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Thread: differentiating spherical surface

  1. #1 differentiating spherical surface 
    Forum Masters Degree bit4bit's Avatar
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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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  3. #2  
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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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  4. #3  
    Forum Masters Degree bit4bit's Avatar
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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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  5. #4  
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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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  6. #5  
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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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  7. #6  
    Forum Masters Degree bit4bit's Avatar
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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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  8. #7  
    Forum Masters Degree bit4bit's Avatar
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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 R<sup>2</sup>)?...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 R<sup>2</sup>cosΦ, 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| = R<sup>2</sup>cosΦ = 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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  9. #8  
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    Quote Originally Posted by bit4bit
    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

    Quote Originally Posted by bit4bit
    ∂P/∂Θ x ∂P/∂Φ, but these vectors are meant to be velocities aren't they
    No, they’re not.
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  10. #9  
    Forum Masters Degree bit4bit's Avatar
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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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  11. #10  
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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)<sup>2</sup>+y'(t)<sup>2</sup>+z'(t)<sup>2</sup>)=1, so x'(t)<sup>2</sup>+y'(t)<sup>2</sup>+z'(t)<sup>2</sup> = 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 cos<sup>2</sup>(t)+sin<sup>2</sup>(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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  12. #11  
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    What is |dΨ/dt|?
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  13. #12  
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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))<sup>2</sup>+(cos(t))<sup>2</sup>+1<sup>2</sup>] = √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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  14. #13  
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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 (x<sub>0</sub>, y<sub>0</sub>, z<sub>0</sub>) is on the curve defined by Ψ, then there is some t<sub>0</sub> with (x<sub>0</sub>, y<sub>0</sub>, z<sub>0</sub>) = Ψ(t<sub>0</sub>). But then (x<sub>0</sub>, y<sub>0</sub>, z<sub>0</sub>) = Φ(f<sup>-1</sup>(t<sub>0</sub>)), so (x<sub>0</sub>, y<sub>0</sub>, z<sub>0</sub>) 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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  15. #14  
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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 = √ [-(d<sup>2</sup>f/dt<sup>2</sup>)sin<sup>2</sup>(f) + (d<sup>2</sup>f/dt<sup>2</sup>)cos<sup>2</sup>(f) + (d<sup>2</sup>f/dt<sup>2</sup>)]

    = √ [(d<sup>2</sup>f/dt<sup>2</sup>)(sin<sup>2</sup>(f) + cos<sup>2</sup>(f) + 1)]

    = √ [2(d<sup>2</sup>f/dt<sup>2</sup>)]

    = (√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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  16. #15  
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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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  17. #16  
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    Hey there, I think it's just a case of the book not wording its answer very well, as thats the answer it expected for the next part.

    I also noticed an error with my calculation above:

    = √ [2(d<sup>2</sup>f/dt<sup>2</sup>)]

    = (√2)(df/dt)

    so, df/dt = 1/√2, so f(t) = t/√2

    therefore, Φ(t) = (cos(t/√2), sin(t/√2), t/√2)
    √(d<sup>2</sup>f/dt<sup>2</sup>) doesn't necessarily = df/dt, so it should be:

    √(d<sup>2</sup>f/dt<sup>2</sup>) = 1/√2
    d<sup>2</sup>f/dt<sup>2</sup>=(1/√2)<sup>2</sup>
    =1/2

    so, df/dt = t/2, so Φ(t) = (cos(t/2), sin(t/2), t/2)

    Ayway, I've been working through my book calculus book, and I've just finished line and surface integrals, and almost finished the section on calculus with parameterizations, so next up in the book is looking at the vector fields and vector operations of divergence/curl.
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  18. #17  
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    I didn't notice that before--why do you have second derivatives floating around? You took a dot product of a vector involving first derivatives with itself--you should get squares of first derivatives, not second derivatives.
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    Quote Originally Posted by bit4bit
    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'.
    They could have created the latitude lines like the longitude lines. It would have created an eastern and western pole.

    Sincerely,


    William McCormick
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  20. #19  
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    Or they could have created longitude lines like latitude lines. Then we would have two equators rather than one.
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  21. #20  
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    Quote Originally Posted by William McCormick
    They could have created the latitude lines like the longitude lines. It would have created an eastern and western pole.
    And that would have been stupid.
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  22. #21  
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    Quote Originally Posted by serpicojr
    I didn't notice that before--why do you have second derivatives floating around? You took a dot product of a vector involving first derivatives with itself--you should get squares of first derivatives, not second derivatives.
    Oh yeh I just realized, for some reason I was confusing the second derivative notation with the power 2, and thinking I could treat it as a power using squares and square roots (Rookie mistake ). It was actually a scalar multiplication of the vector, dΨ/df, and the scalar, df/dt. The whole calculation should be:

    dΦ/dt = (dΨ/df)(df/dt) = <-sin(f), cos(f), f> * df/dt

    1 = √ [-(df/dt)<sup>2</sup>sin<sup>2</sup>(f) + (df/dt)<sup>2</sup>cos<sup>2</sup>(f) + (df/dt)<sup>2</sup>]

    = √ [(df/dt)<sup>2</sup>(sin<sup>2</sup>(f) + cos<sup>2</sup>(f) + 1)]

    = √ [2(df/dt)<sup>2</sup>]

    = (√2)(df/dt)

    so, (df/dt) = 1/√2, and f(t) = t/√2 + C

    therefore, Φ(t) = (cos(t/√2 + C), sin(t/√2 + C), t/√2 + C)

    Please tell me that's right.
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  23. #22  
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    Yup!
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  24. #23  
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    Hoorah!! Got there in the end. Thanks for the help.
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  25. #24  
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    waw i wish i knew these stuff
    what grade r u ?
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  26. #25  
    Forum Masters Degree bit4bit's Avatar
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    Grade drop-out.

    Nah..I learnt up to A-level maths, which was mostly calculus, and some other stuff like vectors, and matrices..(though I have forgotton most of the matrices stuff). Then I dropped out for a year, and just continued learning calculus on my own, as some mental stimulation. I now have a place on a degree programme though, so loving it. As for Serpicojr I think he almost has his Phd for mathematics?
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  27. #26  
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    Quote Originally Posted by bit4bit
    Grade drop-out.

    Nah..I learnt up to A-level maths, which was mostly calculus, and some other stuff like vectors, and matrices..(though I have forgotton most of the matrices stuff). Then I dropped out for a year, and just continued learning calculus on my own, as some mental stimulation. I now have a place on a degree programme though, so loving it. As for Serpicojr I think he almost has his Phd for mathematics?
    interesting and good
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  28. #27  
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    Quote Originally Posted by bit4bit
    As for Serpicojr I think he almost has his Phd for mathematics?
    Yes!
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