I haven't worked it out ...but

[ Follow Ups ] [ Post Followup ] [ Calculus Message Board ] [ FAQ ]

Posted by Subhotosh Khan on September 10, 2002 at 15:37:34:

In Reply to: thank you... but ARRGG I still can't solve it! posted by colin on September 10, 2002 at 14:13:45:

: 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
******************************
2/c*int[g/(g-cu)]du

let g - cu = y
dy = - c du

2/c*int[g/(g-cu)]du

=-2g/c^2*int[dy/y]... there is your "square" term in the denomenator

try to be very explicit when working with "unwieldy" integration/differentiation. If skip steps - you are liable to mess up. It takes little more time - but - no body gives points for "quick" wrong answer.

Name:
E-Mail:

Subject: