Calculating Arc Length for Parametric Equations with Simple Integration

[Seraph404]

Homework Statement

x = 1+3t^2
y=3+2t^3
0<= x <=4

Homework Equations

L = integral from a to b of \sqrt{[dx/dt]^2 + [dy/dt]^2} dx

The Attempt at a Solution

dx/dt = 6t
dy/dt = 6t^2

L = integral from 0 to 4 of \sqrt{(6t)^2 +(6t^2)^2} dx
= " \sqrt{36t^2 +36t^4} dx
= " \sqrt{(36)(t^2 +t^4)} dx
= " 6 \sqrt{(t^2)(1 +t^2)} dx
= " 6t \sqrt{1 +t^2} dx

let u = 1+t^2
du = 2tdt
1/2 du = tdt

= " 3 (u)^(1/2)du
= 3(2/3) [ u^(3/2)] from 0 to 4
= 2[(1+t^2)^(3/2)] from 0 to 4
= 2[(17)^(3/2)-1]

Is that correct? Just want to make sure.

 

[berkeman]

One way you can check is to use an integral table (like the CRC math tables) to check the integration directly. I think the integral on this page:

http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions

with the form x * r may match what you are trying to integrate. Does it give you the same answer?

 

[Seraph404]

Haha. I apologize. The question I asked was actually very simple compared to the other topics being posted in this forum.

The link you sent me looks like cases in which you would use trig substitution (though, that doesn't seem to be how the problems are worked out). But in the problem above, I'm integrating 6 sqrt(1+t^2) t dt, which is easy. I only ask for confirmation because I'm really bad at catching my own mistakes, and had already messed up on the last few steps of this problem three times before getting what I have now.