Results 1 to 4 of 4

Thread: [SOLVED] differentiation integration with respect to a vector

  1. #1
    Member
    Joined
    Jun 2008
    From
    Plymouth
    Posts
    125
    Thanks
    5

    [SOLVED] differentiation integration with respect to a vector

    Let r=xi+yj+zk be a displacement vector. We have:

    $\displaystyle df=r.\nabla f$

    Where f is a scalar field

    What if f is a vector field. Does that mean that:
    $\displaystyle df=r.\nabla f$

    What about integration of scalar/vector fields with respect to vectors. I am interested in how you solve these:

    $\displaystyle \int f \, dr$ Where f is a scalar field and

    $\displaystyle \int g \, dr$ Where g is a vector field.

    Any input is appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by fobos3 View Post
    Let r=xi+yj+zk be a displacement vector. We have:

    $\displaystyle df=r.\nabla f$

    Where f is a scalar field

    What if f is a vector field. Does that mean that:
    $\displaystyle df=r.\nabla f$

    What about integration of scalar/vector fields with respect to vectors. I am interested in how you solve these:

    $\displaystyle \int f \, dr$ Where f is a scalar field and

    $\displaystyle \int g \, dr$ Where g is a vector field.

    Any input is appreciated.
    $\displaystyle \nabla f $ is the gradient of scalar field f. A vector does not have a gradient in this sense. A vector can have curl and divergence, but not gradient. The gradient of a scalar field is a vector field.

    Nabla, is defined as : $\displaystyle \nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz}) $

    If you have a vector $\displaystyle \vec{f}=(F_x,F_y,F_z)$, then you can have only two multiplicative operations with this vector and the nabla vector. Those are the dot product ($\displaystyle \nabla . \vec{f} $)and the cross product ($\displaystyle \nabla \times \vec{f} $). And these represent divergence and curl respectively. It does not make sense to have scalar multiplcation between two vectors, which is what you are proposing with $\displaystyle \nabla \vec{f}$.

    You may have scalar multiplication between two scalars, and scalar multiplcation between a scalar and a vector (a lá, $\displaystyle \vec{grad}(f)$, but you may not have scalar multiplication between two vectors.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    From
    Plymouth
    Posts
    125
    Thanks
    5
    Quote Originally Posted by Mush View Post
    $\displaystyle \nabla f $ is the gradient of scalar field f. A vector does not have a gradient in this sense. A vector can have curl and divergence, but not gradient. The gradient of a scalar field is a vector field.

    Nabla, is defined as : $\displaystyle \nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz}) $

    If you have a vector $\displaystyle \vec{f}=(F_x,F_y,F_z)$, then you can have only two multiplicative operations with this vector and the nabla vector. Those are the dot product ($\displaystyle \nabla . \vec{f} $)and the cross product ($\displaystyle \nabla \times \vec{f} $). And these represent divergence and curl respectively. It does not make sense to have scalar multiplcation between two vectors, which is what you are proposing with $\displaystyle \nabla \vec{f}$.

    You may have scalar multiplication between two scalars, and scalar multiplcation between a scalar and a vector (a lá, $\displaystyle \vec{grad}(f)$, but you may not have scalar multiplication between two vectors.
    Huh? The gradient of a vector field is a second order tensor.

    v=$\displaystyle v_1e_1+v_2e_2+v_3e_3$

    Then $\displaystyle \nabla v=\sum _{i=1}^3 \sum _{j=1}^3 \frac{\partial v_j}{\partial x_i}e_ie_j$

    Where $\displaystyle e_1,e_2,e_3$ are vectors.

    Check this out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2008
    From
    Plymouth
    Posts
    125
    Thanks
    5
    Anyone? Come on guys I'm really stuck here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differentiate with respect to vector
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 10th 2010, 08:09 PM
  2. Replies: 3
    Last Post: Sep 18th 2010, 02:31 PM
  3. [SOLVED] Differentiation / Integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 13th 2009, 07:14 PM
  4. Vector Angle with respect to y axix
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Sep 20th 2009, 02:07 AM
  5. Differentiation of 2^n with respect to n
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 11th 2009, 04:38 AM

Search Tags


/mathhelpforum @mathhelpforum