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Vector Calculus Identites Question

  1. Sep 4, 2014 #1
    1. The problem statement, all variables and given/known data

    let r (vector) =xi+yj+zk and r=sqrt(x^2+y^2+z^2), let f(r) be a C2 scalar function.

    1. Prove that ∇f = dr/df * vector r

    2. Using part 1, calculate ∇ cosh(r^5), check answer by direct calculation

    3. Using Vector Identities, calculate ∇ X (cosh(r^5)*∇f


    2. Relevant equations

    Vector Calculus Basic Identities

    3. The attempt at a solution

    i know ∇f=(df/dx,df/dy,df/dz)

    but i have no idea what df/dr, i assuming its just d/dr * f but i dont even know what d/dr is ?

    as i don't know part 1, i couldnt fully do part 2, but i did try the direct calculation and got

    =(5x(x^2+y^2+z^2)^3/2 * sinh(x^2+y^2+z^2)^5/2,5y(x^2+y^2+z^2)^3/2 * sinh(x^2+y^2+z^2)^5/2,5z(x^2+y^2+z^2)^3/2 * sinh(x^2+y^2+z^2)^5/2)

    for part 3, i'm pretty sure it equals 0 as after doing the determinant everything kind of cancels out, i can't be bothered typing this part it would take too long.

    i'm not after an answer, if someone could just explain what d/dr is and if ∇f=(df/dx,df/dy,df,dz), how am i even meant to find it as there's no f function given in the question.

    Thanks for any help i am really struggling
     
  2. jcsd
  3. Sep 4, 2014 #2

    LCKurtz

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    Don't use the same letter for the vector and its magnitude. Let's use ##\vec R = \langle x,y,z\rangle## and ##r## for its length. You have written dr/df but what you mean is to prove ##\nabla f = \frac{df}{dr}\vec R##.

    Now you have this differentiable function ##f(r)## and you want to calculate its gradient:$$
    \nabla f(r) = \langle \frac \partial {\partial x} f(r),\frac \partial {\partial y} f(r),
    \frac \partial {\partial z} f(r)\rangle$$Now use the chain rule to take those derivatives and see what happens.
     
  4. Sep 5, 2014 #3
    thanks for the reply, doing that i got:

    ∇f(r)=(x/√(x^2+y^2+z^2),y/√(x^2+y^2+z^2),z/√(x^2+y^2+z^2))

    but i am still confused as to prove ∇f=df/drR⃗, what is df/dr is that the f(r) function being derived in terms of r (the length) and then i have to multiply it by R vector.

    Thanks again for any help, i have been trying to get this question for ages now
     
  5. Sep 5, 2014 #4

    LCKurtz

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    [Edit, added] I overlooked that you didn't apply the chain rule to your vector$$
    \nabla f(r) = \langle \frac \partial {\partial x} f(r),\frac \partial {\partial y} f(r),
    \frac \partial {\partial z} f(r)\rangle$$. Remember you are to differentiate a function of ##r## with respect to ##x,y,z## and you must use the chain rule. That's why your work below shows nothing about ##f(r)## in the answer, and it should. Do you understand the chain rule?

    Yes. It is ##f'(r)##. Look at what you have above. Each component has ##\sqrt{x^2+y^2+z^2}## in the denominator. That is ##r##. Factor it out of the vector as ##\frac 1 r##. That gives you ##\frac{\vec R}{r}## which is a unit vector in the ##\vec R## direction, which I would denote as ##\hat R##. Looking at that result, I would guess what you were really asked to show was that$$
    \nabla f = \frac{df}{dr}\hat R$$which wasn't clear from your notation in the original post.
     
    Last edited: Sep 5, 2014
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