Chapter 14: Q. 18 (page 1132)

$Compute the divergence of the vector fields in Exercises 17 - 22 . G (x, y) = <x \cos (xy), y \cos (xy)>$

Short Answer

Expert verified

$div G = 2 \cos (xy) - 2 xy \sin (xy)$

Step by step solution

Step 1. Given information is:

$G (x, y) = <x \cos (xy), y \cos (xy)>$

Step 2. Divergence of vector field

$If G (x, y) = g_{1} (x, y) i + g_{2} (x, y) j, then div G = \frac{\partial g_{1}}{\partial x} + \frac{\partial g_{2}}{\partial y} Here, G (x, y) = <x \cos (xy), y \cos (xy)> Compare this with, G (x, y) = g_{1} (x, y) i + g_{2} (x, y) j Then, g_{1} (x, y) = x \cos (xy), g_{2} (x, y) = y \cos (xy) div G = \frac{\partial g_{1}}{\partial x} + \frac{\partial g_{2}}{\partial y} = \frac{\partial (x \cos (xy))}{\partial x} + \frac{\partial (y \cos (xy))}{\partial y} = 2 \cos (xy) - 2 xy \sin (xy)$

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$Compute the divergence of the vector fields in Exercises 17 - 22 . G (x, y) = <x \cos (xy), y \cos (xy)>$

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Divergence of vector field

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ComputethedivergenceofthevectorfieldsinExercises17–22.G(x,y)=xcos(xy),ycos(xy)

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Divergence of vector field

One App. One Place for Learning.

Most popular questions from this chapter

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$Compute the divergence of the vector fields in Exercises 17 - 22 . G (x, y) = <x \cos (xy), y \cos (xy)>$