Chapter 14: Q. 21 (page 1132)

$Compute the divergence of the vector fields in Exercises 17 - 22 . F (x, y, z) = \sin^{- 1} (xy) i + \ln (y + z) j + \frac{1}{2 x + 3 y + 5 z + 1} k$

Short Answer

Expert verified

$div F = \frac{y}{\sqrt{1 - {(xy)}^{2}}} + \frac{1}{y + z} - \frac{5}{{(2 x + 3 y + 5 z + 1)}^{2}}$

Step by step solution

Step 1. Given information is:

$F (x, y, z) = \sin^{- 1} (xy) i + \ln (y + z) j + \frac{1}{2 x + 3 y + 5 z + 1} k$

Step 2. Divergence of vector field

$If F (x, y, z) = f_{1} (x, y, z) i + f_{2} (x, y, z) j + f_{3} (x, y, z) k, then div F = \frac{\partial f_{1}}{\partial x} + \frac{\partial f_{2}}{\partial y} + \frac{\partial f_{3}}{\partial z} Here, F (x, y, z) = \sin^{- 1} (xy) i + \ln (y + z) j + \frac{1}{2 x + 3 y + 5 z + 1} k Compare this with, F (x, y, z) = f_{1} (x, y, z) i + f_{2} (x, y, z) j + f_{3} (x, y, z) k Then, f_{1} (x, y, z) = \sin^{- 1} (xy), f_{2} (x, y, z) = \ln (y + z), f_{3} (x, y, z) = \frac{1}{2 x + 3 y + 5 z + 1} div F = \frac{\partial f_{1}}{\partial x} + \frac{\partial f_{2}}{\partial y} + \frac{\partial f_{3}}{\partial z} = \frac{\partial (\sin^{- 1} (xy))}{\partial x} + \frac{\partial (\ln (y + z))}{\partial y} + \frac{\partial (\frac{1}{2 x + 3 y + 5 z + 1})}{\partial z} = \frac{y}{\sqrt{1 - {(xy)}^{2}}} + \frac{1}{y + z} - \frac{5}{{(2 x + 3 y + 5 z + 1)}^{2}}$

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$Compute the divergence of the vector fields in Exercises 17 - 22 . F (x, y, z) = \sin^{- 1} (xy) i + \ln (y + z) j + \frac{1}{2 x + 3 y + 5 z + 1} k$

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Divergence of vector field

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ComputethedivergenceofthevectorfieldsinExercises17–22.F(x,y,z)=sin-1(xy)i+ln(y+z)j+12x+3y+5z+1k

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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Applied Mathematics

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Study anywhere. Anytime. Across all devices.

$Compute the divergence of the vector fields in Exercises 17 - 22 . F (x, y, z) = \sin^{- 1} (xy) i + \ln (y + z) j + \frac{1}{2 x + 3 y + 5 z + 1} k$