Curl & Divergence - Calculus 3

1. A vector field $\mathbf{F}(x,y,z)$ satisfies $\text{div}(\mathbf{F}) = x^2 + y^2 + z^2$ and $\text{curl}(\mathbf{F}) \cdot \mathbf{k} = xy$, where $\mathbf{k}$ is the unit vector in the $z$-direction. If $\mathbf{F} = (P, Q, R)$ and $\frac{\partial P}{\partial x} = x^2$, what is $\frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$?
   Answer: $y^2 + z^2$ From the divergence condition: $\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = x^2 + y^2 + z^2$. Given that $\frac{\partial P}{\partial x} = x^2$, we have $x^2 + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = x^2 + y^2 + z^2$. Therefore $\frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = y^2 + z^2$. The information about the $z$-component of the curl equaling $xy$ gives us $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = xy$, but this constraint isn't needed for the current calculation. Choice B incorrectly adds $x^2$ instead of subtracting it. Choice C incorrectly keeps $x^2$ and removes $y^2$. Choice D incorrectly includes an extra $x^2$ term.
2. Consider the vector field $\mathbf{G}(x,y,z) = (xyz, x^2y, z^2)$. If the divergence of $\mathbf{G}$ equals the $z$-component of the curl of $\mathbf{G}$ at some point, which condition must hold at that point?
   Answer: $yz + x^2 + 2z = 2x - xy$ First find divergence: $\text{div}(\mathbf{G}) = \frac{\partial(xyz)}{\partial x} + \frac{\partial(x^2y)}{\partial y} + \frac{\partial(z^2)}{\partial z} = yz + x^2 + 2z$. Next find the $z$-component of curl: $\frac{\partial(x^2y)}{\partial x} - \frac{\partial(xyz)}{\partial y} = 2xy - xz$. Setting divergence equal to $z$-component of curl: $yz + x^2 + 2z = 2xy - xz$. Choice A omits the $2z$ term from divergence. Choice B has the wrong sign pattern. Choice D omits $2z$ and has incorrect equality.
3. The vector field $\mathbf{W}(x,y,z) = (\ln(x^2 + y^2 + 1), \arctan(z), ye^x)$ has divergence equal to:
   Answer: $\frac{2x}{x^2 + y^2 + 1} + 0 + ye^x$ The divergence is $\frac{\partial}{\partial x}\ln(x^2 + y^2 + 1) + \frac{\partial}{\partial y}\arctan(z) + \frac{\partial}{\partial z}(ye^x)$. First term: $\frac{2x}{x^2 + y^2 + 1}$. Second term: $\arctan(z)$ doesn't depend on $y$, so this is $0$. Third term: $ye^x$ doesn't depend on $z$, so this is $0$. Therefore divergence = $\frac{2x}{x^2 + y^2 + 1} + 0 + 0 = \frac{2x}{x^2 + y^2 + 1}$. Choice A incorrectly differentiates $\arctan(z)$ with respect to $z$ instead of $y$ and $ye^x$ with respect to $x$ instead of $z$. Choice C makes the same error as A for the second term. Choice D incorrectly computes the first partial derivative.
4. A vector field $\mathbf{H}(x,y,z) = (f(x,y,z), g(x,y,z), h(x,y,z))$ satisfies $\text{curl}(\mathbf{H}) = (2x, -3y, z^2)$ and $\text{div}(\mathbf{H}) = 6$. If $\frac{\partial f}{\partial x} = x + 1$, what is $\frac{\partial g}{\partial y} + \frac{\partial h}{\partial z}$?
   Answer: $5 - x$ From the divergence formula: $\text{div}(\mathbf{H}) = \frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} + \frac{\partial h}{\partial z} = 6$. Since $\frac{\partial f}{\partial x} = x + 1$, we have $(x + 1) + \frac{\partial g}{\partial y} + \frac{\partial h}{\partial z} = 6$. Therefore $\frac{\partial g}{\partial y} + \frac{\partial h}{\partial z} = 6 - (x + 1) = 5 - x$. The curl information confirms consistency but isn't needed for this calculation. Choice B has wrong sign, Choice C uses 6 instead of 5, Choice D uses 7 instead of 5.

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