How to show the following are true for the two vectors 𝑒̅=4π‘₯π‘¦π‘–Μ…βˆ’2𝑧𝑗̅+(11π‘₯^3)π‘˜Μ… ⁑and 𝑣̅=(π‘₯𝑦/𝑧)𝑖̅+(3π‘₯^2)𝑗̅ +6π‘˜Μ… ? a. βˆ‡Γ—(𝑒̅+𝑣̅)=βˆ‡Γ—π‘’Μ…+βˆ‡Γ—π‘£Μ… b. βˆ‡.(𝑒̅+𝑣̅)=βˆ‡.𝑒̅+βˆ‡.𝑣

1 Answer
Steve M
May 28, 2018

How to show the following are true for the two vectors:

#bb(ul u) = 4xy bb(ul hat i) - 2z bb(ul hat j) +11x^3 bb(ul hat k) \ \ # and
# bb(ul v) = (xy)/z bb(ul hat i) +3x^2 bb(ul hat j) +6 bb(ul hat k) #?

a. #grad xx (bb(ul u) + bb(ul v)) = grad xx bb(ul u) + grad xx bb(ul v) #

b. #grad * (bb(ul u) + bb(ul v)) = grad * bb(ul u) + grad * bb(ul v) #

Explanation:

We have vectors:

# bb(ul u) = 4xy bb(ul hat i) - 2z bb(ul hat j) +11x^3 bb(ul hat k) #

# bb(ul v) = (xy)/z bb(ul hat i) +3x^2 bb(ul hat j) +6 bb(ul hat k) #

And, let us define the vector #bb(ul w) = bb(ul u) + bb(ul v)#, so that:

# bb(ul w) = bb(ul u) + bb(ul v) #

# \ \ \ = (4xy bb(ul hat i) - 2z bb(ul hat j) +11x^3 bb(ul hat k)) #
# \ \ \ \ \ \ \ + ((xy)/z bb(ul hat i) +3x^2 bb(ul hat j) +6 bb(ul hat k)) #

# \ \ \ = (4xy +(xy)/z) bb(ul hat i) +(3x^2- 2z) bb(ul hat j) +(11x^3+6) bb(ul hat k)) #

Part (a):

Then, for the curl, we have:

# LHS = grad xx (bb(ul u) + bb(ul v)) #

# \ \ \ \ \ \ \ \ = grad xx (bb(ul w)) #

# \ \ \ \ \ \ \ \ = | (bb(ul hat i),bb(ul hat j),bb(ul hat k)), ((partial)/(partial x),(partial)/(partial y),(partial)/(partial z)), (4xy +(xy)/z, 3x^2- 2z, 11x^3+6) | #

# \ \ \ \ \ \ \ \ = | ((partial)/(partial y),(partial)/(partial z)), (3x^2- 2z, 11x^3+6) | bb(ul hat i) - | ((partial)/(partial x),(partial)/(partial z)), (4xy +(xy)/z, 11x^3+6) | bb(ul hat j) + | ((partial)/(partial x),(partial)/(partial y)), (4xy +(xy)/z, 3x^2- 2z) | bb(ul hat k) #

# \ \ \ \ \ \ \ \ = (0+2) bb(ul hat i) - (33x^2 + (xy)/z^2) bb(ul hat j) + (6x-4x-x/z) bb(ul hat k) #

# \ \ \ \ \ \ \ \ = 2 bb(ul hat i) - (33x^2 + (xy)/z^2) bb(ul hat j) + (2x-x/z) bb(ul hat k) #

Similarly:

# RHS = grad xx bb(ul u) + grad xx bb(ul v) #

# \ \ \ \ \ \ \ \ = | (bb(ul hat i),bb(ul hat j),bb(ul hat k)), ((partial)/(partial x),(partial)/(partial y),(partial)/(partial z)), (4xy, 3x^2, 11x^3) | + | (bb(ul hat i),bb(ul hat j),bb(ul hat k)), ((partial)/(partial x),(partial)/(partial y),(partial)/(partial z)), ((xy)/z, - 2z, 6) | #

# \ \ \ \ \ \ \ \ = | ((partial)/(partial y),(partial)/(partial z)), (3x^2, 11x^3) | bb(ul hat i) - | ((partial)/(partial x),(partial)/(partial z)), (4xy, 11x^3) | bb(ul hat j) + | ((partial)/(partial x),(partial)/(partial y)), (4xy, 3x^2) | bb(ul hat k) #
# \ \ \ \ \ \ \ \ \ \ + | ((partial)/(partial y),(partial)/(partial z)), (- 2z, +6) | bb(ul hat i) - | ((partial)/(partial x),(partial)/(partial z)), ((xy)/z, 6) | bb(ul hat j) + | ((partial)/(partial x),(partial)/(partial y)), ((xy)/z,- 2z) | bb(ul hat k) #

# \ \ \ \ \ \ \ \ = (0-0)bb(ul hat i) - (33x^2-0) bb(ul hat j) + (6x-4x) bb(ul hat k) #
# \ \ \ \ \ \ \ \ \ \ + (0+2) bb(ul hat i) - (0+(xy)/z^2) bb(ul hat j) + (0-x/z) bb(ul hat k) #

# \ \ \ \ \ \ \ \ = 2 bb(ul hat i) - (33x^2+(xy)/z^2) bb(ul hat j) + (2x-x/z) bb(ul hat k) #

# \ \ \ \ \ \ \ \ = LHS \ \ \ \ # QED

Part (b):

And, for the divergence, we have:

# LHS = grad * (bb(ul u) + bb(ul v)) #

# \ \ \ \ \ \ \ \ = grad * (bb(ul w)) #

# \ \ \ \ \ \ \ \ = (partial)/(partial x) (4xy +(xy)/z ) + (partial)/(partial y) (3x^2- 2z) + (partial)/(partial z) (11x^3+6) #

# \ \ \ \ \ \ \ \ = (4x +(y)/z ) + (0 - 0) + (0+0) #

# \ \ \ \ \ \ \ \ = 4x +y/z #

Similarly:

# RHS = grad * bb(ul u) + grad * bb(ul v) #

# \ \ \ \ \ \ \ \ = (partial)/(partial x) (4xy) + (partial)/(partial y)(3x^2) + (partial)/(partial z)( 11x^3) #
# \ \ \ \ \ \ \ \ \ \ \ \ + (partial)/(partial x)((xy)/z) - (partial)/(partial y)(2z) + (partial)/(partial z)(6) #

# \ \ \ \ \ \ \ \ = 4x + 0 + 0 + y/z - 0 + 0 #

# \ \ \ \ \ \ \ \ = 4x y/z#

# \ \ \ \ \ \ \ \ = 4x y/z#

# \ \ \ \ \ \ \ \ = LHS \ \ \ \ # QED

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