# Help me to prove this statement...

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: $2\pi a\ge C\ge 2 \pi b$ Note by Kho Yen Hong
5 years, 12 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 $2\pi b$

- 5 years, 12 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?

Staff - 5 years, 12 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.

- 5 years, 12 months ago

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

1. radius $a$ vertically by factor $\frac{b}{a}$
2. radius $b$ horizontally by factor $\frac{a}{b}$

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.

- 5 years, 12 months ago

I 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.

- 5 years, 12 months ago

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.

- 5 years, 12 months ago

The perimeter of an ellipse is actually hard to determine exactly. All that we have is approximate formulas. For example, Ramanujan's approximation is

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

The area, on the other hand, is easily obtained, even without calculus.

Staff - 5 years, 12 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?

- 5 years, 12 months ago

Sadly no. We know that if 2 figures are similar, then their perimeter and areas must follow a certain ratio. This works because we have an expansion (with some center). However, the transformation from a circle to an ellipse, is a shearing in one direction. It is easy to convince yourself that when an area element (e.g. any small square) is sheared, the area changes by a proportionate amount (e.g. a rectangle). However, when we look a length element (e.g. line segments), the length changes by a varying amount, depending on how 'tilted' it was. If the line is perpendicular to the direction of shearing, then the length does not increase at all. If the line is parallel to the direction of shearing, then the length increases by $a$ times. What happens for a slanted line? Does it increase by more than $a$, exactly $a$, less than $a$? How about for a curve? (And this is where things get interesting!) This is why we can't make a similar conclusion / derivation.

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, If we know that $\frac{dy}{dx} > 1$ always, then for shear factor $a > 1$, we will have $P^* < a P$.

Staff - 5 years, 12 months ago

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

- 5 years, 12 months ago