Calculate ∇ . q, ∇ x q and ∇ x (∇ x q)

[franky2727]

1(a) For the vector function q(x,y,z)=(xy-z²)i +(yz+x²)j +(xz-y²)k calculate

(i)∇ . q (ii) ∇ . q (iii) ∇ x (∇ x q)

verify that the identity ∇ x (∇ x q) = ∇(∇ . q - ∇²q

i have answers of y+z+x,(-3y,-3z,1x) and (3,-1,3) respectivly for the first part but can't fathom the second part, mainly because subbing back in (which i presume is how this is shown) the part where i sub in y+z+x for (∇ . q) what do i now do with ∇ . (y+z+x). the fact that i end up with this leads me to believe that subbing in is wrong but if it is i really don't know where to go, help please!

 

[Dick]

There's nothing wrong with 'subbing in'. But you want to compute the gradient of (y+z+x). Not the divergence.

 

[franky2727]

the gradient being d/dx, d/dy or d/dz and why?

 

[Dick]

The gradient of a scalar function u is (du/dx,du/dy,du/dz). (y+z+x) is a scalar function. Look up the definition of gradient. Both sides of your identity are vectors. What's the laplacian of q?

 

[franky2727]

so then that gets me down to ∇²q=(-2,2,-2) if I'm not mistaken is this the same as doing ∇(∇ . q) giving me ∇ . (y+z+x)=(-2,2,-2) which gives me (1,1,1)=(-2,2,-2) ? (something tells me I'm missing a scalar here)

 

[franky2727]

i assume that if i have made a mistake that's not going to be clear enough so i'll show my working ∇x(∇xq)=∇(∇.q)-∇²q

(3,-1,3)=∇.(y+z+x)-∇²q
(3,-1,3)=(1,1,1)-∇²q
(2,-2,2)=∇²q
(2,-2,2)=∇.(d/dx,d/dy,d/dz).q=(y+z+x).∇
(2,-2,2)=(1,1,1)

 

[Dick]

You can't take ∇ . (y+z+x). y+z+x is a scalar function. The problem says you should take it's gradient. Like I said before.

 

[Dick]

To find the laplacian of q, take the second derivative of q with respect to x (getting a vector), the second derivative wrt y and the second derivative wrt z and add all three vectors. That's what the ∇^2 is supposed to convey.

 

[franky2727]

so I am left with (3,-1,3)=(2,-2,2)+ ∇(∇.q)

so (3,-1,3)=(2,-2,2)+∇(y+z+x)

so ∇(y+z+x) is just the gradient of (y+z+x) and is just (1,1,1) which makes both sides equal nice on. i am right in thinking this ∇(y+z+x) is dot prod not cros rite?

 

[Dick]

Yes, it's just gradient. ∇(∇ . q)=∇(y+z+x)=(1,1,1). As you said.

 

[franky2727]

orite, so now we have that cleared up I've managed to get almost instantly stuck on the second part of the question. Consider the vector fieldq=yzcos(xy)i+xz(cos(xy)j+(sin(xy)+2z)k show that ∇xq=0

i tried to solve this like i did part 2 of question 1 ((∇x q)) by using the determinant, however i end up with =zcos(xy)+yxz cos(xy) -2 instead of zero?

 

[Dick]

How did you wind up with a scalar instead of a vector?? E.g. the x component of the curl should be d/dy(sin(xy)+2z)-d/dz(xzcos(xy)). What's that? What are the other two components?

 

[franky2727]

i ended up with erm det{(i,j,k),(d/dx,d/dy,d/dz),(yzcos(xy),xzcos(xy),sin(xy)+2z

giving me i((d/dy*sin(xy) +2z)-(d/dz*xzcos(xy)) -j((d/dx(sin(xy)+2z)-(d/dz*yzcos(xy))+k((d/dx*xzcos(xy))-(d/dy*yzcos(xy)

giving me (xcos(xy)-xcos(xy))i - (ycos(xy)-ycos(xy))j + ((zcos(xy)-yzxsin(xy))-(zcos(xy)-(yzxsin(xy))=0

ok so maybe i didnt get that but i do now and that's what matters sorry silly mistake:P i'll be back in 10 minutes when the next question trips me up (joke i hope) thanks for all your help very kind of you :)