LessonNeat Analytic Geometry Problem

MarkFL

La Villa Strangiato
Staff member
Moderator
Math Helper
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 $$x=t$$. Refer to the diagram below:

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 $$x$$?

Here's my solution:

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:

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

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

$$\displaystyle 4\ge a$$

harpazo and anemone

harpazo

Pure Mathematics
Over my head for sure. Is this calculus 3?

MarkFL

La Villa Strangiato
Staff member
Moderator
Math Helper
Over my head for sure. Is this calculus 3?
It was a problem I found online posted as a challenging problem...this is elementary calculus with analytic geometry.

harpazo and anemone

Elementary???

MarkFL

La Villa Strangiato
Staff member
Moderator
Math Helper
Elementary???
Yes. Elementary, as opposed to advanced, such as you would find in analysis courses.

harpazo

Pure Mathematics
What's the difference between analysis and calculus 3?

MarkFL

La Villa Strangiato
Staff member
Moderator
Math Helper
Analysis is more rigorous, and proof based than Calc I-III. In some schools it might be called "Advanced Calculus."

harpazo

harpazo

Pure Mathematics
Thank you. There is a course called Calculus and Analytic Geometry taught at a few of the CUNY colleges in NYC.

Staff member