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integral arclength question need help im stuck?
find the arclength of the curve sqrt(4-x^2) from 0 to 2.
To set this up, i take the derivative and use the arc length equation. For the derivative, i get -x/sqrt(4-x^2). But when I can work out the integral correctly. When I do i end up dividing my zero because of the upper limit of integration. sqrt(4-x^2) @x=2 it will be 0.
I need help ive hit a dead end.
2 Answers
- Astral WalkerLv 71 decade agoFavorite Answer
Formula for arc length is the integral of sqrt[1 + (y')^2]
y' = -x/sqrt(4 - x^2) --> (y')^2 = x^2/(4 - x^2)
1 + (y')^2 = (4 - x^2 + x^2)/(4 - x^2) = 4/(4 - x^2)
sqrt[1 + (y')^2] = 2/sqrt(4 - x^2)
The integral of this is just 2arcsin(x/2)
Evaluating from 0 to 2 is 2[arcsin(1) - 2arcsin(0)]
Since arcsin(1) = pi/2 and arcsin(0) = 0 the answer is pi
- grunfeldLv 71 decade ago
y = sqrt ( 4 - x^2 )
dy / dx = - 2x(1 / 2)( 4 - x^2 )^( - 1 / 2)
dy / dx = - x ( 4 - x^2 )^( - 1 / 2 )
int { sqrt [ 1 + x^2(4 - x^2)^( - 1 ) ] dx }
= 4 * int { dx / (4 - x^2 ) }
Let x = 2 sin t
dx = 2 cos t dt
= 4 * int [ 2 cos t dt / (4 - 4 sin^2 t ) ]
= int [ 2 cos t dt / ( cos^2 t ) ]
= 4 * int ( 2 sec t dt )
= 8 * int ( sec t dt )
= 8* ln ( sec t + tan t ) + C
= 8 * ln ( sec (arcsin ( x / 2 ) ) + tan ( arcsin ( x / 2 ) ) ) + C
Source(s): my brain
