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Dear Scott, I am having a problem with finding the length of the curve.
y=x^(2/3)between x=-1 and x=8.
I know that according to the formula: ds = sqrt(1 + (dy/dx)^2) dx
dy/dx = 2/3 x^-1/3
So ds = sqrt(1 + 4/9 x^(-2/3))
But then how to find the integral of the sqrt(1 + 4/9 x^(-2/3))?
And the range is between (-1, 8) so shall we substitute it (-1,0) and (0,8) and find the sum of it?
Or shall we find the length of inverse function x=y(3/2 between y=1 and y=4? then is it the same calculation?
Thank you.
This took quite a bit of work and I managed to get it wrong for half a dozen pages, but I think I finally got the right answer. See, once I get the answer, I start typing it in off my head. After I typed in a few lines, I check my work. What Im thinking and what I wrote down finally agree.
The problem, as I read it, is ∫√({1 + (2/[3x^(1/3)])}dx.
The 1st thing I did was let u = 2/(3x^(1/3)).
Since x is to the -1/3, the derivative will end up with x to the -4/3.
That means du = -2/(9x^(4/3)) dx.
Now u^4 = 16/(81x^(4/3)), so multiply both sides by -8/9.
This gives -8/9 du = -16/(81x^(4/3)) dx.
Now we can multiply both sides by u^-4.
This gives (-8/9)(1/u^4) du = dx.
The 1st transformation gives us the new problem -∫√(1+u)(8/9)(1/u^4)du.
The 2nd substitution was to go to trig functions.
Let u be the far side of the triangle, 1 be the near side,
and so the hypotenuse is √(1+uČ).
The angle I will call the next letter after u, which is v.
This transforms the integral into (8/9)∫((secv)/(tan^4(v))sec^2(v) dv.
Multiplying the sec^2(v) * sec v gives sec^3(v), so this is
-(8/9)∫sec^3(v)/(tan^4(v)).
I know a whole lot about sin() and cos(), so when the problem confuses me,
I switch to sin() and cos(). That changes this problem into
(8/9)∫(1/cos^3(v))/(sin^4(v)/(cos^4(v)) dv.
Invert the bottom and multiply, giving
(8/9)∫(1/cos^3(v))(cos^4(v)/sin^4(v))dv.
Note that cos^4/cos^3 = cos, so we we have
(8/9)∫cos(v)/sin^4(v) dv.
For my 3rd magical trick no, my 3rd substitution,
let w = sin v, so dw = cos v dv. (I think I see an answer coming)
This transforms the integral into (8/9)∫(1/w^4)dw.
(or at least, before we do all the back substitutuions)
That integral is (8/9)∫w^(-4)dw, which can be seen to be (8/9)w^(-3)/-3 =
(8/(27w^3)). Now all that has to be done is to note w = sin v,
v was from that triangle with u on the far side and 1 on the near side,
so tan(v)=u, and at that start, u=2/(3x^(1/3)).
Using the triangle with angle v, the other trig functions performed on v can be found.
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Scott A Wilson
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