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Anna11 [10]
4 years ago
14

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

y points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Find the Jacobian of the transformation. x = e−r sin(6θ), y = er cos(6θ) Step 1 For the transformation x = e−r sin(6θ), y = er cos(6θ), the Jacobian is ∂(x, y) ∂(r, θ) = ∂x ∂r ∂x ∂θ ∂y ∂r ∂y ∂θ = er cos(6θ) .
Mathematics
1 answer:
sleet_krkn [62]4 years ago
6 0

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) =  δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)]     since  [cos²(6θ) + sin²(6θ)] = 1

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