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- Oct 11th 2008, 06:57 PM #1

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## cos(e^x)*e^2 = 0

a particle moves along the x-axis in such a way that at time t>0 its position is x=sin(e^t)

at what time does the particle have a velocity of zero?

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I take the derivative and get (e^t)*cos(e^t)

so when does that equal zero?

e^t is never zero, right?

so when does cos(e^t) equal zero?

I'm not sure how to solve a trig function with a variable argument.

help?

- Oct 11th 2008, 07:08 PM #2

- Oct 11th 2008, 07:10 PM #3

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- Oct 11th 2008, 07:18 PM #4

- Oct 11th 2008, 07:41 PM #5

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- Oct 11th 2008, 08:25 PM #6
yes.

but some tips. put a \ before the log so it looks nicer. that is, type \log_e ...

also, for the record, we do not write , instead, we write . that is, the natural log is the logarithm to the base

then find the earliest instance ie when

by the way, the problem never asked for the earliest time. so just state the full answer.

- Oct 12th 2008, 02:26 PM #7

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- Oct 12th 2008, 03:40 PM #8