What Does Equating Areas Reveal About Function Behavior?

[MarkFL]

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:

View attachment 1331

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

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

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 $x$?

 

[anemone]

My solution:

Spoiler

Part(b):

Since I have no idea how to prove the part a of the problem, I'd like to borrow here from the given expression in part a so that I could use it to solve for $f(x)$.

$$\int_0^{t^2}\left(\sqrt{y}-\sqrt{\frac{y}{a}} \right)\,dy=\int_0^t (x^2-f(x))\,dx$$

$$\left[\frac{2y^{\frac{3}{2}}}{3}-\frac{2y^{\frac{3}{2}}}{3\sqrt{a}} \right]_0^{t^2}=\left[ \frac{x^3}{3} \right]_0^{t}-\int_0^t f(x)\,dx$$

$$\int_0^t f(x)\,dx=\left[ \frac{x^3}{3} \right]_0^{t}-\left[\frac{2y^{\frac{3}{2}}}{3}-\frac{2y^{\frac{3}{2}}}{3\sqrt{a}} \right]_0^{t^2}$$

$$\int_0^t f(x)\,dx=\frac{t^3}{3} -\frac{2t^3}{3}+\frac{2t^3}{3\sqrt{a}} $$

$$\int_0^t f(x)\,dx=\frac{2t^3}{3\sqrt{a}}-\frac{t^3}{3} $$

Now, if we integrate a function $f$ and then differentiate the integral with respect to its upper endpoint ($t$ as in our case) we get $f$ back again.

$$\frac{d}{dt} \int_0^t f(x)\,dx=f(t)$$

which means

$$\frac{d}{dt} \int_0^t f(x)\,dx=\frac{d \left(\frac{2t^3}{3\sqrt{a}}-\frac{t^3}{3} \right)}{dt} $$

$$f(t)=\frac{d \left(\frac{2t^3}{3\sqrt{a}}-\frac{t^3}{3} \right)}{dt}$$

$$f(t)=\frac{2t^2}{\sqrt{a}}-t^2 $$

and hence $$f(x)=\frac{2x^2}{\sqrt{a}}-x^2 $$

Part (c):

We're told that $f(x) \ge 0$, so we have

$$\frac{2x^2}{\sqrt{a}}-x^2 \ge 0$$

$$x^2\left(\frac{2}{\sqrt{a}}-1 \right)\ge 0$$

and since $x^2 \ge 0$ for all real $x$, the inequality becomes

$$\frac{2}{\sqrt{a}}-1\ge 0$$

$$\frac{2}{\sqrt{a}}\ge 1$$

$$\sqrt{a}\ge 2$$ and therefore $$a \ge 4$$.

The largest value that a can have so that $f(x) \ge 0$ is hence 4.

 

[MarkFL]

Great job with parts b) and c), anemone! (Sun)

Here is my solution:

Spoiler

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:

$$ax^2=t^2$$

Taking the positive root, we have:

$$x=\frac{t}{\sqrt{a}}$$

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

$$A_t=\int_0^{t^2}\sqrt{y}-\sqrt{\frac{y}{a}}\,dy$$

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

$$B_t=\int_0^t x^2-f(x)\,dx$$

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

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

Here's another slightly different approach:

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

$$dA_t=\left(t-\frac{t}{\sqrt{a}} \right)\,dy$$

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

$$dB_t=\left(x^2-f(x) \right)\,dx$$

We require these differentials to be the same, hence:

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

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

$$y=t^2\,\therefore\,t=\sqrt{y}$$

and so we have:

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

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

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

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

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

$$\left(t-\frac{t}{\sqrt{a}} \right)2t=t^2-f(t)$$

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

$$f(t)=\left(\frac{2-\sqrt{a}}{\sqrt{a}} \right)t^2$$

Hence:

$$f(x)=\left(\frac{2-\sqrt{a}}{\sqrt{a}} \right)x^2$$

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

$$\frac{2-\sqrt{a}}{\sqrt{a}}\ge0$$

$$2\ge\sqrt{a}$$

$$4\ge a$$

 

[anemone]

Hey MarkFL,

Thank you for showing us the neat, concise and clear way to prove part a of the problem and if you ask me, this proving of a certain equation is correct is not my cup of tea!(Sun)(Tongueout)