Monday, April 14, 2008

#6b from HW #17

I got the speed but I can't find the acceleration vector. I thought to find the acceleration vector you just find the second derivative, or the derivative of dx/dt and dy/dt, but my answer isn't the same. Does anyone know how to find the acceleration vector? Thanks.
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2 comments:

Anonymous said...

what you're doing is right. you get the derivative of dx/dt and dy/dt and plug in 1 for t.
it might be that you need to use the chain rule when taking the derivatives.
dx/dt=tan(e^-t)
derivative= (-e^-t)[sec^2(e^-t)] and plug in 1 to get -.423
for dy/dt think of sec as cos^-1 and then take derivative with chain rule.

April 14, 2008 at 11:18 PM
Anonymous said...

derivative of dy/dt

-[cos(e^-t)]^-2 [(-e^-t)(-sin(e^-t))]

April 14, 2008 at 11:22 PM

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