Find the divergence of the vector field.
$V = 2 x^{2} i + 5 y^{3} j + 3 z^{4} k a t x = 2, y = 3, z = 4$

1204

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602

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803

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911

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Solution

The correct option is D 911
$\to V = 2 x^{2}^i + 5 y^{3}^j + 3 z^{4}^k$

$\to ▽ \to V = [\frac{\partial}{\partial x}^i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k] [2 x^{2}^i + 5 y^{3}^j + 3 z^{4}^k]$

$= 4 x + 15 y^{2} + 12 z^{3}$

At (2, 3, 4)

$\to ▽ \to V = 8 + 15 (3)^{2} + 12 (4)^{3}$
= 8 + 135 + 768
= 911

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