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Math Help - Approximate values

  1. #1
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    Approximate values

    The table below gives values of T and L which are related by the law T = kL^n where k and n are constants.

    T - __1____1.2___1.4___1.6___1.8___2
    L - __2.4___3.5__ 4.8__ 6.3___7.9___9.8

    Prove that this law is as stated. Determine the approximate values of k and n, and hence state the law.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dean View Post
    The table below gives values of T and L which are related by the law T = kL^n where k and n are constants.

    T - __1____1.2___1.4___1.6___1.8___2
    L - __2.4___3.5__ 4.8__ 6.3___7.9___9.8

    Prove that this law is as stated. Determine the approximate values of k and n, and hence state the law.
    Consider
    T = kL^n

    Take the log (or natural log, or log to your favorite base, whatever) of both sides:
    log(T) = log(kL^n) = log(L^n) + log(k)

    log(T) = n \cdot log(L) + log(k)

    What does this do for you? Well the equation is now in the form
    y = m \cdot x + b
    where log(T) = y, \ n = m, \ log(L) = x, \ log(k) = b

    So what you do is a linear regression with the log(L) values being your x values and the log(T) values being the y values. Then the slope m of the regression will be n, and the intercept b will be the log(k).

    Edit: For reference I get that n \approx 0.493468 and log(k) \approx -0.188847 \implies k \approx 0.827913 with an r^2 \approx 0.999885, so the fit for T = (0.827913)L^{0.493468} is excellent. This is not quite the same as Soroban's answer (which is easier to get) but is probably more accurate. (Sorry Soroban, I'm not trying to pick on your answer! )

    -Dan
    Last edited by topsquark; August 4th 2007 at 08:13 AM.
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  3. #3
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    Hello, Dean!

    The table below gives values of T and L
    which are related by the law T \:= \:kL^n where k and n are constants.

    \begin{array}{cccccccc}T & | & 1 & 1.2 & 1.4 & 1.6 & 1.8 & 2 \\ \hline<br />
L & | & 2.4 & 3.5 & 4.8 & 6.3 & 7.9 & 9.8<br />
\end{array}

    Prove that this law is as stated.
    Determine the approximate values of k and n, and hence state the law.

    Use any two pairs of values: . (L,\,T) \:=\:(2.4,\,1),\;(3,5,\,1.2)


    Substitute into the law:

    . \begin{array}{ccccc}(2.4,\,1): & k\cdot2.4^n & = & 1 & {\color{blue}[1]} \\<br />
(3,5,\,1.2): & k\cdot3.5^n & = & 1.2 & {\color{blue}[2]} \end{array}

    Divide {\color{blue}[2]} by {\color{blue}[1]}: . \frac{k\cdot3.5^n}{k\cdot2.4^n} \:=\:\frac{1.2}{1}\quad\Rightarrow\quad \left(\frac{3.5}{2.4}\right)^n \:=\:1.2

    Take logs: . \ln\left(\frac{3.5}{2.4}\right)^n \:=\:\ln(1.2)\quad\Rightarrow\quad n\cdot\ln\left(\frac{3.5}{2.4}\right) \:=\:\ln(1.2)

    . . Hence: . n \;=\;\frac{\ln(1.2)}{\ln\left(\frac{3.5}{2.4}\righ  t)} \;=\;0.483234414

    Substitute into {\color{blue}[1]}: . k\cdot2.4^{0.483234414} \;=\;1\quad\Rightarrow\quad k \:=\:0.655041542


    Therefore: . \begin{array}{ccc}n & \approx & 0.483 \\ k & \approx & 0.655\end{array} . and the law is: . T \;=\;0.655L^{0.483}

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