Find-the-arc-length-of-the-curve-x-1-4-y-2-1-2-ln-y-between-the-points-with-the-ordinates-y-1-and-y-2-

Question Number 122588 by bramlexs22 last updated on 18/Nov/20

F i n d t h e arc l e n g t h o f t h e c u r v e

x = \frac{1}{4} y^{2} - \frac{1}{2} \ln (y) b e t w e e n t h e p o i n t s

w i t h t h e o r d i n a t e s y = 1 a n d y = 2 .

Answered by MJS_new last updated on 18/Nov/20

arc length of x = f (y) is given by

\int \sqrt{1 + {(x^{'})}^{2}} d y

x = \frac{1}{4} y^{2} - \frac{1}{2} \ln y is defined for y > 0

x^{'} = \frac{y}{2} - \frac{1}{2 y}

\sqrt{1 + {(x^{'})}^{2}} = \sqrt{\frac{{(y^{2} + 1)}^{2}}{4 y^{2}}} = \frac{y^{2} + 1}{2 y} [because y > 0]

\int \frac{y^{2} + 1}{2 y} d y = \int \frac{y}{2} + \frac{1}{2 y} d y = \frac{1}{4} y^{2} + \frac{1}{2} \ln y

\Rightarrow arc length is \frac{3}{4} + \frac{1}{2} \ln 2

Commented by bramlexs22 last updated on 18/Nov/20

t h a n k y o u b o t h s i r

Answered by liberty last updated on 18/Nov/20

ℓ = \underset{1}{\int^{2}} \sqrt{1 + {(x^{'})}^{2}} dy

\frac{dx}{dy} = \frac{1}{2} y - \frac{1}{2 y} \Rightarrow ℓ = \underset{1}{\int^{2}} \sqrt{1 + {(\frac{1}{2} y - \frac{1}{2 y})}^{2}} dy

ℓ = \underset{1}{\int^{2}} \sqrt{1 + \frac{1}{4} y^{2} + \frac{1}{{4 y}^{2}} - \frac{1}{2}} dy

ℓ = \underset{1}{\int^{2}} \sqrt{\frac{1}{4} y^{2} + \frac{1}{2} + \frac{1}{{4 y}^{2}}} dy

ℓ = \underset{1}{\int^{2}} (\frac{1}{2} y + \frac{1}{2 y}) dy = {[\frac{1}{4} y^{2} + \frac{\ln y}{2}]}_{1}^{2}

ℓ = \frac{3}{4} + \frac{\ln 2}{2} . ▴

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