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Math Help - Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

  1. #1
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    Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

    Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

    test y=ax^n and y=Ae^kx
    1). i need to investigate a power model in the form y=ax^n by plotting my raw data and log x vs log y
    2). i need to investigate a exponential modle in the form y=Ae^kx by plotting the raw data and x vs log y

    i now need to use graphs and correlation coefficients to determine the equation of the model that fits your Data.?????

    how do i find the equation... my text book only gose briefly and i missed the class due sports, i asked my teachers and friends but they only confused me more.
    thank you for the fast help
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  2. #2
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    Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

    What is your "raw data"? Presummably, each data point is (x, y). Find the two logarithms, and plot point (log(x), log(y)). Do those points seem to be close to a staright line?
    The point is that if y= ax^n then lpg(y)= log(ax^n)= nlog(x)+ log(a) where log(y) is a linear function of log(x).

    Similarly, if y= ae^(kx) the log(y)= kx+ log(a) where log(y) is a linear function of x.
    Thanks from topsquark
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    Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

    Hi thanks for being fast,
    yes the raw data is in x and y x being time and y the difference in temperature as apart of a table we were given.
    what i don't under stand is how do i find A and n in the fist and A and k in the next??
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    Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

    and every time i graph with my graphics cal it doesn't pass the x axis
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    Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

    Is there any reason why "not passing the x-axis" is a problem? If you are referring to finding the x-intercept, you can extend the line until it does cross.

    As I said before, the whole point is to try to get, in the first case, log(y)= A log(x)+ B, and, in the second case, log(y)= Ax+ B.

    "A", in each case, is the slope of the line: choose two points, (log(x1), log(y1) and (log(x2), log(y2)), to find the slope \frac{log(y2)- log(y1)}{log(x2)- log(x1)} and (x1, log(y1)), (x2, log(y2)) to find the slope \frac{log(y2)- log(y1)}{x2- x1}. You can find "B" by using that A with one point: B= log(y)- Alog(x) or B= log(y)- Ax.
    Thanks from topsquark and BreazaMathsB
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    Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

    Thank you for that
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