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Homework Help: Calc3 vectors, lenght of curve, equation of the sphere.

  1. Aug 29, 2010 #1
    1.Consider the process A[1,1] and B [-1,4]; let Vector U=AB

    1a.Write u using components. Write u using the standard basis.
    1b.Draw u as a position vector. Find u.
    1c.If C [1,1] and D [5,y], then find the value of y so that AB is parallel to CD.

    2. Find the length of the curve 2t2 3t2 , 0<t<[3/4]
    3. Find the equation of the sphere with center [2,3,4] and is tangent to thee xyplane. At what points does the sphere intersect the z axis?
    2.distance formula


    My Answers

    1a. -1-1=-2 4-1=3 u=<-2,3> u=<1,1> <-1,4>
    1b. u=-22+32[1/2] =131/2
    1c. [5-1/2],[y-1/2] [2,y-1/2]

    2. [PLAIN]http://img255.imageshack.us/img255/9178/hillo1.png [Broken]

    3.[PLAIN]http://img408.imageshack.us/img408/9339/hillo2.png [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 29, 2010 #2


    Staff: Mentor

    This doesn't make any sense to me. What is a "process"? What sort of multiplication is done with AB?
    Are these the parametric equations for the curve?
    The radius of this sphere is not sqrt(29).
    Last edited by a moderator: May 4, 2017
  4. Aug 29, 2010 #3
    AB is a Vector.

    Yes those are parametric equations for the curve.

    I don't know how to solve 3, I need some help.
  5. Aug 30, 2010 #4
    When does a position function hit the z axis? When the y and x axis are 0.
  6. Aug 30, 2010 #5


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    It is against forum policy to "bump" a thread and can get you banned.
  7. Aug 30, 2010 #6


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    If x= x(t) and y= y(t) then the arclength is given by
    [tex]\int\sqrt{(x'(t))^2+ (y'(t))^2} dt[/tex]
    [tex]\int\sqrt{(x(t))^2+ (y(t))^2} dt[/tex]
    which is what you seem to be trying.

    For [itex]x= 2t^2[/itex], [itex]y= 3t^2[/itex] that would be
    [tex]\int\sqrt{4t^2+ 36t^2}dt= 2\sqrt{10}\int t dt[/tex]
    not what you have.
  8. Aug 30, 2010 #7


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    3: If a sphere has center (a, b, c) and is "tangent to the xy-plane" then the radius to that plane is perpendicular to it. That means that the radius of the sphere is "c".
  9. Aug 30, 2010 #8
    Sorry about the bump, thanks for checking my work.
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