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Find a function which divides a rectangle into pieces with given properties

Consider a curve whose Cartesian equation is given by y = f(x), and which passes through the origin. A rectangular region is drawn with one corner at the origin, and the other corner on the curve of the graph of f(x). The curve f(x) then divides the rectangle into two pieces A and B such that one of the regions has area n times the area of the other for every such rectangle. Find the equation of f(x).


Incomplete.

2 comments

  1. Evangelos says:

    I messed up the Latex formatting… Maybe this one will work?

    The area of region A can be written as the following integral, since it is the area between the horizontal line f(x) and the function y = f(t) from t = 0 to t = x :

        \begin{align*} \int_{0}^{x} f(x) - f(t) dt \end{align*}

    And likewise, the area of region B can be written as the following integral, since it is the remaining portion of the rectangle, or the area between the function f(t) and the x-axis from t = 0 to t = x:

        \begin{align*} \int_{0}^{x} f(t) dt \end{align*}

    And since the area of A is n times the area of B, we have

        \begin{align*} \int_{0}^{x} f(x) - f(t) dt = n \int_{0}^{x}f(t) dt \end{align*}

    or

        \begin{align*} \int_{0}^{x} f(x) dt = n+1 \int_{0}^{x}f(t) dt \end{align*}

    Now, let F(t) be such that F'(t) = f(t), evaluating the above integral gives us:

        \begin{align*} x f(x) = (n+1)(F(x) - F(0)) \end{align*}

    Taking the derivative with respect to x on both sides

        \begin{align*} f(x) + x f'(x) = (n+1) f(x) \end{align*}

    Written another way

        \begin{align*} x f'(x) = n f(x). \end{align*}

    Now, we can rearrange the terms on both sides to give the following equation of logarithmic derivatives:

        \begin{align*} \frac{f'(x)}{f(x)} = \frac{n}{x} \end{align*}

    Integrating both sides gives us

        \begin{align*}  \log [f(x)] = n \log(x) + C\\ = \log(x^n) + C  \end{align*}

    And taking the exponential (inverse log) of both sides gives us

        \begin{align*} f(x) = Kx^n \end{align*}

    where K = e^C.

  2. Evangelos says:

    OK, first time writing out a full answer!

    The area of region A can be written as the following integral, since it is the area between the horizontal line f(x) and the function y = f(t) from t = 0 to t = x :

    \int_{0}^{x} f(x) – f(t) dt

    And likewise, the area of region B can be written as the following integral, since it is the remaining portion of the rectangle, or the area between the function f(t) and the x-axis from t = 0 to t = x:

    \int_{0}^{x} f(t) dt

    And since the area of A is n times the area of B, we have

    \int_{0}^{x} f(x) – f(t) dt = n \int_{0}^{x}f(t) dt

    or

    \int_{0}^{x} f(x) dt = n+1 \int_{0}^{x}f(t) dt

    Now, let F(t) be such that F'(t) = f(t), evaluating the above integral gives us:

    x f(x) = (n+1)[F(x) – F(0)]

    Taking the derivative with respect to x on both sides

    f(x) + x f'(x) = (n+1) f(x)

    Written another way

    x f'(x) = n f(x).

    Now, we can rearrange the terms on both sides to give the following equation of logarithmic derivatives:

    \frac{f'(x)}{f(x)} = \frac{n}{x}

    Integrating both sides gives us

        \begin{align*} \log [f(x)] = n \log(x) + C\\ = \log(x^n) + C \end{align*}

    And taking the exponential (inverse log) of both sides gives us

    f(x) = Kx^n

    where K = e^C.

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