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Math Help - Find the tangent vector

  1. #1
    Senior Member bkarpuz's Avatar
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    Post Find the tangent vector

    Dear MHF members,

    I have the following problem.

    Problem. Let the curve (ellipse) \alpha be the intersection of the surfaces x^{2}+2y^{2}+4z^{2}=10 and x-2y=0. Find the tangent vector of \alpha at the point (2,1,-1).

    I was just trying to find the direction of the tangent vector. I tried to find it by using the cross product of the gradients of the surfaces at the given point (as this vector should be in the same direction with the tangent vector), and get the result (-16,-8,-12). But when I try to solve it by using the derivative of the parametric form of the curve at the given point, I get a different result, which is (2\sqrt{2/3},\sqrt{2/3},\sqrt{3/2}). Up to me, both of these results should be the same direction. Please help me.

    Thanks a lot.
    bkarpuz
    Last edited by bkarpuz; October 9th 2011 at 12:22 PM.
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  2. #2
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    earboth's Avatar
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    Re: Find the tangent vector

    Quote Originally Posted by bkarpuz View Post
    Dear MHF members,

    I have the following problem.

    Problem. Let the curve (ellipse) \alpha be the intersection of the surfaces x^{2}+2y^{2}+4z^{2}=10 and x-2y=0. Find the tangent vector of \alpha at the point (2,1,-1).

    I was just trying to find the direction of the tangent vector. I tried to find it by using the cross product of the gradients of the surfaces at the given point (as this vector should be in the same direction with the tangent vector), and get the result (-16,-8,-12). But when I try to solve it by using the derivative of the parametric form of the curve at the given point, I get a different result, which is (2\sqrt{2/3},\sqrt{2/3},\sqrt{3/2}). Up to me, both of these results should be the same. Please help me.

    Thanks a lot.
    bkarpuz
    1. Both vectors are collinear:

    Since \frac23 \cdot \frac94 = \frac32 you can re-write \sqrt{\frac32}=\sqrt{\frac23 \cdot \frac94} = \frac32 \cdot \sqrt{\frac23}

    2. Extract the factor (-8) from the 1st vector. You'll get: \langle 2,1,\frac32 \rangle

    Extract the factor \sqrt{\frac23} from the 2nd vector and you'll get: \langle 2,1,\frac32 \rangle

    Thus both vectors have the same direction.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Re: Find the tangent vector

    Quote Originally Posted by earboth View Post
    1. Both vectors are collinear:

    Since \frac23 \cdot \frac94 = \frac32 you can re-write \sqrt{\frac32}=\sqrt{\frac23 \cdot \frac94} = \frac32 \cdot \sqrt{\frac23}

    2. Extract the factor (-8) from the 1st vector. You'll get: \langle 2,1,\frac32 \rangle

    Extract the factor \sqrt{\frac23} from the 2nd vector and you'll get: \langle 2,1,\frac32 \rangle

    Thus both vectors have the same direction.
    Thank you earboth, so my result is right. The book says (2,3,3) for the answer. But it is obviously wrong.
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