Author 
Topic 

dmmathwiz
Average Member
USA
7 Posts 
Posted  09/29/2007 : 15:58:54

I am lost on how to integrate these complicated square roots.
problem 1: r(t) = <12t,6t,6ln(t)>
arclength s is the integral from 1 to 4.
s(t) = (12+(12t)+(6/t))
how do I solve this definite integral?
problem 2: r(t) = <e^(4t)cos(4t),e^(4t)sin(4t),e^(4t)>
integral from 0 to t. as above s(t) = ((dx/dt)+(dy/dt)+(dz/dt))
not sure how to solve these two definite integrals. Any help would be appreciated. 


galactus
Advanced Member
USA
1464 Posts 
Posted  09/29/2007 : 17:21:16

parametric arc length is given by:
((dx/dt)^2+(dy/dt)^2+(dz/dt)^2)dt
dx/dt=12
dy/dt=12t
dz/dt=6/t
Put it all together:
(144+144t^2+(36/t^2))=(6(2t^2+1))/t



dmmathwiz
Average Member
USA
7 Posts 
Posted  09/29/2007 : 18:44:57

r(t) = <e^(4t)cos(4t),e^(4t)sin(4t),e^(4t)>
integral from 0 to t. s(t) = ((dx/dt)+(dy/dt)+(dz/dt))
I understand the formula I just have no idea how to integrate it.
using the product rule I got:
s(t) = ((4e^(4t)sin(4t)+4cos(4t)e^(4t))+(4e^(4t)cos(4t)+4cos(4t)e^(4t))+(4e^(4t)))
Any ideas on how to integrate this from 0 to t? 


galactus
Advanced Member
USA
1464 Posts 
Posted  09/29/2007 : 19:07:06

The first one is easy to integrate. It's the algebra that's the booger.
[12t+(6/t)]dt, t=1..4
This is cumbersome to differentiate, but it boils down to
(48e^(8t))
4(3)e^(4t)dt
You can't integrate with respect to t from 0 to t. That's a dependent limit. 



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