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:

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

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 $f(x)$.

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

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

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

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

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

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.

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

which means

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

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

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

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

Part (c):

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

\(\displaystyle \frac{2x^2}{\sqrt{a}}-x^2 \ge 0\)

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

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

\(\displaystyle \frac{2}{\sqrt{a}}-1\ge 0\)

\(\displaystyle \frac{2}{\sqrt{a}}\ge 1\)

\(\displaystyle \sqrt{a}\ge 2\) and therefore \(\displaystyle a \ge 4\).

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:

\(\displaystyle ax^2=t^2\)

Taking the positive root, we have:

\(\displaystyle x=\frac{t}{\sqrt{a}}\)

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

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

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

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

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

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

Here's another slightly different approach:

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

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

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

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

We require these differentials to be the same, hence:

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

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

\(\displaystyle y=t^2\,\therefore\,t=\sqrt{y}\)

and so we have:

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

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

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

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

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

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

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

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

Hence:

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

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

\(\displaystyle \frac{2-\sqrt{a}}{\sqrt{a}}\ge0\)

\(\displaystyle 2\ge\sqrt{a}\)

\(\displaystyle 4\ge a\)

 

[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)

 

[CellCoree]

a) To show that $A_t=B_t$ for some time $t$ implies the given equality, we can start by finding the area of each region separately. The area $A_t$ is given by the integral:

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

Similarly, the area $B_t$ is given by the integral:

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

Since we are given that $A_t=B_t$, we can equate the two expressions and obtain:

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

Simplifying, we get:

$\int_0^t f(x)\,dx=0$

Since we know that $f(0)=0$, this means that the function $f(x)$ must be equal to zero for all $x$ up to $t$. Therefore, we can rewrite the integral as:

$\int_0^t f(x)\,dx=\int_0^t 0\,dx=0$

Substituting this back into the original equation, we get:

$\frac{t^3}{3}=0$

This is only true when $t=0$, so we can conclude that if $A_t=B_t$ for some time $t$, then $t=0$. Plugging this back into the original expression for $A_t$, we get:

$A_0=\int_0^0 (0-x^2)\,dx=0$

And for $B_0$, we get:

$B_0=\int_0^0 (0-f(x))\,dx=0$

Therefore, we have shown 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) Since we have shown that $t=0$ when $A_t=B_t$, we can substitute this value