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    Vector Calculus, Partial Derivatives, and Polar Cylindrical and Spherical Coordinates

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    Vectors
    1. The points A(2; -3, 3), B(3,5;4), C(3;8;-2) and D(4;4;6) are vertices of a tetrahedron. Find the volume of the tetrahedron.

    2. There are two vectors: a = (2; -6; -4) and b = (3; -4; 2). Calculate:
    (i) a * b
    (ii) (2a - 3b) * (a + 2b) ; (a + b)^2 ; (a - b)^2

    3. There are two vectors: a = (3; -1; -2) and b = (1; 2; -1). Calculate:
    a * b (2a + b) * b and (2a - b) * (2a + b)

    4. A body is being moved from the initial point A(0, 0; 5) till the point B(3; -2; 5) (along a straight line) due to the action of the force F = (3; -2; -5). What is the work done by the force?

    5. Find the plane that passes through the point M(2; 4; -5) and which is parallel to the vectors: a = (3; 1; -1) and b = (1; -2; 1)

    Partial Derivatives
    Please see the attached document.

    Polar Cylindrical and Spherical Coordinates
    Please see the attached document.

    © BrainMass Inc. brainmass.com December 24, 2021, 10:59 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/vector-calculus-partial-derivatives-coordinates-524122

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    SOLUTION This solution is FREE courtesy of BrainMass!

    Please see the attached MS Word document for the full solutions to the questions posed. Included are diagrams and formulas used to calculate the answers.

    1.

    The tetrahedron is defined by three vectors, u, v and w:

    The volume of the tetrahedron is given by:
    (1.1)
    Where S is the base area and h is the altitude to the base..
    The base is half the area of a parallelepiped defined by the vectors u and v:

    The area is:
    (1.2)

    But if we use the identity:
    (1.3)
    Thus:
    (1.4)
    The cross product results in a vector which is perpendicular to S.
    The height of the tetrahedron is the projection of w on n:

    (1.5)
    The angle is the angle between w and n, thus we can use the dot product identity:
    (1.6)
    Therefore the altitude to the base is:
    (1.7)

    The volume of the tetrahedron is therefore:
    (1.8)
    We have to account for the fact that and apply the absolute value since the volume must be a positive number.

    If are the vectors from the origin to points A,B,C and D respectively we have:
    (1.9)
    In our case:


    The triple product can be written in a matrix form:

    Which gives:

    The volume of the tetragon is


    2.

    We have

    Then:
    (2.1)
    The dot product is distributive:
    (2.2)
    And commutative:
    (2.3)
    Therefore:

    (2.4)
    Which gives (see 2.1):

    For we get:
    (2.5)
    Plugging in the numbers:

    By the same token:
    (2.6)
    With the numbers:


    3.

    For cross product we have:
    (3.1)
    It is anti-commutative
    (3.2)
    It is distributive:
    (3.3)
    In our case:

    Then:

    (3.4)

    Now,
    (3.5)
    Plugging in the numbers:

    While:

    (3.6)
    Plugging in the numbers:


    4.

    The work along a straight line is given by:
    (4.1)
    Where is the vector from point A to point B
    (4.2)
    In our case:


    5.

    If two planes are parallel, then a normal vector to one is a normal vector to the other as well:

    The plane M' is defined by the vectors u and v, therefore the normal to this plane is:
    (5.1)
    This is also the normal to plane M, therefore it must be perpendicular to any arbitrary vector w that lies in the plane M as well.

    The condition for this is:
    (5.2)
    We know that the point P is in the plane M. The vector from the origin to this point is
    An arbitrary point on the plane is , and it is denoted by the general vector
    Therefore an arbitrary vector that lies in the plane that contains the point P is:
    (5.3)
    So the equation of the plane M is simply:
    (5.4)
    In our case:

    And:


    Thus the equation of the plane is:


    6.

    When we take a partial derivative with respect to x, we treat y as a constant, and vice versa - when we take the partial derivative with respect to y we keep x constant.
    In our case:
    (6.1)
    Then:
    (6.2)
    Keeping y constant we get:
    (6.3)
    And now keeping x constant, this is a derivative of a product:


    Contd.

    (6.4)

    Now we reverse the order of differentiation, keeping x constant:
    (6.5)
    And again, differentiating with respect to x, keeping y constant

    (6.6)
    And indeed we see that
    (6.7)

    Now our function is:
    (6.8)
    Then:
    (6.9)
    And:

    (6.10)
    Reversing the order of differentiation:
    (6.11)

    And:

    (6.12)
    And indeed:
    (6.13)

    7.

    In polar coordinates:
    (7.1)

    Then the equation
    (7.2)
    Can be converted to Cartesian coordinates:
    (7.3)
    This gives:
    (7.4)
    Squaring both sides:
    (7.5)

    Using (7.1):

    (7.6)
    This is a parabola opening to the left with focus at and it intersects the y-axis at
    It looks like:

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

    © BrainMass Inc. brainmass.com December 24, 2021, 10:59 pm ad1c9bdddf>
    https://brainmass.com/math/calculus-and-analysis/vector-calculus-partial-derivatives-coordinates-524122

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