Calculating Ellipse Length: 9x^2 + 10y^2 = 90 (to 6 decimal places)

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Homework Statement



Find the length of the ellipse 9x^2 + 10y^2 = 90 correct to six decimal places.

Homework Equations



4Larc in the first quadrant = Lellipse

The Attempt at a Solution



Just checking to see if I did this right:

9x^2 + 10y^2 = 90

x^2/10 + y^2/9 = 1

Therefore a = \sqrt{10} and b = 3.

This makes x = \sqrt{10}sin t and y = 3 cos t

Since \int \sqrt{(dx)^2 + (dy)^2} is the formula for arc length, do I just get: 4\int \sqrt{10(cos t)^2 - 9(sin t)^2}?

And are the bounds just from 0 to \pi/2?

Thanks!
 
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wilcofan3 said:

Homework Statement



Find the length of the ellipse 9x^2 + 10y^2 = 90 correct to six decimal places.


Homework Equations



4Larc in the first quadrant = Lellipse

The Attempt at a Solution



Just checking to see if I did this right:

9x^2 + 10y^2 = 90

x^2/10 + y^2/9 = 1

Therefore a = \sqrt{10} and b = 3.

This makes x = \sqrt{10}sin t and y = 3 cos t

Since \int \sqrt{(dx)^2 + (dy)^2} is the formula for arc length, do I just get:
This is the formula: \int \sqrt{(dx/dt)^2 + (dy/dt)^2}dt
wilcofan3 said:
4\int \sqrt{10(cos t)^2 - 9(sin t)^2}?
There's a mistake in the line above. What do you get when you square dy/dt? Also, don't forget your dt in the integrand.
wilcofan3 said:
And are the bounds just from 0 to \pi/2?
Thanks!
Yes.
 
You should be aware that the calculation of the circumference of an ellipse results in what is known as an "elliptic integral" which cannot be done in terms of elementary functions.
 
The instructions provided by the OP imply that numerical integration is to be performed.
Find the length of the ellipse 9x^2 + 10y^2 = 90
correct to six decimal places.
 
Mark44 said:
This is the formula: \int \sqrt{(dx/dt)^2 + (dy/dt)^2}dt

There's a mistake in the line above. What do you get when you square dy/dt? Also, don't forget your dt in the integrand.

Oh, is it just that it should be positive instead of negative? That's the only thing I'm seeing, and yeah, I need to put the dt in there.

So, other than that, is it correct?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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