Consider a curve whose Cartesian equation is given by , 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 . The curve then divides the rectangle into two pieces and such that one of the regions has area times the area of the other for every such rectangle. Find the equation of .

**Incomplete.**

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 :

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:

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

or

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

Taking the derivative with respect to x on both sides

Written another way

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

Integrating both sides gives us

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

where K = e^C.

Thanks so much Evangelos! All your answers in this section are really helpful.

Tyler, I appreciate the kind words, and I’m glad I can connect with other independent learners. I’m thinking of continuing in the spirit of this blog onto the second volume of Apostol’s calculus series, but it might be a while before I get to it since I’m only finishing problems sporadically. Maybe if you reach the finish line before I do, I can comment on your Apostol Vol 2 blog instead :)

Sir, that was absolutely brilliant! Any news on your efforts for Tommy II? I hope you still plan to do those :)

Hi anon, still haven’t finished out all the chapters in Vol 1 yet, been pretty busy with other stuff since I last posted on the site. I’ll have to pick up where I left off on chapter 11 and see if I can finally close the book on the longest first year calc course ever lol. Thanks for the kind words.

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

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

f(x) = Kx^n

where K = e^C.