## Tuesday, April 7, 2015

### Find The Unknown Function...

Let $f(x)$ be an unknown function defined on $[0,\infty)$ with $f(0)=0$ and $f(x)\le x^2$ for all $x$. For each $0\le t$, let $A_t$ be the area of the region bounded by $y=x^2$, $y=ax^2$ (where $1<a$) and $y=t^2$. Let $B_t$ be the area of the region bounded by $y=x^2$, $y=f(x)$ and $x=t$. See the image below:

a) Show that if $A_t=B_t$ for some time $t$, then:

[MATH]\int_0^{t^2}\sqrt{y}-\sqrt{\frac{y}{a}}\,dy=\int_0^t x^2-f(x)\,dx[/MATH]

b) Suppose $A_t=B_t$ for all $0\le t$. Find $f(x)$.

c) What is the largest value that $a$ can have so that $0\le f(x)$ for all $x$?

a) We may take the $y$-coordinate of point $P$, and using the point on the curve $y=ax^2$ having the same $y$-coordinate, state:

[MATH]ax^2=t^2[/MATH]

Taking the positive root, we have:

[MATH]x=\frac{t}{\sqrt{a}}[/MATH]

And so integrating with respect to $y$, we have:

[MATH]A_t=\int_0^{t^2}\sqrt{y}-\sqrt{\frac{y}{a}}\,dy[/MATH]

Now, integrating with respect to $x$, we see we may state:

[MATH]B_t=\int_0^t x^2-f(x)\,dx[/MATH]

So, if $A_t=B_t$, then we may state:

[MATH]\int_0^{t^2}\sqrt{y}-\sqrt{\frac{y}{a}}\,dy=\int_0^t x^2-f(x)\,dx[/MATH]

Here's another slightly different approach:

If we add a horizontal strip to $A_t$, we find the area of this strip is:

[MATH]dA_t=\left(t-\frac{t}{\sqrt{a}} \right)\,dy[/MATH]

And adding a vertical strip to $B_t$, we find the area of this strip to be:

[MATH]dB_t=\left(x^2-f(x) \right)\,dx[/MATH]

We require these differentials to be the same, hence:

[MATH]\left(t-\frac{t}{\sqrt{a}} \right)\,dy=\left(x^2-f(x) \right)\,dx[/MATH]

On the left, we need to express $t$ as a function of $y$, and we know:

[MATH]y=t^2\,\therefore\,t=\sqrt{y}[/MATH]

and so we have:

[MATH]\left(\sqrt{y}-\sqrt{\frac{y}{a}} \right)\,dy=\left(x^2-f(x) \right)\,dx[/MATH]

Now is is a simple matter of adding all of the elements of the areas, to get:

[MATH]\int_0^{t^2}\sqrt{y}-\sqrt{\frac{y}{a}}\,dy=\int_0^t x^2-f(x)\,dx[/MATH]

b) Now, using the results of part a), differentiating with respect to $t$, we find:

[MATH]\frac{d}{dt}\int_0^{t^2}\sqrt{y}-\sqrt{\frac{y}{a}}\,dy=\frac{d}{dt}\int_0^t x^2-f(x)\,dx[/MATH]

[MATH]\left(t-\frac{t}{\sqrt{a}} \right)2t=t^2-f(t)[/MATH]

Solving for $f(t)$, we find:

[MATH]f(t)=\left(\frac{2-\sqrt{a}}{\sqrt{a}} \right)t^2[/MATH]

Hence:

[MATH]f(x)=\left(\frac{2-\sqrt{a}}{\sqrt{a}} \right)x^2[/MATH]

c) In order for $f(x)$ to be non-negative, we require the coefficient of $x^2$ to be non-negative:

[MATH]\frac{2-\sqrt{a}}{\sqrt{a}}\ge0[/MATH]

[MATH]2\ge\sqrt{a}[/MATH]

[MATH]4\ge a[/MATH]