Level Curves & Gradient Orthogonality - Multivariable Calculus

1. Consider the surface $z = f(x,y) = x^3 - 3xy^2 + 2y$. If you project the curve where this surface intersects the plane $z = 5$ onto the $xy$-plane, what is the slope of the resulting level curve at the point where $x = 2$ and $y = 1$?
   Answer: The point $(2,1)$ is not on the level curve $f(x,y) = 5$, so the slope is undefined First check if $(2,1)$ is on the level curve $f(x,y) = 5$: $f(2,1) = 2^3 - 3(2)(1)^2 + 2(1) = 8 - 6 + 2 = 4 \neq 5$. Since the point is not on the specified level curve, the question about the slope at this point is meaningless. The other choices assume the point is on the curve and apply various (some incorrect) formulas for level curve slopes.
2. For the function $F(x,y) = e^{x+y} \sin(xy)$, suppose the level curve $F(x,y) = k$ passes through the origin and has a well-defined tangent line there. What additional information about the level curves can be determined?
   Answer: The level curve through the origin consists of both coordinate axes, and the tangent concept is undefined at the origin First, $F(0,0) = e^0 \sin(0) = 1 \cdot 0 = 0$. So the level curve through the origin is $F(x,y) = 0$, which means $e^{x+y} \sin(xy) = 0$. Since $e^{x+y} > 0$ always, we need $\sin(xy) = 0$, so $xy = n\pi$ for integer $n$. Near the origin, this gives $xy = 0$, which consists of both the $x$-axis ($y = 0$) and $y$-axis ($x = 0$). At the origin, these two curves intersect, so there's no unique tangent line. The gradient $\nabla F = (e^{x+y}(\sin(xy) + y\cos(xy)), e^{x+y}(\sin(xy) + x\cos(xy)))$ equals $(0,0)$ at the origin, confirming the tangent is undefined.

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