Multivariable Calculus Question of the Day

1. $0$
2. $8\pi^2$
3. $16\pi$ (correct answer)
4. $-16\pi$

Explanation: We use the chain rule for partial derivatives: $\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial t}$. First, find the point $(x,y,z)$ corresponding to $(s,t) = (2, \pi/2)$: $x = 2 \cos(\pi/2) = 0$ $y = 2 \sin(\pi/2) = 2$ $z = 2(\pi/2) = \pi$ Next, compute the partial derivatives of $w$ with respect to $x, y, z$ and evaluate them at $(0, 2, \pi)$: $\frac{\partial w}{\partial x} = 3x^2 y \implies 3(0)^2(2) = 0$ $\frac{\partial w}{\partial y} = x^3 + 2yz^2 \implies 0^3 + 2(2)(\pi)^2 = 4\pi^2$ $\frac{\partial w}{\partial z} = 2y^2z \implies 2(2)^2(\pi) = 8\pi$ Then, compute the partial derivatives of $x, y, z$ with respect to $t$ and evaluate them at $(s,t) = (2, \pi/2)$: $\frac{\partial x}{\partial t} = -s \sin(t) \implies -2 \sin(\pi/2) = -2$ $\frac{\partial y}{\partial t} = s \cos(t) \implies 2 \cos(\pi/2) = 0$ $\frac{\partial z}{\partial t} = s \implies 2$ Finally, substitute these values into the chain rule formula: $\frac{\partial w}{\partial t} = (0)(-2) + (4\pi^2)(0) + (8\pi)(2) = 0 + 0 + 16\pi = 16\pi$. Distractor A is incorrect; it results from omitting the $\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}$ term from the chain rule. Distractor B is incorrect; it results from computing $\frac{\partial w}{\partial s}$ instead of $\frac{\partial w}{\partial t}$. Distractor D is incorrect; it can result from swapping the derivative formulas for sine and cosine.