anonymous
  • anonymous
On Pset #9 part II problem 2(b.), their answer is div (g G) = ∂(gM) + ∂(gM) ∂x ∂y = (gxM + gMx) + (gyN + gNy) = = (gxM + gyN) + (gMx + gNy) = G · �g + g div(G). Yet why is div(gG) not (gxxM + gxMx) + (gyyN + gyNy) ? since g= why is the partial of g with respect to either variable not gxx or gyy ?
OCW Scholar - Multivariable Calculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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phi
  • phi
**since g= *** g is a scalar function of x,y,t (represents a scalar, not a vector) g is not a vector, let alone a vector of partial derivatives. we can write \[ g(x,y,t)\textbf{G}(x,y,t)= < g(x,y,t)M(x,y,t), \ g(x,y,t)N(x,y,t)>\]where M and N are the components of vector G. Or, dropping the variables for succinctness, \[ gG= < gM, \ gN> \] Div(gG) can be written as the "dot product" of \[ \left< \frac{\partial}{\partial x} , \ \frac{\partial}{\partial y} \right>\cdot \\ = \frac{\partial}{\partial x} gM + \frac{\partial}{\partial y} gN \]
anonymous
  • anonymous
Ohhh I see.. I clearly didn't read the definition of "g" closely enough XD. That makes sense.. so they're not related, they're just two separate functions being multiplied together, with the divergence of their product (which at that point is a vector) being taken afterward. Thanks Phi!

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