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    A particke moves along the xaxis such that its displacement x metres from the origin O at time t seconds is given by

    x=2+t-(1/10)e^t

    Find correct to 1dp the value of t when the velocity of P is zero.

    Okay i need to do dx/dt obv, but can't do that with e^t, do i ln everything like...

    lnx=ln2+lnt-ln1/10xe^t (and then differentiate)

    and i don't think u can ln like this can you??? when you ln both sides it should be lnx=ln(2+t....)

    Or maybe i differentiate e^t sepperately...

    ARGG help!
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    d/dt{e^t} = e^t

    i.e. dx/dt = 1 - e^t /10
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    (Original post by BossLady)
    A particke moves along the xaxis such that its displacement x metres from the origin O at time t seconds is given by

    x=2+t-(1/10)e^t

    Find correct to 1dp the value of t when the velocity of P is zero.

    Okay i need to do dx/dt obv, but can't do that with e^t, do i ln everything like...

    lnx=ln2+lnt-ln1/10xe^t (and then differentiate)

    and i don't think u can ln like this can you??? when you ln both sides it should be lnx=ln(2+t....)

    Or maybe i differentiate e^t sepperately...

    ARGG help!
    dx/dt = 1 - (1/10)e^t

    dx/dt = 0: 1 - (1/10)e^t = 0

    (1/10)e^t = 1

    e^t = 10

    t = ln 10 = 2.3 s
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    (Original post by elpaw)
    d/dt{e^t} = e^t

    i.e. dx/dt = 1 - e^t /10
    oh god....."e's are easy" i am so stupid!!!
    (thank you!)
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    just wondering now though...if you have a question like

    3=e^t+ A

    you do ln3=ln(e^t +A) ?
    or erm ln(3-A)=tlne
    ie ln (3-A)=t

    so you can't do ln3=lne^t+lnA right?

    my brains seem to be turning into mushy peas....
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    (Original post by BossLady)
    so you can't do ln3=lne^t+lnA right?
    no, you cant do that.
    you do ln3=ln(e^t +A) ?
    or erm ln(3-A)=tlne
    ie ln (3-A)=t
    both are correct
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    (Original post by elpaw)
    no, you cant do that.

    both are correct
    gotcha thanks!
 
 
 

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