Is the circumference of an ellipse equal to 2πa√1-e^2?

In summary, the circumference of an ellipse is not strictly related to the x-coordinate, as the derivative of the area is equal to the circumference.
  • #1
tomz
35
0
Here is the simple question.

When I differentiate the area of a circle, I got the parameter. As the Area is the sum of Circumference.

But why cannot I get the circumference of ellipse by differentiate the area with respect to the radius (I set it as the longer side). The answer I got is 2πa√1-e^2, where e is eccentricity and a is half the longer side.


THank you!
 
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  • #2
hi tomz! :smile:
tomz said:
When I differentiate the area of a circle, I got the parameter.

no, you get the perimeter (from the greek "peri-", meaning "about") :wink:
As the Area is the sum of Circumference.

But why cannot I get the circumference of ellipse by differentiate the area with respect to the radius (I set it as the longer side). The answer I got is 2πa√1-e^2, where e is eccentricity and a is half the longer side.

because the gap between two ellipses of the same shape isn't of equal thickness all the way round :smile:
 
  • #3
tiny-tim said:
hi tomz! :smile:


no, you get the perimeter (from the greek "peri-", meaning "about") :wink:


because the gap between two ellipses of the same shape isn't of equal thickness all the way round :smile:

Ah, thanks. I think i got it... Thanks.
 
  • #4
tiny-tim said:
hi tomz! :smile:
because the gap between two ellipses of the same shape isn't of equal thickness all the way round :smile:

Could you clarify what you mean by this? What is the visual - where is the gap / thickness?
 
  • #5
raxAdaam said:
Could you clarify what you mean by this? What is the visual - where is the gap / thickness?

Look at the graphs of
[tex]\frac{x^2}{4}+ \frac{y^2}{9}= 1[/tex]
and
[tex]\frac{x^2}{4.01}+ \frac{y^2}{9.01}= 1[/tex]

The distance between the two ellipses, as measured along a line through the origin (I started to say "on a line perpendicular to both ellipes" but there is not such a line through any given point on one ellipse), is not the same for all such lines.
 
  • #6
alright - I see what you mean, now - how does this come to bare on the area to circumference issue? Perhaps the more relevant question is why the derivative of the area of the circle is equal to it's circumference ... thanks in advance!
 
  • #7
A circle happens to have the property that if have two circles with the same center but different radii then the distance between the two circles, as measured along a radius, is always the same. That means that if you set up the "area integral" by taking a ring of inner radius r and thickness "[itex]\Delta x[/itex]" it has area approximately [itex]2\pi r[/itex], its circumference, times the thickness [itex]\Delta r[/itex]. Other figures, not having that property, cannot be set up as the integral of the circumference.
 
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  • #8
i.e. [itex]dA = 2\pi x~dx[/itex]? So if it was possible to express the circumference of an ellipse strictly in terms of x, then we could do the same? (not sure 'bout the [itex]\Delta r[/itex] above ... did you means [itex]\Delta x[/itex]?)
 
  • #9
raxAdaam said:
So if it was possible to express the circumference of an ellipse strictly in terms of x, then we could do the same?

but then it wouldn't be ∫ (circumference) dr,

the dr would be a d(f(r)) … totally different formula! :redface:
 
  • #10
Suppose you took an arbitrary ellipse, and then for each point P on the ellipse, you drew the tangent line at that point, and then the normal line (perpendicular to the tangent line) at that point. And you specified the point Q on the normal line which is a distance x from P. What would the locus of all such points Q look like? Would it be an ellipse, or something weirder? I'm pretty sure that it would be continuous, because I think I can prove that doing this procedure for any continuous curve will yield a continuous curve.
 
  • #11
Imagine what happens to the locus of points Q as the radii are reduced equally. The original ellipse appears to shrink in greater proportion along the minor axis than the major axis. This will eventually cause the minor axis to be reduced to zero, while the radii increase as you move from the minor to the major axis. This figure resembles the infinity symbol, definitely not an ellipse.
 

Related to Is the circumference of an ellipse equal to 2πa√1-e^2?

What is the formula for finding the circumference of an ellipse?

The formula for finding the circumference of an ellipse is 2π√((a^2+b^2)/2), where a and b are the lengths of the semi-major and semi-minor axes respectively.

How is the circumference of an ellipse different from a circle?

The circumference of an ellipse is different from a circle because it varies depending on the length of the semi-major and semi-minor axes. In a circle, the circumference is always equal to 2πr, where r is the radius.

Can the circumference of an ellipse be calculated using only the length of one axis?

Yes, the circumference of an ellipse can be approximated using only the length of one axis. This is known as the Ramanujan approximation and is not exact, but it provides a close estimate.

What is the significance of the circumference of an ellipse in real life?

The circumference of an ellipse is important in many real-life applications, such as calculating the perimeter of an elliptical track for running or the circumference of an elliptical orbit for a satellite. It is also used in engineering and design to create ellipses with specific dimensions.

How can the circumference of an ellipse be measured or determined experimentally?

The circumference of an ellipse can be measured or determined experimentally by using a string to measure the perimeter of the ellipse and then measuring the length of the string. This can then be used to calculate the circumference using the formula 2π√((a^2+b^2)/2).

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