thank you... but ARRGG I still can't solve it!

Posted by colin on September 10, 2002 at 14:13:45:

In Reply to: Re: How does the "adding zero" trick work??? posted by Subhotosh Khan on September 10, 2002 at 08:42:04:

thanks, I get the "adding zero" trick. I'm still bungling it somewhere though... Here's what I've got so far.

dv/dt = g - k/m*Sqrt[v], let c = k/m
dt = dv/(g - c*Sqrt[v])
let u = Sqrt[v]
u^2 = v
2u du = dv
t = INT[2u*du/(g-cu)]
t = 2/c*INT[cu/(g-cu)du]
t = 2/c*INT[cu+g-g/(g-cu)du]
t = -2/c*INT[1 + g/g-cu]du
t = -2m/k[ g*ln(g-k/m*Sqrt[v]) + Sqrt[v] ] + c
c = 2mg/k*ln(g-k/m*Sqrt[v0]) + 2m/k*Sqrt[v0]

and based on these results I get the answer:
2m/k(Sqrt[v0]-Sqrt[v]) + 2mg/k[ln(g-k/mSqrt[v0]/ln(g-k/mSqrt[v])]

and this doesn't agree with the book answer which has a 2nd-degree coefficient for the 2m^2g/k^2 part. Where did that come from? Can you please point out my mistake? Thank you.

Colin

Follow Ups:

• I haven't worked it out ...but Subhotosh Khan 15:37:34 09/10/02 (1)
  • thank you so much! Colin 17:56:42 09/10/02 (0)

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