Finding a Function f(x,y,z) from a Given Vector Field Using ∇ x q=0

[terryfields]

consider the vector field q=yzcos(xy)i+xzcos(xy)j+(sin(xy)+2z)k

verify that ∇ x q=0 and concequently determine a function f(x,y,z) such that q=∇f

now i have completed the first part by using the determinant and multiplying out, it is quite obvious that the whole thing equals zero, however i don't know how to do the second part "concequently determine a function f(x,y,z) such that q=∇f"

my first question is does ∇f simply mean the dot product when there is no dot there? i assume it does

and my second question is how do i approach this question? thanks

 

[tiny-tim]

my first question is does ∇f simply mean the dot product when there is no dot there? i assume it does

and my second question is how do i approach this question? thanks

Hi terryfields!

No, ∇f is the gradient …

for an ordinary function f(x) of a vector x, ∇f is the gradient vector (∂f/∂x₁, … ,∂f/∂x_n)

And if ∇ x q = 0, then there must be an f with q = ∇f.

(and basically you find f by integrating each component of q, and using a little common-sense for the "constants"! )

 

[terryfields]

ah ok, so do i just need to integrate each term in respect to x or in respect to x,y and z?

 

[tiny-tim]

ah ok, so do i just need to integrate each term in respect to x or in respect to x,y and z?

The i term with respect to x, the j term with respect to y, and the k term with respect to z.

 

[terryfields]

ok, so by my calculations i would get (zsin(xy))i+(zsin(xy))j+(zsin(xy)+z²)k i assume this is how it's writen? as i now assume i need to do ∇ x f = 0 to find my constants?

 

[HallsofIvy]

And, as tiny tim said, "using a little common-sense for the "constants"!" The reason he put "constants" in quotes is because the partial derivative with respect to one variable treats the other variables as constants- and you have to consider that in taking an "ant-derivative".

For example, if the v were the vector 2xy\vec{i}+ (x^2+ z)\vec{j}+ (1+ y)\vec{k}, then you must have \partial f/\partial x= 2xy, \partial f/\partial y= x^2+ z, and \partial f/\partial z= 1+ y. Integrating \partial f/\partial x= 2xy gives f= x^2y+ \phi(y,z) where \phi can be any function of y and z since its derivative, with respect to x, is 0. If we differentiate that with respect to y, we get \partial f/\partial y= x^2+ \partial \phi/\partial y= x^2+ z or \partial \phi/\partial y= z. (Notice how the "x^2" cancelled- that HAS to happen because \phi is a function of y and z only. That's why not every function is a "gradient".)

Integrating \partial\ph/\partial y= z gives us \phi(y,z)= yz+ \psi(z). Again, the "constant" of integration may be a function of z. That means that we now have f(x,y,z)= x^2y+ yz+ \psi(z) and differentiating that with respect to z, \partial f/\partial z= y+ d\psi/dz= 1+ y and the "y" term cancels leaving d\psi/dz= 1. (That is an ordinary derivative because \psi is a function of z only.) Integrating that, \psi(z)= z+ C where now the "constant" of integration is now really a constant.

That gives f(x,y,z)= x^2y+ yz+ z+ C and you can check that \nabla (x^2y+ yz+ z)= 2xy\vec{i}+ (x^2+ z)\vec{j}+ (y+1)\vec{k} for any constant C.

 

[terryfields]

i think i understand

so when we get zsin(xy)i +zsin(xy)j+z² from integrating we have to check what we would get from differentiating this?

in this case the i term and j term don't need constants because they would differentiate right back to what they were integrated from? however for the k term we do because if we differentiate z² we just get 2z not 2z+ sin(xy)
so we need to add on a term that will differentiate with respects to z to get sin(xy) namely zsin(xy)?

giving our full answer to be zsin(xy)i+zsin(xy)j+z²+zsin(xy)

 

[tiny-tim]

so when we get zsin(xy)i +zsin(xy)j+z² from integrating
…
giving our full answer to be zsin(xy)i+zsin(xy)j+z²+zsin(xy)

Nooo … f is an ordinary function, not a vector …

when you integrate, you're aiming to get the same function f, each of the three times …

that's where the "constants" come in …

how can you combine zsin(xy) and z² to get a single f that fits?

 

[terryfields]

f=z² +zsin(xy)?? is there a method to do this by or is it just inspection to find it?

 

[tiny-tim]

f=z² +zsin(xy)?? is there a method to do this by or is it just inspection to find it?

That's right!

(and the method is what you just did …

didn't you like it? )