Question

$\huge\underbrace {\color {orange}\bf \: Question \:}$

Find the length of the arc x² + y² - 2ax = 0 in the first quadrant​

Answer 1

Solution :-

The given curve is

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 2ax = 0$

which represents a circle. and can be rewritten as

$\rm :\longmapsto\: ({x}^{2} - 2ax) + {y}^{2} = 0$

$\rm :\longmapsto\: ({x}^{2} - 2ax + {a}^{2} - {a}^{2} ) + {y}^{2} = 0$

$\rm :\longmapsto\: {(x - a)}^{2} + {y}^{2} - {a}^{2} = 0$

$\rm :\longmapsto\: {(x - a)}^{2} + {y}^{2} = {a}^{2} - - - (1)$

which represents a Circle with

$\red{\rm :\longmapsto\:Center \: (a ,0)} \\ \red{\rm :\longmapsto\:radius \: = \: a \: }$

On Differentiating equation (1) w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}\bigg( {(x - a)}^{2} + {y}^{2} \bigg) = \dfrac{d}{dx} {a}^{2}$

$\rm :\longmapsto\:\dfrac{d}{dx} {(x - a)}^{2} + \dfrac{d}{dx} {y}^{2} = \dfrac{d}{dx} {a}^{2}$

$\rm :\longmapsto\:2(x - a) + 2y\dfrac{dy}{dx} = 0$

$\: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: \dfrac{d}{dx} {x}^{n} = n {x}^{ n- 1} \: and \: \dfrac{d}{dx}k = 0 \bigg \}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \dfrac{(x - a)}{y} - - - (2)$

For the length of arc of circle, in first quadrant, the values of x varies from x = 0 to x = 2a

Now, we know that

$\rm :\longmapsto\:Length \: of \: arc = \displaystyle\int^{2a}_{0} \sqrt{1 + {\bigg(\dfrac{dy}{dx} \bigg) }^{2} }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0} \sqrt{1 + {\bigg(\dfrac{ - (a - x)}{y} \bigg) }^{2} }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0} \sqrt{1 + {\bigg(\dfrac{a - x}{y} \bigg) }^{2} }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0} \sqrt{1 + \dfrac{ {(a - x)}^{2} }{ {y}^{2} } }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0} \sqrt{1 + \dfrac{ {(a - x)}^{2} }{ 2ax - {x}^{2} } }dx$

$\: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: {y}^{2} + {x}^{2} - 2ax = 0  \bigg \}}$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0} \sqrt{\dfrac{ 2ax - {x}^{2} + {(a - x)}^{2} }{ 2ax - {x}^{2} } }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0} \sqrt{\dfrac{ \cancel{ 2ax } - \: \cancel{{x}^{2}} + \cancel {x}^{2} + {a}^{2} - \cancel{2ax}}{ 2ax - {x}^{2} } }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0}\dfrac{a}{ \sqrt{2ax - {x}^{2} } } dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0}\dfrac{a}{ \sqrt{ - ({x}^{2} - 2ax)} } dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0}\dfrac{a}{ \sqrt{ - ({x}^{2} - 2ax + {a}^{2} - {a}^{2} )} } dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0}\dfrac{a}{ \sqrt{ - ({(x - a)}^{2} - {a}^{2} )} } dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{2a}_{0}\dfrac{a}{ \sqrt{ {a}^{2} - {(x - a)}^{2}} } dx$

$\rm \:  \:  =  \:  \: a\bigg( {sin}^{ - 1}\dfrac{x - a}{a} \bigg)_0^{2a}$

$\: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \:  \displaystyle\int \tt \: \dfrac{dx}{ {a}^{2} - {x}^{2}} = {sin}^{ - 1}\dfrac{x}{a} + c \bigg \}}$

$\rm \:  \:  =  \:  \: a\bigg( {sin}^{ - 1}\dfrac{2a - a}{a} - {sin}^{ - 1}\dfrac{0 - a}{a} \bigg)$

$\rm \:  \:  =  \:  \: a\bigg( {sin}^{ - 1}\dfrac{a}{a} - {sin}^{ - 1}\dfrac{ - a}{a} \bigg)$

$\rm \:  \:  =  \:  \: a( {sin}^{ - 1}1 + {sin}^{ - 1} 1)$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \:  {sin}^{ - 1} ( - x) = - {sin}^{ - 1}x\bigg \}}$

$\rm \:  \:  =  \:  \: a\bigg(\dfrac{\pi}{2} + \dfrac{\pi}{2} \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: {sin}^{ - 1} 1 = \dfrac{\pi}{2}\bigg \}}$

$\rm \:  \:  =  \:  \: a\pi \: units$

$\: \: \: \: \: \: \: \boxed{\bf\implies \:Length \: of \: arc \: = \: a\pi \: units}$

Additional Information :-

1. Length of arc of parametric curve is

$\rm :\longmapsto\:Length \: of \: arc = \displaystyle\int^{t_2}_{t_1} \sqrt{ {\bigg(\dfrac{dx}{dt} \bigg) }^{2} + {\bigg(\dfrac{dy}{dt} \bigg) }^{2} }dt$

2. Length of arc of polar curves is

$\rm :\longmapsto\:Length \: of \: arc = \displaystyle\int^{ \beta }_{ \alpha } \sqrt{ {r}^{2} + {\bigg(\dfrac{dr}{d \theta} \bigg) }^{2} }d \theta$