That_GuyMember Posts: 30 From: TN,USA Registered: 02-26-2005 |
I need to know what the anti-derivative of: v(t) = 1 - arctan (e^t) i know at t=1 the particle position (anti-derivative of v(t)) is at y=-1. I need to kneed to know where it is at t=2. I can also solve for the answer using a TI-83 calculatur. I would like to know how to do it both analliticly(forgive my spelling) and with the calculatur. ------------------ |

ArchAngelMember Posts: 3450 From: SV, CA, USA Registered: 01-29-2002 |
check this one: V(t) = t - (e^t)arctan(e^t) - (1/2)ln( 1 + (e^t)^2 ) + C the antiderivative for the arctan is a crazy formula, something usually you want to check up on a referce chart for. if this is wrong, correct me. ------------------ |

That_GuyMember Posts: 30 From: TN,USA Registered: 02-26-2005 |
Thanks. ------------------ |