Physics Help Forum Resistive Force Calculus Help

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

Dec 21st 2013, 06:46 PM   #21
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 Originally Posted by topsquark Let v = sqrt(g/k) * u. Then dv = sqrt(g/k) du. The integral becomes: So Which you can finish from here. (The solution for v can be put into the form of an inverse hyperbolic tangent for a "cleaner" solution, but I think the log format is easier to grasp.) -Dan
I thought it would be helpful to show how that integral was evaluated. The method is called expansion by patial fractions. We write 1/(1 - u^2) as

1/(1 - u^2) = (1/2)[ 1/(1 - u) - 1/(1 + u) ]

When this is substituted its easy to see how to evaluate it in terms of natural logarithms.

 Tags calculus, force, resistive

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