Solution: ▽ · F = −3xy / (x2 + y2)^5/2

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The divergence of the vector field F, defined as F = x i / (x² + y²)^(3/2) + y j (x² + y²)^(3/2), is calculated to be ▽ · F = −1 / (x² + y²)^(3/2). This result is derived by finding the partial derivatives of F with respect to x and y, applying the product rule, and simplifying the expressions. The final expression for the divergence confirms the behavior of the vector field in relation to its spatial variables.

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Find ▽ · F if F = x i / (x2 + y2)^3/2 + y j (x2 + y2)^3/2 .

Solution: ▽ · F = −1 / (x2 + y2)^3/2
 
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Find the partial derivatives of F with respect to x, y and z and then add them up.
 
scholesmu said:
Find ▽ · F if F = x i / (x2 + y2)^3/2 + y j (x2 + y2)^3/2 .

Solution: ▽ · F = −1 / (x2 + y2)^3/2
The derivative of [itex]x/(x^2+ y^2)^{3/2}= x(x^2+ y^2)^{-3/2}[/itex] with respect to x, by the product rule, is [itex](x^2+ y^2)^{-3/2}- (3/2)(x)(2x)(x^2+ y^2)^{-5/2}[/itex]
[itex]= (x^2+ y^2)(x^2+ y^2)^{-5/2}- 3x^2(x^2+ y^2)^{-5/2}[/itex][itex]= (y^2- 2x^2)/(x^2+ y^2)^{5/2}[/itex].

Now, what is the derivative of [itex]y/(x^2+ y^2)^{3/2}[/itex]?
 

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