Arc Length
Solved Problems
Example 5.
Find the arc length of the curve \[y = \ln \sec x\] from \(x = 0\) to \(x = \frac{\pi }{3}.\)
Solution.
Figure 5. 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
\[\int {\sec xdx} = \ln \left| {\sec x + \tan x} \right| + C,\]
we get the following answer:
\[L = \left. {\left[ {\ln \left| {\sec x + \tan x} \right|} \right]} \right|_0^{\frac{\pi }{3}} = \ln \left| {\sec \frac{\pi }{3} + \tan \frac{\pi }{3}} \right| - \ln \left| {\sec 0 + \tan 0} \right| = \ln \left( {2 + \sqrt 3 } \right) - \underbrace {\ln \left( {1 + 0} \right)}_0 = \ln \left( {2 + \sqrt 3 } \right).\]
Example 6.
Find the arc length of the curve \[y = {e^x}\] from \(x = 0\) to \(x = 1.\)
Solution.
Figure 6. 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
\[L = \left. {\left[ {\sqrt {1 + {e^{2x}}} + \frac{1}{2}\ln \left| {\frac{{\sqrt {1 + {e^{2x}}} - 1}}{{\sqrt {1 + {e^{2x}}} + 1}}} \right|} \right]} \right|_0^1 = \left( {\sqrt {1 + {e^2}} + \frac{1}{2}\ln \frac{{\sqrt {1 + {e^2}} - 1}}{{\sqrt {1 + {e^2}} + 1}}} \right) - \left( {\sqrt 2 + \frac{1}{2}\ln \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \right).\]
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.
Figure 7. 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,
\[L = \int\limits_0^{2\pi } {\sqrt {{{\left( {1 - \cos t} \right)}^2} + {{\sin }^2}t} \,dt} = \int\limits_0^{2\pi } {\sqrt {1 - 2\cos t + \underbrace {{{\cos }^2}t + {{\sin }^2}t}_1} \,dt} = \int\limits_0^{2\pi } {\sqrt {2 - 2\cos t} \,dt} = \int\limits_0^{2\pi } {\sqrt {4{{\sin }^2}\frac{t}{2}} dt} = 2\int\limits_0^{2\pi } {\sin \frac{t}{2}dt} = 4\left. {\left( { - \cos \frac{t}{2}} \right)} \right|_0^{2\pi } = 4\left( { - \cos \pi + \cos 0} \right) = 8.\]
Example 8.
Find the arc length of the astroid \[x\left( t \right) = {\cos ^3}t,\,y\left( t \right) = {\sin^3}t.\]
Solution.
Figure 8. As the curve is given in parametric form, we use the formula
\[L = \int\limits_a^b {\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .\]
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
\[L = 4\int\limits_0^{\frac{\pi }{2}} {\sqrt {9\,{{\cos }^4}t\,{{\sin }^2}t + 9\,{{\sin }^4}t\,{{\cos }^2}t} \,dt} = 4\int\limits_0^{\frac{\pi }{2}} {\sqrt {9\,{{\sin }^2}t\,{{\cos }^2}t\underbrace {\left( {{{\cos }^2}t + {{\sin }^2}t} \right)}_1} dt} = 12\int\limits_0^{\frac{\pi }{2}} {\sin t\cos tdt} = 6\int\limits_0^{\frac{\pi }{2}} {\sin 2tdt} = - 3\left. {\cos 2t} \right|_0^{\frac{\pi }{2}} = - 3\left( {\cos \pi - \cos 0} \right) = 6.\]
Example 9.
Find the length of the cardioid \[r\left( \theta \right) = 1 + \cos \theta .\]
Solution.
Figure 9. 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.
Figure 10. 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
\[\theta = \tan t,\;\;\sqrt {1 + {\theta ^2}} = \sec t,\;\;d\theta = {\sec ^2}t\,dt,\]
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
\[L = \left. {\left[ {\frac{{\theta \sqrt {1 + {\theta ^2}} }}{2} + \frac{1}{2}\ln \left| {\sqrt {1 + {\theta ^2}} + \theta } \right|} \right]} \right|_0^{2\pi } = \left( {\frac{{\cancel{2}\pi \sqrt {1 + 4{\pi ^2}} }}{\cancel{2}} + \frac{1}{2}\ln \left| {\sqrt {1 + 4{\pi ^2}} + 2\pi } \right|} \right) - \underbrace {\left( {0 + \frac{1}{2}\ln 1} \right)}_0 = \pi \sqrt {1 + 4{\pi ^2}} + \frac{1}{2}\ln \left( {\sqrt {1 + 4{\pi ^2}} + 2\pi } \right) \approx 21.26\]
