I don't really know this statement is true or false, can anyone help me on this? For the equation of ellipse below: \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] , where \[a\ge b\] Let C be the circumference of the ellipse, show that: \[2\pi a\ge C\ge 2 \pi b\]

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TopNewestI believe you can use the Squeeze Theorem on this problem. First, we can set \(a\) or \(b\) as a constant because we can scale the entire figure to a certain size. Let's say that we set \(b=1\). Now we can represent \(2\pi a\) and \(C\) as functions of \(a\); then applying the squeeze theorem via some clever argument (like algebraic manipulation, etc.) will suffice. – Daniel Liu · 3 years, 4 months ago

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In a more conceptual and less rigorous sense, we would arrive at your ellipse either if we distorted a circle with...

If \(a = b\), we have a circle with radius \(r = a = b\),

Otherwise, since \(a >b\), in case (1) we are shrinking the circle vertically and therefore shrinking the circumference.

In case (2) we are stretching the circle horizontally and therefore increasing the circumference. – Shreyas Balaji · 3 years, 4 months ago

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case 1: when \( a >b\)\( 2\pi a \) is the circumference of circle with radius \( a \), which will enclose the whole ellipse by touching at two points\( (a,0) \) and \( (-a,0) \) so we can say that \( C<2 \pi a \)

\( 2\pi b \) is the circumference of circle with radius \( b \), which will be enclosed by the whole ellipse touching at two points\( (0,b) \) and \( (0,-b) \) so we can say that \( C>2 \pi b \)

case 2: when \( a=b \)when \( a=b \), the given equation represents circle. so the circumference will be equal to either \( 2\pi a \) or \( 2\pi b \) – Crazy Singh · 3 years, 4 months ago

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– Calvin Lin Staff · 3 years, 4 months ago

Why is case 1 true? So what if the circle encloses the entire ellipse? Why does that immediately imply that the perimeter of the circle is greater than the ellipse?Log in to reply

– Crazy Singh · 3 years, 4 months ago

by seeing the graph case 1 is absolutely clear, even the one who does not know mathematics. can tell which curve is bigger one after looking at the graph, well I will try more to express this in mathematical equations.Log in to reply

well it is quite an easy problem. well i am not formatting it as you can easily understand my method. we know that length of a function (i call it that as i myself figured it out last year)is integration of (1+(d(f(x))/dx)^2)^0.5) from x=a to x=b ,where f(x) is the function. you can integrate it yourself as it is elementary and would come to conclusion that your statement is true. – Rishabh Shukla · 3 years, 4 months ago

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\[ \pi [ 3 (a+b) - \sqrt{ (3a+b)(a+3b) }] \]

The area, on the other hand, is easily obtained, even without calculus. – Calvin Lin Staff · 3 years, 4 months ago

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– Daniel Liu · 3 years, 4 months ago

Is it not true that the perimeter of an ellipse is simply a scale factor of the perimeter of a circle? The area can be easily "derived" by letting an ellipse be a distorted circle; can't the perimeter also be derived as so?Log in to reply

For a more rigorous treatment of the above paragraph:

Say the figure that we're interested is given by the differentiable curve \( y = f(x) \), and we are shearing in the y-direction, to get the new curve \( y^* = a f(x) \). How does the area change? The initial area is \( A = \int f(x) \, dx \), and the new area is \( A^* = \int a f(x) \, dx \). Now, since constants pull out, we get \( A^* = a A \).

How about the perimeter? As mentioned by Rishabh, the perimeter is given by \( P = \int \sqrt{ 1 + \left( \frac{ dy} { dx} \right)^2 }\, dx \). The new perimeter is now \[ P^* = \int \sqrt{ 1 + \left( \frac{ dy^*} { dx} \right)^2 }\, dx = \int \sqrt{ 1 + a^2\left( \frac{ dy} { dx} \right)^2 }\, dx \] It is now clear why the constant \(a\) doesn't pull out, and as such we need not have \( P^* = a P \).

It is also hard to determine the exact relationship between \( P^* \) and \( aP \). As an example,

Ifwe know that \( \frac{dy}{dx} > 1 \) always, then for shear factor \( a > 1 \), we will have \( P^* < a P \). – Calvin Lin Staff · 3 years, 4 months agoLog in to reply

– Kho Yen Hong · 3 years, 4 months ago

Can you further show out your working? Because I try to use this formula before, but I failed to integrate it!Log in to reply