Home » Blog » Find a curve bisecting the curves x2 and x2/2

Find a curve bisecting the curves x2 and x2/2

Consider the figure

Rendered by QuickLaTeX.com

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

    \[ y = \frac{16}{9} x^2. \]


  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).

    • Artem says:

      Exactly! the assumption of quadratic function c2 is not needed at all. It is discovered by integrating some unknown inverse of it. Then using the second fundamental theorem of calculus we can remove lower bound at 0 for the unknown integral since in the picture the second function starts at (0, 0). Then using the 1st theorem of calculus, we differentiate, and invert, to come back to c2. Then by variable substitution (u = 3/4x) we get the final result.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):