Mendicant_Bias
  • Mendicant_Bias
Another Calculus II question regarding arc length, posted below in a minute. The only thing I actually need help with is a final step of integration involving u-substitution.
Mathematics
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jamiebookeater
  • jamiebookeater
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Mendicant_Bias
  • Mendicant_Bias
\[y = (1/3)(x ^{2}+2)^{(3/2)}\] (Finding the length of this curve on the interval x = 0 to x = 3)
Mendicant_Bias
  • Mendicant_Bias
\[\frac{ dy }{ dx } = (\frac{ 1 }{ 3 })(\frac{ 3 }{ 2 })\sqrt{x ^{2}+2}(2x) = x \sqrt{x ^{2}+2}\] The formula for finding the length of a curve is:\[\int\limits_{a}^{b}\sqrt{1+f'(x)^{2}}dx\]
anonymous
  • anonymous
so wats problem in this ???

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Mendicant_Bias
  • Mendicant_Bias
\[(\frac{ dy }{ dx })^{2} = x ^{4}+2x ^{2}\]\[\int\limits_{0}^{3}\sqrt{1+x ^{4}+2x ^{2}}dx\] u-substitution right here is where stuff gets confusing, there's some intermediate steps of simplifying, but it goes from that to:
Psymon
  • Psymon
No need to u-sub. factor it like a quadratic : )
Mendicant_Bias
  • Mendicant_Bias
\[u = x ^{2}, du = 2xdx\]\[\frac{ 1+u }{ \sqrt{u} }du\] Psymon: I will, but if I could, I'd like to see how I can simplify this in order to do it with u-subsitution.
Mendicant_Bias
  • Mendicant_Bias
(Forgot the integral sign & limits of integration.)
anonymous
  • anonymous
if u subsictute u=x2+2 than how will it work
Mendicant_Bias
  • Mendicant_Bias
What? I said u was x^2, not x^2 + 2.
anonymous
  • anonymous
so u r given u=x2
anonymous
  • anonymous
u hv to suppose
Mendicant_Bias
  • Mendicant_Bias
Oh, and WHOAH is factoring it like a quadratic way easier, thanks for mentioning that, Psymon. \[\sqrt{1 + x ^{4}+2x ^{2}} = \sqrt{(x ^{2}+1)(x ^{2}+1)} = x ^{2}+1\]\[\int\limits_{0}^{3}(x ^{2}+1)dx = \frac{ x ^{3} }{ 3 }+x |||\]
Mendicant_Bias
  • Mendicant_Bias
But, with the u-sub, what do I have to suppose?
anonymous
  • anonymous
1+u/√u du= (1+x2/x )2x.dx =(1+x2)2.dx
Mendicant_Bias
  • Mendicant_Bias
Oh, I forgot a two in the u-sub.
anonymous
  • anonymous
nd then integrate (1+x2)2.dx will give u ur required answer
Mendicant_Bias
  • Mendicant_Bias
\[\int\limits_{?}^{?}\frac{ 1 + u}{ 2\sqrt{u} }du\]
anonymous
  • anonymous
oh man u give me that ok go with this as u soppoed now now also put limits
anonymous
  • anonymous
r u satisfield @Mendicant_Bias
Mendicant_Bias
  • Mendicant_Bias
Yeah, thanks, I got it now.
Mendicant_Bias
  • Mendicant_Bias
\[\int\limits_{0}^{9}\frac{ 1+u }{ 2\sqrt{u} }du = \int\limits_{0}^{3}(1+x ^{2})dx\]\[\frac{ 1 + u }{ 2\sqrt{u} }du = \frac{ 1 + x ^{2} }{ 2x }2xdx = \frac{ 2x }{ 2x }+ \frac{ 2x ^{3} }{ 2x } = 1+x ^{2}\]
Mendicant_Bias
  • Mendicant_Bias
(Oh, and of course, tacking on that dx at the end, but you got the point.)
anonymous
  • anonymous
u missed 2 1st ... now its complete i think
Mendicant_Bias
  • Mendicant_Bias
And evaluate either at their respective limits with respect to their variables, integrands, and integration variables. Yep, I did miss 2 at first.
Psymon
  • Psymon
Usually with these problems they make it very convenient to factor. Either it factors pretty obviously, like we just had, we may need to make common denominators with fractions and factor something out of the square root, too. A lot of the time, the simplification of what isinside of the square root is merely a sign change difference from what was before you plugged it into the square root and squared it. For example, here's a quick worked out problem where this becomes the case. So let the function be: \[y = \frac{ x ^{3} }{ 6 }+\frac{ 1 }{ 2x }\]taking the derivative I get: \[\frac{ 3x ^{2} }{ 6 }-\frac{ 1 }{ 2x ^{2} }=\frac{ 1 }{ 2 }(x ^{2}-\frac{ 1 }{ x ^{2} })\]So this is what I would plug in, square underneath the square root and then add 1 to. Doing that I get this: \[\sqrt{1+\frac{ 1 }{ 4 }(x ^{4}-2+\frac{ 1 }{ x ^{4} })}=\sqrt{\frac{ 1 }{ 4 }(x ^{4}+2+\frac{ 1 }{ x ^{4} })}=\frac{ 1 }{ 2 }(x ^{2}+\frac{ 1 }{ x ^{2} })\] Sorry for the wait. I just wanted to show something to look out for. Apart from factoring and messing around with whats inside of the square root so you can factor things outside, a lot of the time the difference in what you finish with is merely a sign change. What I squared was x^2 - 1/x^2, what came out was the same but with a +
anonymous
  • anonymous
yup its too easy .... i hv had a lot of prctice with this kind of questions for past many years
Mendicant_Bias
  • Mendicant_Bias
Have you graduated? I'm an Electrical Engineering/Physics undergrad, just starting my second year, I'm not really good with math but I'm certainly improving.
anonymous
  • anonymous
m doing electrical engineering now in 3rd year ....from a school of uni of bradford uk.
Mendicant_Bias
  • Mendicant_Bias
Nice! Well, good luck with everything.
anonymous
  • anonymous
thanks man.
anonymous
  • anonymous
u too

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