Lesson Neat Analytic Geometry Problem

MarkFL

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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:

unknownfunction.png


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:

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

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

\(\displaystyle 4\ge a\)
 
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harpazo

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Over my head for sure. Is this calculus 3?
 

MarkFL

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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.
 
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harpazo

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Elementary???
 

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harpazo

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What's the difference between analysis and calculus 3?
 

MarkFL

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Analysis is more rigorous, and proof based than Calc I-III. In some schools it might be called "Advanced Calculus."
 
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harpazo

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

MarkFL

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The Calc I-III courses I took also had "with analytic geometry" in their titles.
 
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harpazo

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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?