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Math Help - cos(e^x)*e^2 = 0

  1. #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?
    *************
    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?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dented42 View Post
    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?
    \cos x is zero whenever we have an odd multiple of \frac {\pi}2. that is, \cos x = 0 \implies x = (2n + 1) \frac {\pi}2 for n \in \mathbb{Z}

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

    now continue
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    yay!

    thank you so much! I think I can take it from here.
    I just needed a kick in the right direction.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dented42 View Post
    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
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    we can use latex?
    yay!

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

    and, if cos(f(t))=0

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

    in this case f(u)=e^t

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

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

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

    then find the earliest instance ie when t>1

    log_e\frac{\pi(2*1+1)}{2}=log_e\frac{\pi}{2}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dented42 View Post
    we can use latex?
    yay!

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

    and, if cos(f(t))=0

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

    in this case f(u)=e^t

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

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

    simplify: 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 \log_e x, instead, we write \ln x. that is, the natural log is the logarithm to the base e

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

    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.
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    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 \log_e x, instead, we write \ln x. that is, the natural log is the logarithm to the base 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 \log_e, or something.

    thanks again!
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dented42 View Post
    for some reason I thought it would be more formal to write \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
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