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Pre-Calculus and Calculus

Arc Length Integrals

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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) = <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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