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    Radius of Curvature

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    (a) Find the point of intersection of the line containing the points (3; 2; 1) and (4; 3; 3), and the plane containing the points (5; 4; 2), (3; 1; 6) and (6; 5; 3).

    (b) Find the radius of curvature at (1; 0; 1) on the three dimensional curve given by
    r(t) = cos ti + sin tj + e^tk. Sketch this curve, and briefly describe its shape.

    © BrainMass Inc. brainmass.com December 24, 2021, 9:45 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/radius-curvature-409880

    SOLUTION This solution is FREE courtesy of BrainMass!

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    (i) As is known the equation of the line which contains the points ( ) and ( ) is

    , (1)

    Here we have = 3, = 2, = 1 , and 4, 3, 3. Hence, from (1) we obtain

    , , .

    Therefore, the equation of the line, given in the parametric form is

    , . (2).

    Let us now write the equation of the plane which contains points ( ), ( ) and . It is known that it has a form the determinant

    =0, (3)

    Here = 5, = 4, = 2, 3, 1, 6 and 6, =5, =3. Let us substitute these values in (3), and then evaluate the determinant. For that let us factorize it by first row. Then we have

    Therefore, we obtain that the equation of the plane is

    . (4)

    Substituting values of x,y, z given by (2) in (4), we obtain the following equation

    Solving it, we obtain that

    .

    Substituting this value of t into (2), we obtain that

    , 1, .

    Therefore, the intersection of the given line and the given plane is ( ).

    (ii) Recall that the curvature of a curve can be calculated by the following formula

    , (5)
    while it's radius is equal to

    .

    Since the point (1,0,1) is located on this curve, we have

    and .

    This implies that Further, we have

    and .

    Substituing , we have

    . (6)

    Therefore, the vector product is equal to

    =
    Therefore

    = = .

    Similarly, from (6), we have

    .

    Now applying (5), we obtain

    .

    Therefore the radius of the curvature is

    =2 .

    To sketch the curve , let us observe that its projection to the plane is a circle with the radius and the center at (0,0,0). Moreover, when tends to , the third coordinate of the point in the curve tends to . Besides, when tends to , the third coordinate approaches . Therefore, the curve has a form of the spiral reeling on the vertical cylinder of the radius 1, and such that one of its ends approaches the plane, but never touches it. While the second end goes up infinitely.

    Remark. The picture is taken from

    http://www.google.com/search?q=helix+curve&hl=en&biw=1440&bih=809&prmd=ivns&tbm=isch&tbo=u&source=univ&sa=X&ei=dOwLTtaAIuHy0gGew9z9DQ&sqi=2&ved=0CFAQsAQ

    (However, this curve is not the so-called helix, though looks like it in some sense).The lower end of the curve shown in the picture (is very close , but) does not touch the plane, and all the more does not touch axis.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:45 pm ad1c9bdddf>
    https://brainmass.com/math/calculus-and-analysis/radius-curvature-409880

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