# Length of a Curve

   

## Formula of Length of a Curve

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    $\int_{a}^{b} \; \sqrt{1+\left( \dfrac{df}{dx} \right)^2 }\; dx$ 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 $(10,12)$.

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. $L \approx 4.29$