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

[wilcofan3]

Homework Statement

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

Homework Equations

$$4L$$arc in the first quadrant = $$L$$ellipse

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!

 

[Mark44]

Homework Statement

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

Homework Equations

$$4L$$arc in the first quadrant = $$L$$ellipse

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: $$\int \sqrt{(dx/dt)^2 + (dy/dt)^2}dt$$

$$4\int \sqrt{10(cos t)^2 - 9(sin t)^2}$$?

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.

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

Yes.

 

[HallsofIvy]

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.

 

[Mark44]

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.

 

[wilcofan3]

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?