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Grid Curves

  1. Jul 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider the parametric surface r(u,v)=<vsinu, vcosu, v^2>
    a) Identify the shape of the surface

    b) The point (1,1,2) is on the surface. Find:

    i) A grid curve wit hv constant that contains this point

    ii) A grid curve with u constant that contains this point

    c) Find the tangent vector to both grid curves you just found at the point (1,1,2)

    d) Find the angle between the grid curves at the point (1,1,2).

    e) Find a parametric equation for the plane containing the wo tangent vectors from part c, and containing the point (1,1,2).

    f) Describe the relationship between the plane and the surface.




    2. Relevant equations

    Dot product. And n[tex]\cdot[/tex](r-r[tex]_{0}[/tex])=0.


    3. The attempt at a solution

    For a-c I am pretty confident that my answers are right, but d-f is where I need some help.

    a. Paraboloid

    b.
    i. <[tex]\sqrt{2}[/tex]sinu, [tex]\sqrt{2}[/tex]cosu, 2>

    ii. <v[tex]\frac{\sqrt{2}}{2}[/tex], v[tex]\frac{\sqrt{2}}{2}[/tex], v[tex]^{2}[/tex]>

    c. <[tex]\sqrt{2}[/tex],0,2> and <[tex]\frac{\sqrt{2}}{2}[/tex], [tex]\frac{\sqrt{2}}{2}[/tex], 2>

    d. I used the definition of the dot product, and dotted i and ii, using [tex]\frac{\pi}{4}[/tex] as u, and [tex]\sqrt{2}[/tex] as v. I got that [tex]\theta[/tex]=0, which doesn't seem right to me.

    e. I'm not sure on this one either. I could perhaps the second equation that I posted, but I don't know what I'd use for n.

    f. Since I'm not sure on e, I'm not sure of f. As soon as I know e though, I'm sure I can do f.
     
    Last edited: Jul 26, 2008
  2. jcsd
  3. Jul 26, 2008 #2

    tiny-tim

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    Hi BoundByAxioms! :smile:

    (have a square-root: √ and a squared: ² and a pi: π :smile:)

    Both your tangents are wrong.

    The second one, I think you've just made a error of arithmetic.

    But I don't see how you got the first one at all. :confused:

    Try again! :smile:
     
  4. Jul 26, 2008 #3
    Hmm, then I fear that I am lost. Could you give me a hint? Thanks.
     
  5. Jul 26, 2008 #4

    HallsofIvy

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    To find the tangent vector in the "u" direction, differentiate r(u,v)= <vcos u, vsin(u), v2> with respect to u. To find the tangent vector in the "v" direction, differentiate r(u,v)= <vcos u, vsin(u), v2> with respect to v. To find them at the given point, substitute [itex]u= \pi/4[/itex] and [itex]v= \sqrt{2}[/itex].
     
  6. Jul 27, 2008 #5
    So, when r(u,v)=<vcos(u), vsin(u), v2>, the derivative with respect to u is <-vsin(u), vcos(u), 0>. And the derivative with respect to v is <cos(u), sin(u), 2v>. Now, I plug in [tex]\sqrt{2}[/tex] for v and [tex]\pi/4[/tex] for u. So I get, <-1, 1, 0> and <[tex]\sqrt{2}/2[/tex], [tex]\sqrt{2}/2[/tex], 2[tex]\sqrt{2}[/tex]>. Is this correct so far?
     
  7. Jul 27, 2008 #6

    tiny-tim

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    So far … so good! :smile:

    (except … I don't know whether they expect you to give the unit vector … I note that the question asks for "the" rather than "a" tangent vector. :confused:)
     
  8. Jul 28, 2008 #7
    How would you find d? Do you uese the tangent lines or do you find the angle of the grid curve with the points plugged in or the tangent line?
     
  9. Jul 28, 2008 #8

    tiny-tim

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    Welcome to PF!

    Hi Leon! Welcome to PF! :smile:

    As BoundByAxioms said, you use the definition of the dot product (so you can find cosθ).

    And yes, you use the tangent lines … the angle between two curves is the angle between their tangent lines. :smile:
     
  10. Jul 28, 2008 #9
    Oh thx. So the answer should be 90 unless i did something wrong.
     
  11. Jul 29, 2008 #10
    I think so:

    <-1,1,0> [tex]\bullet[/tex] <[tex]\frac{\sqrt{2}}{2}[/tex], [tex]\frac{\sqrt{2}}{2}[/tex], 0> = 0. And since the dot product is zero, the angle between them is [tex]\frac{\pi}{2}[/tex]. Remember to always answer questions in radians.
     
  12. Jul 29, 2008 #11
    Thank you very much. I'll remember that.
     
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