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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) = 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

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)dtdx/dt=12dy/dt=12tdz/dt=6/tPut 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) = 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

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..4This is cumbersome to differentiate, but it boils down to(48e^(8t))4(3)e^(4t)dtYou can't integrate with respect to t from 0 to t. That's a dependent limit.
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