For a function \( f \) that is continuous on the \( [a,b] \), the length of the curve \( y = f(x) \) from \( a \) to \( b \) is given by [1] [2] [3]
\[ \int_{a}^{b} \; \sqrt{1+\left( \dfrac{df}{dx} \right)^2 }\; dx \]
Fig.1 - Length of a Curve From the Point \( (a,f(a)) \) to the Point \( (b,f(b)) \)
Examples and Solutions
Example 1
Find the length of the arc of the parabola \( y = 0.1 x^2 + 2 \) between the points \( (-15,24.5) \) and \( (10,12) \).
Fig.2 - Arc Length of a Parabola
Solution to Example 1
We first calculate the derivative
\[ \dfrac{dy}{dx} = 0.2 x \]
Use the formula for the arc length given above
\[ L = \int_{-15}^{10} \; \sqrt{1+\left( 0.2 x \right)^2 }\; dx \]
Use trigonometric substitution \( \quad \tan u = 0.2 x \)
Take the derivative of both sides of the above substitution
\( \sec ^2 (x) \dfrac{du}{dx} = 0 .2 \) which gives \( dx = 5 \sec ^2 u du \)
Solving \( \quad \tan u = 0.2 x \) for \( u \) gives \( u = \arctan (0.2 x) \)
Limits of integration after substitution
\( u_1 = \arctan (0.2(-15)) \approx -1.24904 \) when \( x = - 15 \) the lower limit of the integral
\( u_2 = \arctan (0.2(10)) \approx 1.10714 \) when \( x = 10 \) the upper limit of the integral
\[ L = 5 \int_{\arctan (0.2(-15))}^{\arctan (0.2(10))} \; \sqrt{1+\tan^2 u }\; dx \]
Use the identity \( \sqrt{1 + \tan^2 u } = |\sec u| \) and make the substitution \( \quad dx = 5 \sec ^2 u du \) to obtain
\[ 5 \int_{\arctan (0.2(-15))}^{\arctan (0.2(10))} \; |\sec u| \sec ^2u \; du \]
For \( u \) the interval \( \left[ \arctan (0.2(-15)) , \arctan (0.2(10)) \right] \), \( \sec u \) is positive. Hence the integral becomes
\[ L = 5 \int_{\arctan (0.2(-15))}^{\arctan (0.2(10))} \; \sec ^3u \; du \]
Use the indefinite integral of \( \sec ^3u \) given by
\[ \int \sec^3 x \; dx = \dfrac{1}{2} \left( \tan x \; \sec x + \ln |\tan x + \sec x| \right) + c \]
to evaluate the arc length
\[ L = \left[ \dfrac{1}{2} \left( \sec u \tan u + \ln| \sec u + \tan u| \right) \right]_{\arctan (0.2(-15))}^{\arctan (0.2(10))} \\\\ \approx 43.05 \]
Example 2
Find the length of the arc along the curve \( f(x) = \ln(\sin x) \) between the points \( (\dfrac{\pi}{4}, f(\dfrac{\pi}{4})) \) and \( (\dfrac{\pi}{2}, f(\dfrac{\pi}{2})) \).
Fig.3 - Arc Length Along the Curve \( y = \ln (\sin x)) \)
Solution to Example 2
Calculate the derivative
\[ \dfrac{dy}{dx} = \dfrac{\ln(\sin x)}{dx} \\ = \cot x\]
Applying the formula for the arc alength
\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \; \sqrt{1+( \cot x)^2 } \; dx \]
Use the trigonometric identity \( 1+(\cot x)^2 = csc^2 x \), \( L \) becomes
\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \; |\csc x| \; dx \]
\( \csc x \) is positive on the the closed interval of integration \( [ \dfrac{\pi}{4} , \dfrac{\pi}{2} ] \) and therefore \( |\csc x| = \csc x \) and \( L \) becomes
\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \; \csc x \; dx \]
Use the common integral of \( \csc x \) : \( \displaystyle \int \csc x \; dx = \ln \left|\tan \left(\dfrac{x}{2}\right)\right|+C \) to write
\[ L = \left[ \ln \left|\tan \left(\dfrac{x}{2}\right)\right| \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \approx 0.88137 \]
Example 3
Find the length of the arc along the half circle given by \( f(x) = 2 + \sqrt {9 - (x+2)^2} \) between the points \( (-4, f(-4) \) and \( (0, f(0) \).
Fig.4 - Arc Length Along the Half Circle \( y = 2 + \sqrt {9 - (x+2)^2} \)
Solution to Example 3
Calculate the derivative
\[ \dfrac{df}{dx} = -\dfrac{x+2}{\sqrt{9 - (x+2)^2}} \]
Applying the formula for the arc alength
\[ L = \int_{-4}^{0} \; \sqrt{1+( -\dfrac{x+2}{\sqrt{9 - (x+2)^2}} )^2 } \; dx \]
The above integral is a challenging one and therefore Symbolab software was used to approximate the above integral numerically.
\[ L \approx 4.29 \]
More References and Links
University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8
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