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Math Help - two ln problems

  1. #1
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    two ln problems

    1. The rate of transmission in a telegraph cable is observed to be proportional to x^2\ln (1/x), where x is the ratio of the radius of the core to the thickness of the insulation ( 0<x<1 ). What value of x gives the maximum rate of transmission?
    I overlooked this simplicity of this problem. I just needed to find the deriv of the problem where it = 0. This was .60653019.

    2. f(x)=ln(x)/(1+ln(x)^2). Find lim as x->0+ and x-> inf
    nevermind, it's 0 for both limits
    Last edited by thedoge; January 25th 2007 at 09:04 PM. Reason: solved both
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  2. #2
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    Quote Originally Posted by thedoge View Post
    1. The rate of transmission in a telegraph cable is observed to be proportional to x^2\ln (1/x), where x is the ratio of the radius of the core to the thickness of the insulation ( 0<x<1 ). What value of x gives the maximum rate of transmission?...
    Hello,

    I presume the function is:

    t(x)=x^2 \cdot \ln(x). You are looking for a maximum of t(x). This will happen if t'(x) = 0. Thus calculate the first derivative. Use the product rule:

    t'(x) = \ln(x) \cdot 2x + x^2 \cdot \frac{1}{x}=x(2 \ln(x) + 1)

    Now t'(x) = 0. Solve for x:

    x(2 \ln(x) + 1)=0 \Longleftrightarrow \underbrace{x = 0}_{\text{not allowed here}} \text{ or } \ln(x)=- \frac{1}{2}

    \ln(x)=- \frac{1}{2} \Longleftrightarrow x=\sqrt{\frac{1}{e}} \approx 0.606530659...

    EB
    Last edited by earboth; January 26th 2007 at 10:19 PM.
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  3. #3
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    Quote Originally Posted by thedoge View Post
    ...

    2. f(x)=ln(x)/(1+ln(x)^2). Find lim as x->0+ and x-> inf
    nevermind, it's 0 for both limits
    Hello,

    you are supposed to know

    (A) ---> \lim_{x \rightarrow \infty}{(\ln(x))}=\infty

    and because \lim_{y \rightarrow \infty}{\frac{1}{y}}=0

    (B) ---> \lim_{x \rightarrow 0}{(\ln(x))}=\lim_{y \rightarrow \infty}{\left( \ln \left( \frac{1}{y} \right) \right)}=\lim_{y \rightarrow \infty}{(-\ln(y))}=-\infty

    f(x)=\frac{\ln(x)}{1+(\ln(x))^2}

    \lim_{x \rightarrow \infty}{f(x)} is undefined. Therefore use the rule de l'H˘pital:

    \lim_{x \rightarrow \infty}{\frac{\ln(x)}{1+(\ln(x))^2}}=\lim_{x \rightarrow \infty}{\frac{\frac{1}{x}}{2\ln(x) \cdot \frac{1}{x}}}=\lim_{x \rightarrow \infty}{\frac{1}{2\ln(x) \cdot }}=0

    According to (A) the limit is zero because the denominator approaches infinity.

    \lim_{x \rightarrow 0}{\frac{\ln(x)}{1+(\ln(x))^2}}=\lim_{y \rightarrow \infty}{\frac{\frac{1}{\frac{1}{y}}}{2\ln(\frac{1}  {y}) \cdot \frac{1}{\frac{1}{y}}}}=\lim_{y \rightarrow \infty}{\frac{1}{2\ln(\frac{1}{y})}}=\lim_{y \rightarrow \infty}{\frac{1}{-2\ln(y) }} = 0

    EB
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