Question

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.

Answer 1

\[y = (1/3)(x ^{2}+2)^{(3/2)}\] (Finding the length of this curve on the interval x = 0 to x = 3)

Answer 2

\[\frac{ dy }{ dx } = (\frac{ 1 }{ 3 })(\frac{ 3 }{ 2 })\sqrt{x ^{2}+2}(2x) = x \sqrt{x ^{2}+2}\]
The formula for finding the length of a curve is:\[\int\limits_{a}^{b}\sqrt{1+f'(x)^{2}}dx\]

Answer 3

so wats problem in this ???

Answer 4

\[(\frac{ dy }{ dx })^{2} = x ^{4}+2x ^{2}\]\[\int\limits_{0}^{3}\sqrt{1+x ^{4}+2x ^{2}}dx\]
u-substitution right here is where stuff gets confusing, there's some intermediate steps of simplifying, but it goes from that to:

Answer 5

No need to u-sub. factor it like a quadratic : )

Answer 6

\[u = x ^{2}, du = 2xdx\]\[\frac{ 1+u }{ \sqrt{u} }du\] 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.

Answer 7

(Forgot the integral sign & limits of integration.)

Answer 8

if u subsictute u=x2+2 than how will it work

Answer 9

What? I said u was x^2, not x^2 + 2.

Answer 10

so u r given u=x2

Answer 11

u hv to suppose

Answer 12

Oh, and WHOAH is factoring it like a quadratic way easier, thanks for mentioning that, Psymon. \[\sqrt{1 + x ^{4}+2x ^{2}} = \sqrt{(x ^{2}+1)(x ^{2}+1)} = x ^{2}+1\]\[\int\limits_{0}^{3}(x ^{2}+1)dx = \frac{ x ^{3} }{ 3 }+x |||\]

Answer 13

But, with the u-sub, what do I have to suppose?

Answer 14

1+u/√u du= (1+x2/x )2x.dx

=(1+x2)2.dx

Answer 15

Oh, I forgot a two in the u-sub.

Answer 16

nd then integrate (1+x2)2.dx
will give u ur required answer

Answer 17

\[\int\limits_{?}^{?}\frac{ 1 + u}{ 2\sqrt{u} }du\]

Answer 18

oh man u give me that ok go with this as u soppoed now now also put limits

Answer 19

r u satisfield @Mendicant_Bias

Answer 20

Yeah, thanks, I got it now.

Answer 21

\[\int\limits_{0}^{9}\frac{ 1+u }{ 2\sqrt{u} }du = \int\limits_{0}^{3}(1+x ^{2})dx\]\[\frac{ 1 + u }{ 2\sqrt{u} }du = \frac{ 1 + x ^{2} }{ 2x }2xdx = \frac{ 2x }{ 2x }+ \frac{ 2x ^{3} }{ 2x } = 1+x ^{2}\]

Answer 22

(Oh, and of course, tacking on that dx at the end, but you got the point.)

Answer 23

u missed 2 1st ... now its complete i think

Answer 24

And evaluate either at their respective limits with respect to their variables, integrands, and integration variables. Yep, I did miss 2 at first.

Answer 25

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 +

Answer 26

yup its too easy .... i hv had a lot of prctice with this kind of questions for past many years

Answer 27

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.

Answer 28

m doing electrical engineering now in 3rd year ....from a school of uni of bradford uk.

Answer 29

Nice! Well, good luck with everything.

Answer 30

thanks man.

Answer 31

u too