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Find a curve bisecting the curves x2 and x2/2

Consider the figure

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We say the curve C bisects the region between C_1 and C_2 in area if for every point P on the curve C the area of regions A and B are equal. Given the equations

    \[ C: \ y = x^2 \qquad C_1: \ y = \frac{1}{2}x^2 \]

find an equation for C_2 such that the curve C bisects the region between C_1 and C_2 in area.


First, we can calculate the area of the region A. This is the difference in the integrals from 0 to the x-coordinate of P of x^2 and \frac{1}{2}x^2. Since P lies on the curve C defined by the equation y = x^2 we may write P = (t,t^2) for some t \geq 0. Then the area of A is given by

    \begin{align*}  \int_0^t x^2 \, dx - \int_0^t \frac{x^2}{2} \, dx &= \int_0^t \left( x^2 - \frac{x^2}{2} \right) \, dx \\  &= \int_0^t \frac{x^2}{2} \, dx \\  &= \left. \frac{x^3}{6} \right|_0^t \\  &= \frac{t^3}{6}. \end{align*}

Now, we make the assumption that the equation for C_2 is of the form kx^2 for some positive real number k. (I don’t know a good way to justify this assumption other than it’s the most obvious first thing to try, and it happens to work.) Then to find the area of region B first we find equations for the curves C and C_2 in terms of y (so that we may integrate along the y-axis which is somewhat easier). So we have

    \begin{align*}  C: \ y &= x^2 &\implies && x = \sqrt{y} \\  C_2: \ y &= kx^2 & \implies && x = \frac{1}{\sqrt{k}} \sqrt{y}. \end{align*}

Then, we integrate along the y-axis from 0 to t^2 (since this is the y-coordinate of P) the difference between these two curves. The area of B is then

    \begin{align*}  \int_0^{t^2}\left( \sqrt{y} - \frac{1}{\sqrt{k}} \sqrt{y} \right) \, dy &= \left( 1 - \frac{1}{\sqrt{k}} \right) \int_0^{t^2} \sqrt{y} \, dy \\  &= \left(1 - \frac{1}{\sqrt{k}} \right) \left( \frac{2}{3} y^{\frac{3}{2}} \Big \rvert_0^{t^2} \right) \\  &= \left( 1 - \frac{1}{\sqrt{k}} \right) \left( \frac{2t^3}{3} \right) \\  &= \frac{2t^3}{3} - \frac{2t^3}{3 \sqrt{k}}. \end{align*}

Now we set the areas of the regions equal and solve for k,

    \begin{align*}  \frac{t^3}{6} = \frac{2t^3}{3} - \frac{2t^3}{3 \sqrt{k}} && \implies && \frac{1}{6} &= \frac{2}{3} - \frac{2}{3 \sqrt{k}} \\  && \implies && \frac{2}{3 \sqrt{k}} &= \frac{1}{2} \\  && \implies && 4 &= 3 \sqrt{k} \\  && \implies && k &= \frac{16}{9}. \end{align*}

Therefore, the equation describing C_2 is given by

    \[ y = \frac{16}{9} x^2. \]

One comment

  1. Zerong Xi says:

    A new function x=g(y) can be defined as the reverse of C2: y=f(x). With a similar procedure, we can obtain the expression of g(y) and then reverse it to have f(x).

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