Vector Calculus Identites Question

[jamesdocherty]

Homework Statement

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

Homework Equations

Vector Calculus Basic Identities

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 don't even know what d/dr is ?

as i don't know part 1, i couldn't 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

 

[LCKurtz]

Homework Statement

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

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.

 

[jamesdocherty]

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

 

[LCKurtz]

[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?

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.

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.