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| bit4bit |
 Posted: Wed Jun 04, 2008 5:38 am Post subject: differentiating spherical surface |
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I'm having some trouble understanding this:
On the Earth's surface, your lattitude is Φ, and your longitude is Θ. Assuming the Earth is a perfect sphere, your posisiton on the Earth's surface is given by: (R is radius of Earth)
P(Θ,Φ) = ( Rcos(Φ)cos(Θ), Rcos(Φ)sin(Θ), Rsin(Φ) )
so, ∂P/∂Φ = < -Rsin(Φ)cos(Θ), -Rsin(Φ)sin(Θ), Rcos(Φ) >
so, |∂P/∂Φ| = R (I won't write the working)
Now, ∂P/∂Θ = < -Rcos(Φ)sin(Θ), Rcos(Φ)cos(Θ), 0 >
so, |∂P/∂Θ| = Rcos(Φ)
Now what I don't get is if P is your position on the surface of the Earth (or any uniform sphere), then why should the rate of change of your position with respect to your latitude..
|∂P/∂Φ| = R
... be any different to the rate of change of position with respect to your longitude?..
|∂P/∂Θ| = Rcos(Φ)
I must be missing something obvious here, because it doesn't make sense that the rate of change of position on the Earths surface should be different when going 'over' the sphere, to going 'around' the sphere....not if the sphere is a perfect sphere.
Cheers
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| Harold14370 |
| Posted: Wed Jun 04, 2008 5:50 am Post subject: |
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| At the poles you can walk in a circle and cover 360 degrees of longitude in a few steps. |
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| bit4bit |
| Posted: Wed Jun 04, 2008 5:57 am Post subject: |
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True, but at the equator, you could walk in a circle, and cover 360' of lattitude in a few steps, so basically I can't understand why |∂P/∂Φ| is not equal to Rcos(Θ), just as |∂P/∂Θ| = Rcos(Φ).
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| Harold14370 |
| Posted: Wed Jun 04, 2008 6:53 am Post subject: |
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| No. Go get your globe out of the basement and dust it off, then look at the latitude and longitude lines. To change latitude you have to go north or south. |
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| serpicojr |
| Posted: Wed Jun 04, 2008 7:45 am Post subject: |
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The important thing is this. All longitude lines are great circles--they have maximal circumference among circles on the sphere, which is 2πR. Latitude lines, however, are circles that vary in circumference as the latitude varies: when latitude is ±90º, we're at the poles, and the circles have circumference 0; when latitude is 0º, we're at the equator, and the circle has circumference 2πR. In general, the circumference happens to be 2πRcos(Φ) for any latitude -90º ≤ Φ ≤ 90º.
So if you move at a constant angular velocity around a circle on the surface of a sphere, your linear speed depends on the circumference of the circle. In fact, it depends linearly. And since circumference depends linearly on R times the cosine of latitude, so does your linear speed. So your speed is kRcos(Φ) for some k, and you've shown that k = 1. |
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| bit4bit |
| Posted: Wed Jun 04, 2008 8:10 am Post subject: |
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Thanks both, I had a look at my globe and totally got it.  Think I forgot that there's a difference between 'latitude' and 'longitude'.
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| bit4bit |
| Posted: Fri Jun 06, 2008 3:08 pm Post subject: |
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I've got a couple more questions about this.
1.) If ∂P/∂Φ represents your angular velocity,w, as you move around the circumference of a circle, then is it equal to R because your displacement,s, around the circumference of a circle is related to your angle,Φ, by s=RΦ, and w=ds/dΦ=R? So then linear velocity is <RcosΦ,RsinΦ> (in R2)?...just wondering where the 'k' came from in your post, Sepicojr.
2. If you take the cross product of ∂P/∂Θ and ∂P/∂Φ:
∂P/∂Θ x ∂P/∂Φ
.. you get a vector that is perpendicular to the Earth's surface. The magnitude of this vector is R2cosΦ, and the book says it means that "if you stake out a parcel of land that is 1 radian of longitude by 1 radian of latitude, then its area is proportional to the cosine of your latitude"...so |∂P/∂Θ x ∂P/∂Φ| = R2cosΦ = Area of surface enclosed by 1 radian of logitude by 1 radian of latitude. I noticed that the result is equal to R x RcosΦ = ∂P/∂Θ x ∂P/∂Φ, but these vectors are meant to be velocities aren't they, and an Area is a distance multiplied by another distance.
I'm getting a bit mixed up with this can anyone help?
Cheers
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| JaneBennet |
| Posted: Fri Jun 06, 2008 4:28 pm Post subject: |
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| bit4bit wrote: |
| and w=ds/dΦ=R? |
Your angular velocity is ω = dΦ/dt. If s is distance along a circle of longitude, then s = RΦ, so your linear velocity along that circle is ds/dt = Rω.
If both your latitude and your longitude are changing, then I suppose your linear velocity can be found by using the chain rule for partial derivatives:
dP/dt = ∂P/∂Φ·dΦ/dt + ∂P/∂Θ·dΘ/dt
| bit4bit wrote: |
| ∂P/∂Θ x ∂P/∂Φ, but these vectors are meant to be velocities aren't they |
No, they’re not.
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| bit4bit |
| Posted: Sun Jun 08, 2008 4:22 am Post subject: |
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Thanks Jane, I've thought about it a bit more, and I think I understand it now.
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| bit4bit |
| Posted: Sun Jun 08, 2008 11:53 am Post subject: |
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I've got another question about parameterizations, and thought I might as well drop it in here instead of making a new thread.
Let Ψ(t) = (cos(t), sin(t), t)
1. Find a parameterization, Φ, of the curve parameterized by Ψ, in which |dΦ/dt| = 1.
I don't really understand what I have to do. dΦ/dt = <x'(t), y'(t), z'(t)> so √(x'(t)2+y'(t)2+z'(t)2)=1, so x'(t)2+y'(t)2+z'(t)2 = 1, but I don't know how I'm supposed to relate this to Ψ(t).
I thought about the parameterization Φ(Ψ)=(x(Ψ),y(Ψ),z(Ψ)), having a parameterization of a parameterization, but then x, y and z, are each related to cos(t), sin(t), and t, but I don't know how. I also know that cos2(t)+sin2(t) = 1, and I'm sure this'll be helpful.
I just want someone to give me a nudge, so I can have a go. Is any of what I've written along the right lines? Thanks alot.
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| serpicojr |
| Posted: Sun Jun 08, 2008 12:12 pm Post subject: |
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| bit4bit |
| Posted: Sun Jun 08, 2008 2:27 pm Post subject: |
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I've got it now, dΦ/dt is supposed to be a unit vector of dΨ/dt.
So, |dΨ/dt| = √[(-sin(t))2+(cos(t))2+12] = √2
So, dΦ/dt = (dΨ/dt) / |dΨ/dt| = <-sin(t)/√2, cos(t)/√2, t/√2>
= <-√2 sin(t)/2, √2 cos(t)/2, √2/2>
So |dΦ/dt| = 1
Then Φ(t) = ∫ (dΦ/dt) dt = (√2 cos(t)/2, √2 sin(t)/2, t*√2/2)
Cheers
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| serpicojr |
| Posted: Sun Jun 08, 2008 2:38 pm Post subject: |
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Now hold on, partner. Does this parameterize the same curve as the original equation? (Hint: no... you figure out why.) So you're going to have to think about this a little more carefully.
One way is this (this may be the only way, and I wouldn't be surprised if it were): an easy way to find a new parameterization is to take an invertible function f: R -> R and then to define:
Φ(t) = Ψ(f(t))
This certainly parameterizes the same curve: if (x0, y0, z0) is on the curve defined by Ψ, then there is some t0 with (x0, y0, z0) = Ψ(t0). But then (x0, y0, z0) = Φ(f-1(t0)), so (x0, y0, z0) is also on the curve defined by Φ. A symmetric argument shows that any point on the curve defined by Φ is on the curve defined by Ψ. Now, assuming that f is differentiable, how can you relate the derivatives of Φ to those of Ψ and f? |
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| bit4bit |
| Posted: Mon Jun 23, 2008 6:08 am Post subject: |
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Hi Serpicojr, sorry I've not replied yet, I've been moving house and have no internet for a few weeks. Currently using the local library when I can.
Anyway, It doesn't parameterize the original curve because it is the original curve scaled in x,y, and z, by √2/2. So another way to change the rate of change of position with respect to time is to simply change how fast the curve is parameterized, by having time as a new function, f(t), then using, Φ(t) = Ψ(f(t)). Is that what you're saying? So then we need to find f(t), such that |dΦ/dt| = 1.
Doing that, I got:
dΦ/dt = (dΨ/df)(df/dt) = <-sin(f), cos(f), f> * df/dt
1 = √ [-(d2f/dt2)sin2(f) + (d2f/dt2)cos2(f) + (d2f/dt2)]
= √ [(d2f/dt2)(sin2(f) + cos2(f) + 1)]
= √ [2(d2f/dt2)]
= (√2)(df/dt)
so, df/dt = 1/√2, so f(t) = t/√2
therefore, Φ(t) = (cos(t/√2), sin(t/√2), t/√2)
I didn't understand the last part of what you said about the invertable function, but Ive checked the books answer and it was:
Φ(t)= (√2 cos(t)/2, √2 sin(t)/2, t*√2/2)
The question is the first part in a question about 'torsion', and so this answer leads on to the later parts of the question. I can't understand why their answer is this, because as you pointed out it parameterizes a different curve, but i'll post the other parts of the question if you need.
Cheers.
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| serpicojr |
| Posted: Mon Jun 23, 2008 8:22 am Post subject: |
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| Yeah, I'm not sure what your book is thinking, but I agree with your new answer--you trace out the same curve, and you get the correct magnitude of the derivative. |
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