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Problem of the Week #79 - September 30th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: If the ellipse $x^2/a^2 + y^2/b^2 = 1$ is to enclose the circle $x^2+y^2=2y$, what values of $a$ and $b$ minimize the area of the ellipse?

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's question was correctly answered by anemone (you just made it in at the last minute (Sun)) and MarkFL. I'm going to post both of their fine solutions below.

anemone's solution:
Since the ellipse is to enclose the circle, we know the ellipse will intersect the circle for only one value of $x$(this means also for two different $y$ values since the ellipse is to enclose the circle), that is, if we substitute the equation of the ellipse into the equation of the circle, the discriminant of the quadratic equation that we got must be zero.

So, if we substitue \(\displaystyle x^2=\left( 1-\frac{y^2}{b^2} \right) \left(a^2 \right)\) into the equation \(\displaystyle x^2+y^2=2y\), we get

\(\displaystyle x^2+y^2=2y\)

\(\displaystyle \left( 1-\frac{y^2}{b^2} \right) \left(a^2 \right)+y^2=2y\)


\(\displaystyle (b^2-y^2)a^2 +b^2y^2=2b^2y\)

\(\displaystyle (b^2-a^2)y^2 -2b^2y+a^2b^2\)

Set the discriminant equals to zero we get

\(\displaystyle (-2b^2)^2-4(b^2-a^2)(a^2b^2)=0\)

\(\displaystyle 4b^2(b^2-a^2(b^2-a^2)=0\)

\(\displaystyle 4b^2(b^2-a^2b^2+a^4)=0\)

Since \(\displaystyle b^2 \ne 0\), it must be \(\displaystyle b^2-a^2b^2+a^4=0\) or \(\displaystyle b^2=\frac{a^4}{a^2-1}\)

Now, bear in mind that we're asked to find the values of [FONT=MathJax_Math]a[/FONT] and [FONT=MathJax_Math]b[/FONT] so that the area of the ellipse is at its minimum, this simply means we need to find an equation the defines the area of the ellipse(A) in terms of $a$ and the area of the ellipse could be found using calculus technique. Since the ellipse has 4 symmetric quarter, we can find the area of any one quarter and multiply it by 4 in order to get the total area.

\(\displaystyle A=4\int_0^b \frac{a}{b}\sqrt{b^2-y^2} dy\) [SIZE=+0]
[/SIZE]

We now make the substitution \(\displaystyle y=b\sin \theta\) so that \(\displaystyle dy=b\cos \theta d \theta\)and the area is now given by

\(\displaystyle A=4\frac{a}{b}\int_0^{\frac{\pi}{2}}(\sqrt{b^2-b^2\sin^2 \theta}) b\cos \theta d \theta\) [SIZE=+0]
[/SIZE]

\(\displaystyle A=4\frac{a}{b}(b^2)\int_0^{\frac{\pi}{2}} \cos^2 \theta d \theta\) [SIZE=+0]
[/SIZE]

\(\displaystyle A=4ab\int_0^{\frac{\pi}{2}} \left( \frac{\cos 2\theta}{2}+\frac{1}{2} \right) d \theta\) [SIZE=+0]
[/SIZE]

\(\displaystyle A=4ab \left( \frac{\sin 2\theta}{4}+\frac{\theta}{2} \right)_0^{\frac{\pi}{2}} \) [SIZE=+0]
[/SIZE]

\(\displaystyle A=4ab \left( 0+\frac{\pi}{4}-0 \right) \) [SIZE=+0]
[/SIZE]

\(\displaystyle A=ab\pi \)[SIZE=+0]
[/SIZE]

or \(\displaystyle A^2=a^2b^2\pi^2=a^2(\frac{a^4}{1-a^2})\pi^2 =\left(\frac{a^6}{a^2-1} \right)\pi^2\)

To minimize A, we differentiate the equation of $A$ with respect to $a$ and obtain

\(\displaystyle 2A\frac{dA}{da}=\left(\frac{(1-a^2)(6a^5)-(a^6)(-2a)}{(1-a^2)^2} \right)\pi^2\)

\(\displaystyle 2A\frac{dA}{da}=\frac{a^5\pi^2(4a^2-6)}{(1-a^2)^2} \)

Since $A \ne 0$, \(\displaystyle \frac{dA}{da}=0\) is possible iff \(\displaystyle a^2=\frac{6}{4}=\frac{3}{2}\) and since $a>0$, we must have \(\displaystyle a=\sqrt{\frac{3}{2}}\) and when \(\displaystyle a=\sqrt{\frac{3}{2}}\), \(\displaystyle b^2=\frac{(\frac{3}{2})^2}{\frac{3}{2}-1}=\frac{9}{2}\) which gives \(\displaystyle b=\frac{3}{\sqrt{2}}\).

Now, to determine the nature of the inflexion point where $(a, b)=\left(\sqrt{\frac{3}{2}}, \frac{3}{\sqrt{2}} \right)$ gives the minimum/maximum area of the ellipse, we consider the gradient of the graph of $A$ versus $a$ from left to right and we see that

a1.1$\sqrt{\dfrac{3}{2}}\approx 1.2247$2
A9.12033.5
$\dfrac{dA}{da}$-17.205.23
sign of $\dfrac{dA}{da}$-0+
change of slope of tangent\_/

Therefore, we can conclude that the minimal area of the ellipse is attained at the point $(a, b)=\left(\sqrt{\frac{3}{2}}, \frac{3}{\sqrt{2}} \right)$.



MarkFL's solution:
First, we want to define our objective function $A(a,b)$, the area of the ellipse, in terms of the parameters $a$ and $b$:

We know the ellipse is centered at the origin, and will pass through the points $(0,b),\,(a,0)$ and so by symmetry, we may state:

\(\displaystyle A(a,b)=4b\int_0^a \sqrt{1-\left(\frac{x}{a} \right)^2}\,dx\)

using the substitution:

\(\displaystyle \frac{x}{a}=\sin(\theta)\,\therefore\,dx=a\cos( \theta)\,d \theta\)

we now have:

\(\displaystyle A(a,b)=4ab\int_0^{\frac{\pi}{2}}\sqrt{1-\sin^2(\theta)}\cos(\theta)\,dx\)

\(\displaystyle A(a,b)=4ab\int_0^{\frac{\pi}{2}}\cos^2(\theta)\,dx\)

Using a double-angle identity for the cosine function, we obtain:

\(\displaystyle A(a,b)=2ab\int_0^{\frac{\pi}{2}}1+\cos(2\theta)\,dx\)

Applying the FTOC, there results:

\(\displaystyle A(a,b)=2ab\left[\theta+\frac{1}{2}\sin(2\theta) \right]_0^{\frac{\pi}{2}}=\pi ab\)

Thus, our objective function is:

\(\displaystyle A(a,b)=\pi ab\)

Now, for the constraint, lets begin with the equations of the circle and the ellipse:

\(\displaystyle x^2+y^2=2y\)

\(\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Obviously there will have to be at least one point where the two curves are tangent. At these tangent points, the slope of the two curves will have to be the same as well.

If we multiply the second equation by $-a^2$ and then add the two equations, we eliminate $x$ to obtain:

\(\displaystyle y^2-\frac{a^2}{b^2}y^2=2y-a^2\)

Writing this equation in standard quadratic form in $y$, we have:

\(\displaystyle \left(b^2-a^2 \right)y^2-2b^2y+a^2b^2=0\)

Next, implicitly differentiating the equation of the circle, we find:

\(\displaystyle 2x+2y\frac{dy}{dx}=2\frac{dy}{dx}\)

\(\displaystyle \frac{dy}{dx}=\frac{x}{1-y}\)

And doing the same for the equation of the ellipse, we obtain:

\(\displaystyle \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0\)

\(\displaystyle \frac{dy}{dx}=-\frac{b^2x}{a^2y}\)

Equating the two slopes, we then get:

\(\displaystyle \frac{x}{1-y}=-\frac{b^2x}{a^2y}\)

\(\displaystyle y=\frac{b^2}{b^2-a^2}\)

Substituting this into the quadratic we found in $y$, we now have:

\(\displaystyle \left(b^2-a^2 \right)\left(\frac{b^2}{b^2-a^2} \right)^2-2b^2\left(\frac{b^2}{b^2-a^2} \right)+a^2b^2=0\)

And this gives us our constraint function:

\(\displaystyle g(a,b)=b^2\left(a^2-1 \right)-a^4=0\)

Using Lagrange multipliers, we obtain the system:

\(\displaystyle \pi b=\lambda\left(2ab^2-4a^3 \right)\)

\(\displaystyle \pi a=\lambda\left(2a^2b-2b \right)\)

Solving both equations for $\lambda$ and equating, we obtain:

\(\displaystyle \lambda=\frac{\pi b}{2a\left(b^2-2a^2 \right)}=\frac{\pi a}{2b\left(a^2-1 \right)}\)

\(\displaystyle \frac{b}{a\left(b^2-2a^2 \right)}=\frac{a}{b\left(a^2-1 \right)}\)

\(\displaystyle b^2\left(a^2-1 \right)=a^2\left(b^2-2a^2 \right)\)

\(\displaystyle a^2b^2-b^2=a^2b^2-2a^4\)

\(\displaystyle b^2=2a^4\)

Substituting for $b^2$ into the constraint function, we find:

\(\displaystyle 2a^4\left(a^2-1 \right)-a^4=0\)

Taking $0<a$ and dividing through by $a^4$, we obtain:

\(\displaystyle 2\left(a^2-1 \right)-1=0\)

\(\displaystyle a^2=\frac{3}{2}\)

\(\displaystyle a=\sqrt{\frac{3}{2}}\)

And thus:

\(\displaystyle b=\sqrt{2}a^2=\frac{3}{\sqrt{2}}\)

And so, we may conclude that the area of the ellipse is minimized for:

\(\displaystyle (a,b)=\left(\sqrt{\frac{3}{2}},\frac{3}{\sqrt{2}} \right)\)

Here is a plot of the circle and the ellipse:

 
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