# LessonNeat Analytic Geometry Problem

#### MarkFL

##### La Villa Strangiato
Staff member
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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
Banned
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

Banned
Elementary???

#### MarkFL

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

#### harpazo

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

#### MarkFL

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

• harpazo

#### harpazo

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

#### MarkFL

##### La Villa Strangiato
Staff member
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The Calc I-III courses I took also had "with analytic geometry" in their titles.

• anemone and harpazo

#### harpazo

##### Pure Mathematics
Banned
The Calc I-III courses I took also had "with analytic geometry" in their titles.
What is the hardest topic to grasp in calculus 1, 2, and 3, in your opinion?