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Math Help - Finding a Vector Function

  1. #1
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    Finding a Vector Function

    Hey guys,
    I am having a bit of a problem with vector functions. This problem says: A circular cylinder of radius 1 and the parabolic cylinder z=2x^2 intersect. What is the vector function if the curve of intersection is x(0)=0?

    I first set x = cos(t) and y = sin(t); then solved for z = 2 cos^2(t). Would this be correct? Or am I approaching this problem incorrectly?

    Any feed and help is appreciated.

    Thanks
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  2. #2
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    Re: Finding a Vector Function

    That problem doesn't make much sense to me as stated. Have you included the full statement of the problem?

    Quote Originally Posted by Beevo View Post
    A circular cylinder of radius 1 and the parabolic cylinder z=2x^2 intersect.
    There are numerous "circular cylinder of radius 1" in 3-space. Is that freedom of positioning such a circlar cylinder what this problem is about?

    Quote Originally Posted by Beevo View Post
    What is the vector function if the curve of intersection is x(0)=0?
    That reads that the "vector function" you're asked to find is a vector form for a curve.

    \text{It's asking you to find } F : \mathbb{R} \rightarrow \mathbb{R}^3.

    But the meaning of "the curve of intersection is x(0)=0" is very unclear. Is it the constant curve at the origin? Is it *some* curve satisfying x = 0 at time 0? That isn't even a point in 3 space. If it's *some* curve, then why the word "the". Saying "x(0)=0" is simply not describing "THE curve of intersection" for those two objects in 3 space.

    ---------------------------------

    What is this circular cylinder of radius 1? It seems to me that both of the below satisfy "the curve of intersection is x(0)=0". I really just don't get it.

    x^2 + (z+1)^2 = 1 \text{ implies } F(t) = (0, t, 0) = t\hat{j} \in \mathbb{R}^3 \ \forall \ t \in \mathbb{R}.

    y^2 + (z+1)^2 = 1 \text{ implies } F(t) = (0, 0, 0) = \vec{0} \in \mathbb{R}^3 \ \forall \ t \in \mathbb{R}.
    Last edited by johnsomeone; October 26th 2012 at 02:23 PM.
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  3. #3
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    Re: Finding a Vector Function

    Yes, that looks good to me. The parabolic cylinder and the circular cylinder intersect in a curve given by x= cos(t), y= sin(t), z= 2cos^2(t)
    Thanks from Beevo
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  4. #4
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    Re: Finding a Vector Function

    Well the problem provided a graph. First, the graph of the cylinder has the z-axis as the central axis, so I am assuming the equation would be x^2 + y^2 = 1. Then the other equation is z=2x^2, which intersects with the cylinder. Then the main question states, which vector function has the curve of intersection as its graph if x(0)=0.

    Also, does the x(0)=0 matter? What if it was x(0)=1.
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  5. #5
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    Re: Finding a Vector Function

    It is x(0) = 1 if x(t) = cos(t).
    Although it's not how to usually do it, in this case, if that's basically the form you're seeking, make it x(t) = sin(t), y(t) = cos(t) in order to have x(0) = 0.
    Though again, who knows what this problem is asking, so who knows if that's a reasonable answer. Surely not me.

    Edit:
    Quote Originally Posted by Beevo View Post
    Well the problem provided a graph. First, the graph of the cylinder has the z-axis as the central axis, so I am assuming the equation would be x^2 + y^2 = 1.
    Yes - that's right. That makes the problem much more clear.

    F(t) = (\sin(t), \cos(t), 2\sin^2(t)), \text{ or } \vec{F}(t) = \sin(t)\hat{i} + \cos(t)\hat{j} + 2\sin^2(t)\hat{k}, \text{ is what I bet they're looking for.}
    Last edited by johnsomeone; October 26th 2012 at 03:20 PM.
    Thanks from Beevo
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  6. #6
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    Re: Finding a Vector Function

    Quote Originally Posted by johnsomeone View Post
    It is x(0) = 1 if x(t) = cos(t).
    Although it's not how to usually do it, in this case, if that's basically the form you're seeking, make it x(t) = sin(t), y(t) = cos(t) in order to have x(0) = 0.
    Though again, who knows what this problem is asking, so who knows if that's a reasonable answer. Surely not me.

    Edit:

    Yes - that's right. That makes the problem much more clear.

    F(t) = (\sin(t), \cos(t), 2\sin^2(t)), \text{ or } \vec{F}(t) = \sin(t)\hat{i} + \cos(t)\hat{j} + 2\sin^2(t)\hat{k}, \text{ is what I bet they're looking for.}
    Yeah that is it. Thanks for your help.
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