📝 Level curves and contour plots (14 MCQs)

📖 Explanation: The gradient vector $\nabla f$ is always perpendicular to level curves and points in the direction of steepest ascent. Dense packing indicates large gradient magnitude. Moving perpendicular to level curves toward higher values maximizes the directional derivative $D_{\mathbf{u}}f = \|\nabla f\|\cos\theta$, which peaks when $\theta = 0$. Other options confuse contour orientation with optimization criteria or misapply gradient direction.

📖 Explanation: For $f(x,y)=x^2-y^2$, solving $x^2=y^2$ yields lines $y=\pm x$ intersecting at $(0,0)$. Computing partials gives $f_x=2x$, $f_y=-2y$, so $\nabla f(0,0)=(0,0)$. The gradient exists and equals zero; it is not undefined. Level curves may intersect precisely at critical points where the gradient vanishes, contradicting the misconception that crossing implies non-differentiability or discontinuity.

📖 Explanation: Concentric elliptical level curves indicate a local extremum at the center. Spacing between successive level curves reflects gradient magnitude: wider spacing means smaller $\|\nabla T\|$. As ellipses become more circular and spaced farther apart radially, the temperature gradient weakens. Thus, the rate of change along any radial path decreases monotonically. This connects geometric contour density to analytical gradient behavior without requiring explicit functional form.

📖 Explanation: Level curves satisfy $x^2 + 4y^2 = k$, forming ellipses with semi-axes $\sqrt{k}$ and $\sqrt{k}/2$. Curvature of an ellipse is maximal at endpoints of the minor axis $(0,\pm b)$ where bending is sharpest. Since fuel cost scales with curvature, throttle reduction is needed there. Option A confuses major/minor axes; C incorrectly assumes uniform curvature; D invokes irrelevant torsion for planar curves.

📖 Explanation: For $g(x,y)=xy$, $\nabla g = (y,x)$. In QI both components are positive, so the gradient lies in QI but its direction depends on the ratio $y/x$. Only along $y=x$ does it point exactly northeast. Elsewhere, e.g., near x-axis ($y \ll x$), it points mostly eastward. Correct level curves don't guarantee uniform gradient direction; this tests understanding that gradient orientation varies pointwise even within a single quadrant.

📖 Explanation: While both surfaces have circular level curves, $\nabla S_1 = (2x,2y)$ and $\nabla S_2 = (4x,4y)$, so $\|\nabla S_2\| = 2\|\nabla S_1\|$. Heat flux (proportional to gradient magnitude) is twice as large for $S_2$. Identical contour shapes indicate similar qualitative behavior but not quantitative equivalence. This distinguishes geometric similarity from analytical scaling, addressing the misconception that level curve topology fully determines physical quantities like flux.

📖 Explanation: Uniformly spaced parallel lines indicate constant gradient (linear function) for $x<0$. Exponentially converging lines for $x>0$ suggest $h$ grows exponentially with $x$, as level sets $h=c$ solve $x = \ln(c)/k$, yielding logarithmic spacing in $x$ but exponential convergence in the plot. Continuity at $x=0$ is plausible, but derivative may jump. Option B incorrectly invokes saddle points; C overstates singularity; D contradicts observed smooth convergence pattern.

📖 Explanation: Flying along a level curve means $dP/dt = 0$, so no work is done by pressure forces. However, airspeed changes relate to acceleration tangential to the path. If level curves diverge (spacing increases), $\|\nabla P\|$ decreases, reducing centripetal-like constraints on curved trajectories. In atmospheric dynamics, weaker gradients correlate with reduced geostrophic imbalance effects, allowing inertial acceleration. Option B reverses causality; C invokes irrelevant inflection points; D dismisses valid dynamical interpretation.

📖 Explanation: Level curve shape alone doesn't dictate differentiability; a smooth bump function can interpolate between circular and square isocontours while preserving $C^1$ regularity. The key is whether the defining function transitions smoothly, not the visual geometry. Option A falsely equates shape change with non-smoothness; C misunderstands that non-intersection doesn't guarantee differentiability; D incorrectly generalizes square contours as inherently non-smooth. This Olympiad-style question tests deep understanding that level set geometry and function regularity are distinct concepts.

📖 Explanation: Although level curves of $\sin x \cos y = c$ form rectangular grids, $\nabla f = (\cos x \cos y, -\sin x \sin y)$ is perpendicular to these curves but not necessarily axis-aligned. For example, at $(\pi/4, \pi/4)$, $\nabla f = (0.5, -0.5)$, pointing diagonally. Perpendicularity to axis-aligned curves yields vertical/horizontal gradients only if curves are perfectly horizontal/vertical lines; general rectangular grids have varying normal directions. This exposes confusion between curve orientation and gradient direction.

📖 Explanation: The Lagrange condition $\nabla f = \lambda \nabla g$ geometrically requires level curves to be tangent (gradients parallel). Transverse intersection means gradients are linearly independent, violating the necessary condition for constrained extrema (assuming $\nabla g \neq \mathbf{0}$). Thus, such points cannot be extrema. Option B incorrectly suggests $\nabla g = \mathbf{0}$ permits transverse intersection; actually $\nabla g = \mathbf{0}$ makes tangency undefined. Option C overreaches by claiming saddle; D abandons valid geometric insight.

📖 Explanation: Level curve spacing estimates gradient magnitude, but linear interpolation between contours assumes gradient constancy. Sharp bending implies rapid change in gradient direction/magnitude (high curvature), violating linearity assumption. Second-order terms in Taylor expansion become significant, causing prediction errors. This applies broadly to modeling scenarios where contour geometry signals nonlinearity. Option B blames resolution unnecessarily; C overgeneralizes to invalidate all methods; D dismisses legitimate physical phenomena.

📖 Explanation: While both have circular level curves, $f$ increases quadratically while $g$ increases linearly with radius. Gradient $\nabla f = 2(x,y)$ grows with distance, whereas $\nabla g = (x,y)/\sqrt{x^2+y^2}$ has constant unit magnitude away from origin. Level curve geometry determines gradient direction and relative spacing, but absolute rate of change depends on how function values scale between contours. Identical shapes mask fundamentally different sensitivities, revealing misunderstanding that contours capture topology, not metric properties.

📖 Explanation: Level curves define sets where $D=c$, but not which $c$ corresponds to which set. Reassigning density values to existing oval contours (e.g., mapping higher $c$ to western ovals) preserves geometry while relocating the peak. This exploits the fact that level curves constrain only preimages, not the codomain ordering. Rotation alters geometry; adding trends distorts shapes; D ignores reparameterization freedom. This integrates conceptual understanding of level sets as equivalence classes versus functional representation.