Register to reply 
Deriving cross product and dot product, stuck at beginning. 
Share this thread: 
#1
Nov2411, 05:29 PM

P: 53

1. The problem statement, all variables and given/known data
Assuming that ∅ is a differentiable scalar valued function and F a differentiable vector field, derive the following identities. a)∇(dotted with)(∅F) = ∇∅(dotted with)F + ∅∇(dotted with)F b)∇(crossed with)(∅F) = ∇∅(crossed with)F + ∅∇(crossed with)F 2. Relevant equations 3. The attempt at a solution Honestly don't know where to start. 


#2
Nov2411, 05:32 PM

P: 53

Nevermind, delete this, I've got it, just didn't put the initial effort into it.



#3
Nov2911, 07:14 PM

P: 53

Would this be the correct derivation for part a)
So far all I see is: ∅F is the vector field ∅ = ∅(x,y,z) F = <P,Q,R> ∇(dotted with)F = x partial P + y partial Q + z partial R ∇∅ = <x partial ∅, y partial ∅, z partial ∅> ∅F = <∅P, ∅Q, ∅R> a)∇(dotted with)(∅F) = [∇∅](dotted with)F + ∅∇(dotted with)F ∇(dotted with)(∅F) = [∇∅](dotted with)F + ∅[∇(dotted with)F] ∇(dotted with)(∅F) = [∇∅](dotted with)F + ∅(∂/∂xP + ∂/∂yQ + ∂/∂zR) ∇(dotted with)(∅F) = [∇∅](dotted with)F + ∅(∂/∂xP + ∂/∂yQ + ∂/∂zR) ∇(dotted with)(∅F) = <∂/∂x∅, ∂/∂y∅, ∂/∂z∅>(dotted with)<P,Q,R> + ∅(∂/∂xP + ∂/∂yQ + ∂/∂zR) ∇(dotted with)(∅F) = <0,0,0>(dotted with)<P,Q,R> + ∅(∂/∂xP + ∂/∂yQ + ∂/∂zR) ∇(dotted with)[∅F] = ∅(∂/∂xP + ∂/∂yQ + ∂/∂zR) <∂/∂x,∂/∂y,∂/∂z>(dotted with)<∅P, ∅Q, ∅R> = ∅(∂/∂xP + ∂/∂yQ + ∂/∂zR) ∅(∂/∂xP + ∂/∂yQ + ∂/∂zR) = ∅(∂/∂xP + ∂/∂yQ + ∂/∂zR) ∇(dotted with)(∅F) = [∇∅](dotted with)F + ∅∇(dotted with)F 


#4
Dec111, 02:32 PM

P: 53

Deriving cross product and dot product, stuck at beginning.
Hello? I did the work out and nobody can spot anything I did wrong, or if i did it right?



#5
Dec111, 02:40 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,722

No, that isn't correct. You can't treat [itex]\phi[/itex] like a constant.



#6
Dec111, 02:46 PM

P: 53

Thank you for the reply.
My professor wrote all of those on the board: ϕ = ϕ(x,y,z) ϕF is the vector field ϕ = ϕ(x,y,z) F = <P,Q,R> ∇(dotted with)F = x partial P + y partial Q + z partial R ∇ϕ = <x partial ϕ, y partial ϕ, z partial ϕ> ϕF = <ϕP, ϕQ, ϕR> ϕ is a function of x,y, and z. If I can't treat it as a constant in this situation, what can I do with it? 


#7
Dec111, 03:47 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,722

Just start with the definition of the divergence and apply it to ϕF = (ϕP, ϕQ, ϕR):
[tex]\nabla\cdot(\phi \mathbf{F}) = \frac{\partial}{\partial x} (\phi P) + \frac{\partial}{\partial y} (\phi Q) + \frac{\partial}{\partial z} (\phi R)[/tex]Now use the product rule on each of the three terms. 


#8
Dec111, 08:55 PM

P: 53

∇⋅(ϕF)=∂/∂x(ϕP)+∂/∂y(ϕQ)+∂/∂z(ϕR)
so =(ϕ'P + P'ϕ) + (ϕQ' + ϕ'Q) + (ϕR' + ϕ'R) = ϕ'(P+Q+R) + ϕ(P'+Q'+R') So it looks like ϕ'(P+Q+R) = (∇ϕ)⋅F and ϕ(P'+Q'+R') = ϕ(∇⋅F) and that is the end of the proof? 


#9
Dec111, 10:29 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,722

You're on the right track, but you need to keep track of the fact that the derivatives are with respect to different variables so you can't, for example, simply collect terms and factor ϕ' out to get the first term.



Register to reply 
Related Discussions  
Angle between 2 vectors using 1) Dot product and 2) cross product gives diff. answer?  Calculus & Beyond Homework  8  
Find theta from the cross product and dot product of two vectors  Calculus & Beyond Homework  11  
Cross product and dot product of forces expressed as complex numbers  Introductory Physics Homework  4  
Dot product and cross product evaluation questions  Calculus & Beyond Homework  4  
Stuck on proof! Proving cross product derivative!  Introductory Physics Homework  10 