How do I prove the circumference of an ellipse
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I'm trying to find the circumference of an ellipse with a horizontal radius of $h$ and a vertical radius of $k$. The equation for such an ellipse centered at the origin would by $(x/h)^2 + (y/k)^2 = 1$. I've tried the arc-length formula in both cartesian form and parametric form, and got stuck on both of them.
First, I set $f(x) = dfrac{k}{h}sqrt{(h^2 - x^2)}$. I tried to solve $2displaystyleint_{-h}^h sqrt{f'(x))^2 + 1} :dx$, and managed to get it down to the following:
$displaystyledfrac{2k}{h}int_{-h}^hsqrt{dfrac{x^2}{(h^2 - x^2) + (h/k)^2}}dx$
But I didn't know how to go from there either.
Next, I set $f(t) = hcos(t)$ and $g(t) = ksin(t)$. I tried to solve $displaystyleint_{-pi}^{pi}(f'(t))^2 + (g'(t))^2 dx$, and managed to get it down to the following.
$displaystyleint_{-pi}^{pi} sqrt{hsin^2(t) + kcos^2(t)}dt$
Once again, I didn't know how to go on.
arc-length
edited Dec 15 '18 at 3:49
Yadati Kiran
1,7891619
asked Dec 9 '18 at 7:52
Wire BowlWire Bowl
113
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add a comment |
$begingroup$
I'm trying to find the circumference of an ellipse with a horizontal radius of $h$ and a vertical radius of $k$. The equation for such an ellipse centered at the origin would by $(x/h)^2 + (y/k)^2 = 1$. I've tried the arc-length formula in both cartesian form and parametric form, and got stuck on both of them.
First, I set $f(x) = dfrac{k}{h}sqrt{(h^2 - x^2)}$. I tried to solve $2displaystyleint_{-h}^h sqrt{f'(x))^2 + 1} :dx$, and managed to get it down to the following:
$displaystyledfrac{2k}{h}int_{-h}^hsqrt{dfrac{x^2}{(h^2 - x^2) + (h/k)^2}}dx$
But I didn't know how to go from there either.
Next, I set $f(t) = hcos(t)$ and $g(t) = ksin(t)$. I tried to solve $displaystyleint_{-pi}^{pi}(f'(t))^2 + (g'(t))^2 dx$, and managed to get it down to the following.
$displaystyleint_{-pi}^{pi} sqrt{hsin^2(t) + kcos^2(t)}dt$
Once again, I didn't know how to go on.
arc-length
edited Dec 15 '18 at 3:49
Yadati Kiran
1,7891619
asked Dec 9 '18 at 7:52
Wire BowlWire Bowl
113
$endgroup$
$begingroup$
see answers of this.
$endgroup$
– achille hui
Dec 15 '18 at 4:20
add a comment |
$begingroup$
I'm trying to find the circumference of an ellipse with a horizontal radius of $h$ and a vertical radius of $k$. The equation for such an ellipse centered at the origin would by $(x/h)^2 + (y/k)^2 = 1$. I've tried the arc-length formula in both cartesian form and parametric form, and got stuck on both of them.
First, I set $f(x) = dfrac{k}{h}sqrt{(h^2 - x^2)}$. I tried to solve $2displaystyleint_{-h}^h sqrt{f'(x))^2 + 1} :dx$, and managed to get it down to the following:
$displaystyledfrac{2k}{h}int_{-h}^hsqrt{dfrac{x^2}{(h^2 - x^2) + (h/k)^2}}dx$
But I didn't know how to go from there either.
Next, I set $f(t) = hcos(t)$ and $g(t) = ksin(t)$. I tried to solve $displaystyleint_{-pi}^{pi}(f'(t))^2 + (g'(t))^2 dx$, and managed to get it down to the following.
$displaystyleint_{-pi}^{pi} sqrt{hsin^2(t) + kcos^2(t)}dt$
Once again, I didn't know how to go on.
arc-length
edited Dec 15 '18 at 3:49
Yadati Kiran
1,7891619
asked Dec 9 '18 at 7:52
Wire BowlWire Bowl
113
$endgroup$
I'm trying to find the circumference of an ellipse with a horizontal radius of $h$ and a vertical radius of $k$. The equation for such an ellipse centered at the origin would by $(x/h)^2 + (y/k)^2 = 1$. I've tried the arc-length formula in both cartesian form and parametric form, and got stuck on both of them.
First, I set $f(x) = dfrac{k}{h}sqrt{(h^2 - x^2)}$. I tried to solve $2displaystyleint_{-h}^h sqrt{f'(x))^2 + 1} :dx$, and managed to get it down to the following:
$displaystyledfrac{2k}{h}int_{-h}^hsqrt{dfrac{x^2}{(h^2 - x^2) + (h/k)^2}}dx$
But I didn't know how to go from there either.
Next, I set $f(t) = hcos(t)$ and $g(t) = ksin(t)$. I tried to solve $displaystyleint_{-pi}^{pi}(f'(t))^2 + (g'(t))^2 dx$, and managed to get it down to the following.
$displaystyleint_{-pi}^{pi} sqrt{hsin^2(t) + kcos^2(t)}dt$
Once again, I didn't know how to go on.
arc-length
arc-length
edited Dec 15 '18 at 3:49
Yadati Kiran
1,7891619
asked Dec 9 '18 at 7:52
Wire BowlWire Bowl
113
edited Dec 15 '18 at 3:49
Yadati Kiran
1,7891619
asked Dec 9 '18 at 7:52
Wire BowlWire Bowl
113
edited Dec 15 '18 at 3:49
Yadati Kiran
1,7891619
edited Dec 15 '18 at 3:49
Yadati Kiran
1,7891619
edited Dec 15 '18 at 3:49
Yadati Kiran
1,7891619
1,7891619
asked Dec 9 '18 at 7:52
Wire BowlWire Bowl
113
asked Dec 9 '18 at 7:52
Wire BowlWire Bowl
113
asked Dec 9 '18 at 7:52
Wire BowlWire Bowl
113
113
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see answers of this.
$endgroup$
– achille hui
Dec 15 '18 at 4:20
add a comment |
$begingroup$
see answers of this.
$endgroup$
– achille hui
Dec 15 '18 at 4:20
$begingroup$
see answers of this.
$endgroup$
– achille hui
Dec 15 '18 at 4:20
$begingroup$
see answers of this.
$endgroup$
– achille hui
Dec 15 '18 at 4:20
add a comment |
1 Answer
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The perimeter of an ellipse cannot be in general calculated analytically, you can solve it numerically. Here you can see the wikipedia article for elliptic integrals. What you are concretly looking for is the complete elliptic integral forms.
I said in general because we know analytically the perimeter of a circunference, which is a particular case of an ellipse.
answered Dec 9 '18 at 8:05
AlexAlex
63
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1 Answer
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1 Answer
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active
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$begingroup$
The perimeter of an ellipse cannot be in general calculated analytically, you can solve it numerically. Here you can see the wikipedia article for elliptic integrals. What you are concretly looking for is the complete elliptic integral forms.
I said in general because we know analytically the perimeter of a circunference, which is a particular case of an ellipse.
answered Dec 9 '18 at 8:05
AlexAlex
63
$endgroup$
add a comment |
$begingroup$
The perimeter of an ellipse cannot be in general calculated analytically, you can solve it numerically. Here you can see the wikipedia article for elliptic integrals. What you are concretly looking for is the complete elliptic integral forms.
I said in general because we know analytically the perimeter of a circunference, which is a particular case of an ellipse.
answered Dec 9 '18 at 8:05
AlexAlex
63
$endgroup$
add a comment |
$begingroup$
The perimeter of an ellipse cannot be in general calculated analytically, you can solve it numerically. Here you can see the wikipedia article for elliptic integrals. What you are concretly looking for is the complete elliptic integral forms.
I said in general because we know analytically the perimeter of a circunference, which is a particular case of an ellipse.
answered Dec 9 '18 at 8:05
AlexAlex
63
$endgroup$
The perimeter of an ellipse cannot be in general calculated analytically, you can solve it numerically. Here you can see the wikipedia article for elliptic integrals. What you are concretly looking for is the complete elliptic integral forms.
I said in general because we know analytically the perimeter of a circunference, which is a particular case of an ellipse.
answered Dec 9 '18 at 8:05
AlexAlex
63
answered Dec 9 '18 at 8:05
AlexAlex
63
answered Dec 9 '18 at 8:05
AlexAlex
63
answered Dec 9 '18 at 8:05
AlexAlex
63
63
add a comment |
add a comment |
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see answers of this.
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– achille hui
Dec 15 '18 at 4:20