Rotational and Conservative vector field.

Discussion in 'Differentiation and Integration' started by rorl_, Jun 7, 2021.

  1. rorl_

    rorl_

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    Hello everyone !

    Could someone help me do this? I even don't know where to begin.

    1. Be F(x,y,z) = (y², 2xy + y^3z, 3ye^3z) calculate rot F and find a potential function for F.
    2. Calculate the line integral along any curve from (0,0,0) to the point (1,-1,2).
     
    rorl_, Jun 7, 2021
    #1
  2. rorl_

    HallsofIvy

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    HOW DO I ACCESS LATEX ON THIS BOARD?
    I have tried \(, [itex], [math], and $.

    If you really "have no idea where to begin", I would suggest that you begin by taking a course in, or at least read a book on, vector Calculus, typically a "Calculus III course.

    There you will learn that "rot F" is short for "nabla cross F" or "$\nabla\times F$". [math]\nabla[/math] is the "vector differential operator" [itex]\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}[/itex].

    Just as there are three kinds of "vector product", the scalar product, the dot product, and the cross product, so there are three differential operations we can perform on scalar or vector functions with nabla:

    With a scalar function, the "gradient", grad, which is like multiplying a scalar times a vector and produces a vector function:
    \(grad \phi= \nabla \phi(x,y,z)= \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial z}\vec{k}\).

    With a vector function, the "Divergence", div, which is like the dot product of two vectors and produces a scalar function:
    \(div \vec{v}= \nabla\cdot \vec{v}= \frac{\partial v_x}{\partial x}+ \frac{\partial v_y}{\partial y}+ \frac{\partial v_z}\{artial z}\).

    With a vector function, the "curl", rot, which is like the cross product of two vectors and produces a vector function:
    \(rot \vec{v}= \nabla\times \vec{v}= \left(\frac{\partial v_z}{\partial y}- \frac{\partial v_y}{\partial z}\right)\vec{i}- \left(\frac{\partial v_z}{\partial x}- \frac{\partial v_x}{\partial z}\right)\vec{j}+ \left(\frac{\partial v_y}{\partial x}- \frac{\partial v_x}{\partial y}\right)\vec{k}\).

    That rather complicated formula can be remembered as a determinant:

    \(\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \vec{i} & \vec{j} & \vec{k} \end{array}\right|\).\)
     
    Last edited: Nov 8, 2021
    HallsofIvy, Nov 8, 2021
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  3. rorl_

    HallsofIvy

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    Now, as for these specific problems,
    rot(y[sup]2[/sup], 2xy+ y[sup]3[/sup]z, 3ye[sup]3z[/sup])

    is given by (partial(3ye^(3z))/partial y- partial(2xy+ y[sup]3[/sup]z)/partial z)i- (partial(3ye[sup]3z[/sup]/partial x- partial(y[sup]2[/sup])/partial z)j+ (partial 2xy+ y[sup]3[/sup]z/partial x- (2xy+ y[sup]3[/sup]z)/partial y)k

    = (3e^(3z)- y[sup]3[/sup])i+ (2y- 3e[sup]3z[/sup])k
     
    HallsofIvy, Nov 8, 2021
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  4. rorl_

    nycmathguy

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    I don't this site accept LaTex. Contact the owner of the site. Be advised, he does not respond right away.
     
    nycmathguy, Nov 8, 2021
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