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Math Help - Rate of change

  1. #1
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    Post Rate of change

    How do you do the following rate of change questions

    when t = 1.2 for all the following questions:

    1) y = 4/t + 4lnt

    2) x = sin(t^2 +1) , y = cos(2t -3)

    3) x = 1 / 1+2t , y = t / 1+t

    4) q = 2e^-t/2.cos2t

    5) x = e^2t.t^3(2-t)^4

    Please could you explain the process as im quite lost. Thanks
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  2. #2
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    Post

    e for 4) and 5) is the exponent
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  3. #3
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    Quote Originally Posted by dadon
    How do you do the following rate of change questions

    when t = 1.2 for all the following questions:

    1) y = 4/t + 4lnt

    2) x = sin(t^2 +1) , y = cos(2t -3)

    3) x = 1 / 1+2t , y = t / 1+t

    4) q = 2e^-t/2.cos2t

    5) x = e^2t.t^3(2-t)^4

    Please could you explain the process as im quite lost. Thanks
    The rate of change is the derivative with respect to time. So for 1):

    <br />
y = 4/t + 4 \ln(t)<br />

    so:

    <br />
\frac{dy}{dt}=\frac{-4}{t^2} + \frac{4}{t}<br />
,

    so when t=1.2:

    <br />
\frac{dy}{dt}=\frac{-4}{1.2^2} + \frac{4}{1.2}=0.5555..<br />

    RonL
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  4. #4
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    Post Re:

    Thanks! On that question (1) I was confused about the 4lnt part. You make it seem so simple!
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