Arc Length

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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 \(L:\)

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