calc, arc length problem?
two lanes of a running track are modeled as semi ellipses.
the equation for lane 1 is: y=Sqrt(100-0.2x^2)
and the equation for lane 2 is: y=Sqrt(150-0.2x^2)
the starting point for lane 1 is at the negative x intercept. The finish points for both lanes are the positive x intercepts. where should the starting point be placed on lane 2 so that the two lane lengths will be equal? (assume the runners run clockwise)
Determine the starting point coordinates for lane 2.
OK so far I have that the x and y intercepts are:
Lane 1: (10,0) (0, sqrt(500)) and (0, -sqrt(500))
Lane 2: (sqrt(150),0) (0, sqrt(750)) and (0, -sqrt(750))
and the formula for the arc length for both is:
1: the integral of sqrt[1+(1/2(100-0.2(x^2))^(1/2)(0.4x))] dx
2: the integral of sqrt[1+(1/2(150-0.2(x^2))^(1/2)(0.4x))] dx
sorry if its a bit messy.... and thats all the further I can go! I have no idea how to integrate this. I'm completely lost.
Comments
You need to evaluate the first integral from -sqrt(500) to sqrt(500).
Don't forget to square f'(x) since it simplifies the equation to the point where it makes it a lot easier to integrate.
After you solve the definite integral of the first integral, set it equal to the second integral from -sqrt(750) to b and solve for b.
You don't need the y-intercepts at all.
After you find b plug it back into the original equation for lane 2 to find your y value where x = b
First you have your x-intercepts and y-intercepts mixed up:
Lane 1: y-intercept (0, 10) x-intercepts (√500, 0) and (-√500, 0)
Lane 2: y-intercept (0, √150) x-intercepts (√750, 0) and (-√750, 0)
Second, your integrals are a little off
(100-0.2(x^2))^(1/2) should be (100-0.2(x^2))^(-1/2)
(0.4x) should be (-0.4x)
and y'(x) should be squared
1: the integral of sqrt[1 + y'(x)^2]
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Calculate arc length of lane 1, using following formula:
.. √500
s = ∫ √(1+(y')²) dx
. -√500
y = √(100-0.2x²)
y' = 1/2 * (100-0.2x²)^(-1/2) * (-0.4x)
y' = -0.2x / √(100-0.2x²)
(y')² = 0.04x² / (100-0.2x²)
.. √500
s = ∫ √(1+(y')²) dx = ∫ √(1 + 0.04x²/(100-0.2x²)) dx ≈ 52.7037
. -√500
Note: There is no easy way to calculate this integral. I just used a graphing calculator that can calculate definite integrals
Similarly we find integral for second lane:
.. √750
s = ∫ √(1+(y')²) dx = ∫ √(1 + 0.04x²/(150-0.2x²)) dx ≈ 52.7037
.... n
Using definite integral calculator, and trying different values of n, I found that n ≈ -19.9
So starting point of lane 2 is at x ≈ -19.9
y ≈ √(150-0.2(-19.9)²) ≈ 8.4
Approximate starting point for lane 2: (-19.9, 8.4)
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EDIT:
This solution may seem a little messy, but a quick search of the internet shows that there is no nice formula or easy way to calculate the perimeter of an ellipse, let alone an arc length of an ellipse.
That's not too hard actually. I'll give you the directions, you follow, okay?
So, first change the equations to parametic. This will make them is to integrate. Put limit in radians, -pi to zero. Then integrate the second one from theta to zero. Equate. Find theta. Put in the parametric equations and find out x and y
i'm sorry but...wtf?!
i'm in calc ab and have never seen this in my life
sorry man but i have no idea