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Math Help - Rates-Change with lns and sin/cos

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    Rates-Change with lns and sin/cos

    I can't get the right answer for these two, mainly because they include trigonometry and lns.

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I can't get the right answer for these two, mainly because they include trigonometry and lns.

    For the first:

    (a)
    we want S(12)

    S(12) = 72 - 15 \ln (12 + 1) = 72 - 15 \ln (13) \approx 33.53

    (b)
    we want S'(4)

    Now S'(t) = - \frac {15}{t + 1}

    \Rightarrow S'(4) = - \frac {15}{5} = -3

    Do you know how to find the derivative of ln ?

    (c)
    I assume the original score is to be taken when t = 0

    This means the original score is 72

    we want to know when does the average score becomes less than \frac {3}{4} 72 = 54

    So we must solve for S(t) < 54

    we want t such that:

    S(t) = 72 - 15 \ln(t + 1) < 54

    \Rightarrow -15 \ln(t + 1) < -18

    \Rightarrow ln(t + 1) > \frac {6}{5}

    \Rightarrow e^{ \frac {6}{5}} > t + 1

     \Rightarrow t > e^{ \frac {6}{5}} - 1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I can't get the right answer for these two, mainly because they include trigonometry and lns.

    The second

    The rate of change of voltage is v'(t)

    Now, v'(t) = -2 \sin(t) - 2 \sin(2t) ........By the Chain rule

    we want t such that v'(t) = 0

    That is, we want to solve -2 \sin(t) - 2 \sin(2t) = 0

    \Rightarrow \sin(t) + \sin(2t) = 0

    \Rightarrow \sin(t) + 2 \sin(t) \cos(t) = 0

    \Rightarrow \sin(t) (1 + 2 \cos(t)) = 0

    \Rightarrow \sin(t) = 0 \mbox { or} 1 + 2 \cos(t) = 0

    \Rightarrow t = 0 \mbox { or} \cos(t) = - \frac {1}{2}

    \Rightarrow t = 0 \mbox { , } \pi \mbox { , } 2 \pi \mbox { or } t = \frac {2 \pi}{3} \mbox { , } \frac {4 \pi}{3}


    EDIT 1: added  \pi and 2 \pi to the solutions

    EDIT 2: Thanks for looking out behemoth100...would you believe that I intentionally left the answer incomplete to see if SportfreundeKeaneKent would catch it?
    Last edited by Jhevon; May 27th 2007 at 02:12 PM.
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    Quote Originally Posted by Jhevon View Post
    The second

    The rate of change of voltage is v'(t)

    Now, v'(t) = -2 \sin(t) - 2 \sin(2t) ........By the Chain rule

    we want t such that v'(t) = 0

    That is, we want to solve -2 \sin(t) - 2 \sin(2t) = 0

    \Rightarrow \sin(t) + \sin(2t) = 0

    \Rightarrow \sin(t) + 2 \sin(t) \cos(t) = 0

    \Rightarrow \sin(t) (1 + 2 \cos(t)) = 0

    \Rightarrow \sin(t) = 0 \mbox { or} 1 + 2 \cos(t) = 0

    \Rightarrow t = 0 \mbox { or} \cos(t) = - \frac {1}{2}

    \Rightarrow t = 0 \mbox { or } t = \frac {2 \pi}{3} \mbox { , } \frac {4 \pi}{3}
    Ok well given that its only 7am in the morning this may be a stupid comment but...
    Isnt sin(t) = 0 at 0 pi and 2pi? The interval is 0<t<2pi (except they are equal to or less than signs).
    So isn't your answer incomplete?

    If I have made a mistake on BASIC DIFFERENTIATION AND TRIG THEN I SHOULD BE SHOT!!! I'm scared agaist going up against the almighty Jhevon
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by behemoth100 View Post
    Ok well given that its only 7am in the morning this may be a stupid comment but...
    Isnt sin(t) = 0 at 0 pi and 2pi? The interval is 0<t<2pi (except they are equal to or less than signs).
    So isn't your answer incomplete?

    If I have made a mistake on BASIC DIFFERENTIATION AND TRIG THEN I SHOULD BE SHOT!!! I'm scared agaist going up against the almighty Jhevon
    "almighty Jhevon"? where did that come from?

    You are correct sir! my solution is incomplete, I shall add your answers to mine
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  6. #6
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    Nah, I wouldn't have caught that, I'm new to this stuff. Thanks anyways, with the first one.
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