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

1. ## 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?
*************
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?

2. Originally Posted by dented42
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?
*************
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?
$\displaystyle \cos x$ is zero whenever we have an odd multiple of $\displaystyle \frac {\pi}2$. that is, $\displaystyle \cos x = 0 \implies x = (2n + 1) \frac {\pi}2$ for $\displaystyle n \in \mathbb{Z}$

thus, we need $\displaystyle e^t = (2n + 1) \frac {\pi}2$

now continue

3. ## yay!

thank you so much! I think I can take it from here.
I just needed a kick in the right direction.

4. Originally Posted by dented42
thank you so much! I think I can take it from here.
I just needed a kick in the right direction.
no problem

whenever you need a kick from someone, i am here for you. i'll be glad to hand out punches as well

5. we can use latex?
yay!

so we want to solve for t in the following equation $\displaystyle cos(e^t)*e^t=0$
$\displaystyle e^t$ is never zero, therefore what we really want is to solve: $\displaystyle cos(e^t)=0$

and, if $\displaystyle cos(f(t))=0$

then $\displaystyle f(t)=\frac{\pi(2n+1)}{2}$

in this case $\displaystyle f(u)=e^t$

so $\displaystyle e^t=\frac{\pi(2n+1)}{2}$

take the log of both sides: $\displaystyle log_ee^t=log_e\frac{\pi(2n+1)}{2}$

simplify: $\displaystyle t=log_e\frac{\pi(2n+1)}{2}$

then find the earliest instance ie when $\displaystyle t>1$

$\displaystyle log_e\frac{\pi(2*1+1)}{2}=log_e\frac{\pi}{2}$

6. Originally Posted by dented42
we can use latex?
yay!

so we want to solve for t in the following equation $\displaystyle cos(e^t)*e^t=0$
$\displaystyle e^t$ is never zero, therefore what we really want is to solve: $\displaystyle cos(e^t)=0$

and, if $\displaystyle cos(f(t))=0$

then $\displaystyle f(t)=\frac{\pi(2n+1)}{2}$

in this case $\displaystyle f(u)=e^t$

so $\displaystyle e^t=\frac{\pi(2n+1)}{2}$

take the log of both sides: $\displaystyle log_ee^t=log_e\frac{\pi(2n+1)}{2}$

simplify: $\displaystyle t=log_e\frac{\pi(2n+1)}{2}$
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 $\displaystyle \log_e x$, instead, we write $\displaystyle \ln x$. that is, the natural log is the logarithm to the base $\displaystyle e$

then find the earliest instance ie when $\displaystyle t>1$
you mean t = 1.

$\displaystyle log_e\frac{\pi(2*1+1)}{2}=log_e\frac{\pi}{2}$
yes, but again, use "ln" instead

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

7. Originally Posted by Jhevon
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 $\displaystyle \log_e x$, instead, we write $\displaystyle \ln x$. that is, the natural log is the logarithm to the base $\displaystyle e$

you mean t = 1.

yes, but again, use "ln" instead

by the way, the problem never asked for the earliest time. so just state the full answer.
yes, I mean t=1.
for some reason I thought it would be more formal to write $\displaystyle \log_e$, or something.

thanks again!

8. Originally Posted by dented42
for some reason I thought it would be more formal to write $\displaystyle \log_e$, or something.
nope, "ln" is good enough. in advanced math they may just even write "log" to mean the natural log, instead of the base 10 logarithm it represents in "elementary" math

thanks again!
don't mention it