-{ \sqrt 2 }+{ \ln \frac{{\sqrt {1 + {e^2}} – 1}}{e} }-{ \ln \frac{1}{{\sqrt 2 + 1}} }={ \sqrt {1 + {e^2}} }-{ \sqrt 2 }+{ \ln \left[ {\left( {\sqrt {1 + {e^2}} – 1} \right)\left( {\sqrt 2 + 1} \right)} \right] }-{ \underbrace {\ln e}_1 }={ \sqrt {1 + {e^2}} }-{ \sqrt 2 – 1 }+{ \ln \left[ {\left( {\sqrt {1 + {e^2}} – 1} \right)\left( {\sqrt 2 + 1} \right)} \right]}\approx {2.004}$

### Example 7.

Find the length of one arc of the cycloid given in parametric form by the equations $$x\left( t \right) = t – \sin t,$$ $$y\left( t \right) = 1 – \cos t.$$

Solution.

The arc length of a parametric curve is expressed by the integral

$L = \int\limits_{{t_1}}^{{t_2}} {\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .$

For the first arc of cycloid, $$0 \le t \le 2\pi .$$ Hence,

${L \text{ = }}\kern0pt{\int\limits_0^{2\pi } {\sqrt {{{\left( {1 – \cos t} \right)}^2} + {{\sin }^2}t} \,dt} }={ \int\limits_0^{2\pi } {\sqrt {1 – 2\cos t + \underbrace {{{\cos }^2}t + {{\sin }^2}t}_1} \,dt} }={ \int\limits_0^{2\pi } {\sqrt {2 – 2\cos t} \,dt} }={ \int\limits_0^{2\pi } {\sqrt {4{{\sin }^2}\frac{t}{2}} dt} }={ 2\int\limits_0^{2\pi } {\sin \frac{t}{2}dt} }={ 4\left. {\left( { – \cos \frac{t}{2}} \right)} \right|_0^{2\pi } }={ 4\left( { – \cos \pi + \cos 0} \right) }={ 8.}$

### Example 8.

Find the arc length of the astroid $$x\left( t \right) = {\cos ^3}t,$$ $$y\left( t \right) = {\sin^3}t.$$

Solution.

As the curve is given in parametric form, we use the formula

${L \text{ = }}\kern0pt{\int\limits_a^b {\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .}$

Find the derivatives:

${x^\prime\left( t \right) = \left( {{{\cos }^3}t} \right)^\prime }={ 3{\cos ^2}t\left( { – \sin t} \right) }={ – 3{\cos ^2}t\sin t,}$

${y^\prime\left( t \right) = \left( {{{\sin }^3}t} \right)^\prime }={ 3{\sin ^2}t\cos t.}$

We calculate the length of one arc of the astroid lying in the first quadrant and then multiply the result by $$4.$$ So, we have

${L \text{ = }}\kern0pt{4\int\limits_0^{\frac{\pi }{2}} {\sqrt {9\,{{\cos }^4}t\,{{\sin }^2}t + 9\,{{\sin }^4}t\,{{\cos }^2}t} \,dt} }={ 4\int\limits_0^{\frac{\pi }{2}} {\sqrt {9\,{{\sin }^2}t\,{{\cos }^2}t\underbrace {\left( {{{\cos }^2}t + {{\sin }^2}t} \right)}_1} dt} }={ 12\int\limits_0^{\frac{\pi }{2}} {\sin t\cos tdt} }={ 6\int\limits_0^{\frac{\pi }{2}} {\sin 2tdt} }={ – 3\left. {\cos 2t} \right|_0^{\frac{\pi }{2}} }={ – 3\left( {\cos \pi – \cos 0} \right) }={ 6.}$

### Example 9.

Find the length of the cardioid $$r = 1 + \cos \theta .$$

Solution.

The cardioid is given in polar coordinates. Therefore we use the formula

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

Due to symmetry, we calculate the arc length of the upper half of the cardioid (with $$\theta$$ ranging from $$0$$ to $$\pi$$) and multiply the result by $$2.$$ This yields

${L = \int\limits_\alpha ^\beta {\sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r’\left( \theta \right)} \right]}^2}} d\theta } }={ 2\int\limits_0^\pi {\sqrt {{{\left( {1 + \cos \theta } \right)}^2} + {{\left( { – \sin \theta } \right)}^2}} d\theta } }={ 2\int\limits_0^\pi {\sqrt {1 + 2\cos \theta + \underbrace {{{\cos }^2}\theta + {{\sin }^2}\theta }_1} \,d\theta } }={ 2\int\limits_0^\pi {\sqrt {2 + 2\cos \theta } \,d\theta } }={ 2\sqrt 2 \int\limits_0^\pi {\sqrt {1 + \cos \theta } \,d\theta } }={ 2\sqrt 2 \int\limits_0^\pi {\sqrt {2{{\cos }^2}\frac{\theta }{2}} d\theta } }={ 4\int\limits_0^\pi {\cos \frac{\theta }{2}d\theta } }={ 8\left. {\sin \frac{\theta }{2}} \right|_0^\pi }={ 8\left( {\sin \frac{\pi }{2} – \sin 0} \right) }={ 8.}$

### Example 10.

Find the length of the first turn of Archimedean spiral $$r\left( \theta \right) = \theta.$$

Solution.

The arc length of a curve in polar coordinates is given by the equation

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

Here we have $$\alpha = 0,$$ $$\beta = 2\pi,$$ $$r\left( \theta \right) = \theta,$$ $$r^\prime\left( \theta \right) = 1,$$ so the arc length is expressed by the integral

$L = \int\limits_0^{2\pi } {\sqrt {1 + {\theta ^2}} d\theta } .$

We use the trig substitution $$\theta = \tan t$$ to evaluate the integral. Substituting

${\theta = \tan t,\;\;}\kern0pt{\sqrt {1 + {\theta ^2}} = \sec t,\;\;}\kern0pt{d\theta = {\sec ^2}t\,dt,}$

we rewrite the indefinite integral in the form

${I = \int {\sqrt {1 + {\theta ^2}} d\theta } }={ \int {{{\sec }^3}tdt} .}$

Now we use the reduction formula

${\int {{{\sec }^3}tdt} }={ \frac{{\sec t\tan t}}{2} }+{ \frac{1}{2}\int {\sec tdt} .}$

Recall that the integral of secant is common and is given by

$\int {\sec tdt} = \ln \left| {\sec t + \tan t} \right|.$

Then

${\int {{{\sec }^3}tdt} }={ \frac{{\sec t\tan t}}{2} }+{ \frac{1}{2}\ln \left| {\sec t + \tan t} \right|.}$

Returning back to the variable $$x,$$ we obtain:

${I = \frac{{\theta \sqrt {1 + {\theta ^2}} }}{2} }+{ \frac{1}{2}\ln \left| {\sqrt {1 + {\theta ^2}} + \theta } \right|.}$

Hence, the arc length is

${L \text{ = }}\kern0pt{\left. {\left[ {\frac{{\theta \sqrt {1 + {\theta ^2}} }}{2} + \frac{1}{2}\ln \left| {\sqrt {1 + {\theta ^2}} + \theta } \right|} \right]} \right|_0^{2\pi } }={ \left( {\frac{{\cancel{2}\pi \sqrt {1 + 4{\pi ^2}} }}{\cancel{2}} }\right.}+{\left.{ \frac{1}{2}\ln \left| {\sqrt {1 + 4{\pi ^2}} + 2\pi } \right|} \right) }-{ \underbrace {\left( {0 + \frac{1}{2}\ln 1} \right)}_0 }={ \pi \sqrt {1 + 4{\pi ^2}} + \frac{1}{2}\ln \left( {\sqrt {1 + 4{\pi ^2}} + 2\pi } \right)} \approx {21.26}$