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REALLY hard integral?
Seriously.
y^2 = 4x
Find the arc length of this curve for y within [0,2]
I tried so many approaches... I know the formula, but I've gotten to integral[0,2](sqrt((1/4)+y^2)), and I've already tried trig substitution. I got 1.161695941, but that's apparently not right, according to my online homework system...
Just in case anyone needs it, and they probably will, I've figured out the indefinite integral of sec^3(theta) before, and it is:
(1/2)[sec(theta)tan(theta)+ln|sec(theta)+tan(theta)|]+C
That cut off...
(1/2)[sec(theta)tan(theta)]
+
(1/2)ln|sec(theta)+tan(theta)|+C
Never mind, I got the answer by developing a calculator program to solve integrals like this, and the answer (only in decimal form, since it's a TI-83) is 2.295587149. Now I just need to know how to do it...
Scratch that, I figured it out.
In that case, I might as well write out how I did the problem! :D
Arc Length = integral[0,2](sqrt(1+(y^2/4)))
= integral[0,2]((1/2)*sqrt(4+y^2))),
y = 2tan(theta)
dy = 2sec^2(theta)dtheta
sqrt(4+y^2) = 2sec(theta)
= (1/2)*integral[0,2](2sec(theta)
*2sec^2(theta)dtheta)
= 2*integral[0,2](sec^3(theta)dtheta)
Haha, now this is hard...
= 2((1/2)[sec(theta)tan(theta)]+
(1/2)ln|sec(theta)+tan(theta)|)
= {[sec(theta)tan(theta)]+
ln|sec(theta)+tan(theta)|}
eval @ [0,2]
And from there, that's the answer. :D Wow, it's so epic it's cutting off without linebreaks.
3 Answers
- δοτζοLv 79 years agoFavorite Answer
s = ∫ ds = ∫[0,2] √(1 + (dx/dy)²) dy = ∫[0,2] √(1 + (y/2)²) dy
u = y/2 ⇒ u(0) = 0, u(2) = 1
du = dy/2
2 ∫[0,1] √(1 + u²) du
And I'm betting you can take it from there. It's always best to get rid of fractions whenever possible. They're your friends, but sometimes you just want to be alone.
- NickLv 69 years ago
LOOK AGAIN, I JUST EDITED MY ANSWER, MADE A STUPID CALCULATION ERROR.
Have you tried substituting hyperbolic sine:
ds = (1 + (dx/dy)^2)^(1/2) dy (1)
dx/dy = y/2
ds = (1 + (y/2)^2)^(1/2) dy (2)
y/2 = sinh(u) => dy/du = 2cosh(u):
ds = (1 + sinh^2(u))^(1/2) 2cosh(u) du
use the hyperbolic identity cosh^2(u) - sinh^2(u) = 1 to give:
ds = 2cosh^2(u) du
also there is another hyperbolic identity 2cosh^2(u) = cosh(2u) + 1:
ds = (cosh(2u) + 1) du
The limits are found since u = arsinh(y/2) = ln((y/2) + ((y/2)^2 + 1)^(1/2)), so for y=0 u=0 and for y=2 u=ln(1 + (2)^(1/2)) = 0.881.
Int(ds) = [(1/2)sinh(2u) + u]_0^0.881
arc length = Ds = 2.294
All the identities and information of hyperbolic functions is here:
- 9 years ago
y^2 = 4x
2y * dy = 4 * dx
dy/dx = 2/y
(dy/dx)^2 = 4/y^2
(dy/dx)^2 = 4/(4x)
(dy/dx)^2 = 1/x
sqrt(1 + (dy/dx)^2) * dx =>
sqrt(1 + (1/x)) * dx =>
sqrt(1 + x) * dx / sqrt(x)
x = tan(t)^2
dx = 2 * tan(t) * sec(t)^2 * dt
sqrt(1 + tan(t)^2) * 2 * tan(t) * sec(t)^2 * dt / sqrt(tan(t)^2) =>
sqrt(sec(t)^2) * 2 * tan(t) * sec(t)^2 * dt / tan(t) =>
sec(t) * 2 * sec(t)^2 * dt =>
2 * sec(t)^3 * dt
Integrate
2 * (1/2) * (sec(t) * tan(t) + ln|sec(t) + tan(t)|) + C =>
sec(t) * tan(t) + ln|sec(t) + tan(t)|
x = tan(t)^2
0 = tan(t)^2
tan(t) = 0
sec(t) = sqrt(1 + tan(t)^2) = 1
2 = tan(t)^2
tan(t) = sqrt(2)
sec(t) = sqrt(1 + tan(t)^2) = sqrt(1 + 2) = sqrt(3)
sec(t) * tan(t) + ln|sec(t) + tan(t)|
(sqrt(3) * sqrt(2) + ln|sqrt(3) + sqrt(2)|) - (1 * 0 + ln|1 + 0|) =>
sqrt(6) + ln|sqrt(2) + sqrt(3)| - 0 =>
sqrt(6) + ln|sqrt(2) + sqrt(3)| =>
3.5957055775637669420976777303799
