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# Gradient Field with Trigonometric Function

MVCALC-XCXXEE

Find the gradient field $\vec{F}=\nabla \phi$ when the potential function $\phi$ is given by:

$$\phi=\arctan \frac{y}{2x}+\sin(x+y^2)$$

A

$\vec{F}=\langle \cos(x+y^2)-\frac{2y}{4x^2+y^2}, 2y\cos(x+y^2)+\frac{2x}{4x^2+y^2} \rangle$

B

$\vec{F}=\langle \cos(x+y^2)-\frac{2x}{4x^2+y^2}, 2y\cos(x+y^2)+\frac{2y}{4x^2+y^2} \rangle$

C

$\vec{F}=\langle \cos(x+y^2)+\frac{2y}{4x^2+y^2}, 2y\cos(x+y^2)-\frac{2x}{4x^2+y^2} \rangle$

D

$\vec{F}=\langle 2y\cos(x+y^2)+\frac{2x}{4x^2+y^2}, \cos(x+y^2)-\frac{2y}{4x^2+y^2} \rangle$

E

$\vec{F}=\langle 2y\cos(x+y^2)-\frac{2y}{4x^2+y^2}, \cos(x+y^2)+\frac{2x}{4x^2+y^2} \rangle$