Re: Integration Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg102033] Re: [mg102004] Integration Problem
- From: "David Park" <djmpark at comcast.net>
- Date: Mon, 27 Jul 2009 05:54:58 -0400 (EDT)
- References: <6165421.1248595398258.JavaMail.root@n11>
Generally things go better in Mathematica if you can keep them strictly
symbolic as long as possible - or go directly to numeric routines. (You can
copy and paste this stuff into a Mathematica notebook.)
x[t_] := Exp[2 t]
y[t_] := Exp[3/2 t]
\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(1\), \(2\)]\(
SqrtBox[\(
\*SuperscriptBox[\(\(x'\)[t]\), \(2\)] +
\*SuperscriptBox[\(\(y'\)[t]\), \(2\)]\)] \[DifferentialD]t\)\)
% // N
giving
1/512 (-128 E^(3/2) Sqrt[9 + 16 E] - 36 Sqrt[E (9 + 16 E)] + (324 E)/
Sqrt[9 + 16 E^2] + (1728 E^3)/Sqrt[9 + 16 E^2] + (2048 E^5)/Sqrt[
9 + 16 E^2] + 81 ArcSinh[(4 Sqrt[E])/3] - 81 ArcSinh[(4 E)/3])
49.7621
Notice that I used an exact expression, 3/2 instead of 1.5. (Also, by
writing formal definitions for x and y and using the standard definition of
arc length we eliminate some "hand" steps that might introduce errors - but
you didn't make any errors there.) Or we can go straight to a numerical
routine:
NIntegrate[Sqrt[(2 Exp[2*m])^2 + (1.5 Exp[1.5*m])^2], {m, 1, 2}]
49.7621
But the hybrid statement doesn't work:
N[Integrate[Sqrt[(2 Exp[2*m])^2 + (1.5 Exp[1.5*m])^2], {m, 1, 2}]]
79.6261
Or the same statement without the N:
Integrate[Sqrt[(2 Exp[2*m])^2 + (1.5 Exp[1.5*m])^2], {m, 1, 2}]
79.6261
Looks like a bug to me.
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
From: JerrySpock [mailto:liquidsolids at hotmail.com]
Hello, everyone.
I'm having a problem integrating to find an arc length.
I have two parametric equations:
x=e^(2t)
and
y=e^(1.5t)
I'm looking for the arc length from 1 to 2.
N[
Integrate[
Sqrt[
(2Exp[2*m])^2 + (1.5Exp[1.5*m])^2
],{m, 1, 2}]]
I keep getting the answer 79.6, but my TI-83 says the answer is 49.8. I've
been playing with this for hours, and I can't get it to work. Any ideas
what I'm doing wrong?
[Edited by: admin on Jul 25, 2009 7:22 AM]