📝 Directional Derivatives and Gradients (14 MCQs)

📖 Explanation: This multi-step application requires computing the gradient $\nabla T = \langle -2x, -4y \rangle$, evaluating at (1,1) to get $\langle -2, -4 \rangle$, finding velocity $\mathbf{v} = \langle 1, 2t \rangle = \langle 1, 2 \rangle$, and applying the chain rule $dT/dt = \nabla T \cdot \mathbf{v} = -10$. Option D incorrectly normalizes; the chain rule uses actual velocity, not unit direction, distinguishing physical rate from geometric directional derivative.

📖 Explanation: This error analysis question targets the misconception that positive partial derivatives imply positive directional derivatives in all first-quadrant directions. While $\nabla f$ points into the first quadrant, a unit vector $\mathbf{u}$ nearly perpendicular to $\nabla f$ can still yield a negative dot product if the angle exceeds 90 degrees, demonstrating that component signs alone don't determine directional behavior.

📖 Explanation: This graph-based conceptual question tests understanding that level curves represent constant function values, so movement along them produces no change. The gradient's orthogonality to level curves is a fundamental geometric property ensuring $D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = 0$ for any tangent vector $\mathbf{u}$, regardless of gradient magnitude or ellipse shape, reinforcing the geometric interpretation over computational approaches.

📖 Explanation: This challenging error analysis emphasizes that vanishing first-order directional derivatives at critical points provide no classification information since $\nabla f = \mathbf{0}$ implies $D_{\mathbf{u}}f = 0$ universally. Classification requires examining second-order behavior through $D_{\mathbf{u}}^2 f = \mathbf{u}^T H \mathbf{u}$; an indefinite Hessian indicates saddle behavior where some second-order directional derivatives are positive and others negative, contradicting the minimum conclusion.

📖 Explanation: This mixed-concepts problem combines gradients, constraints, and directional derivatives. Unconstrained steepest ascent follows $\nabla h$, but restriction to curve C requires projecting $\nabla h$ onto C's tangent space. The tangent direction is orthogonal to both surface normals' projections, and the constrained steepest ascent is the component of $\nabla h$ within this tangent space, illustrating how constraints modify gradient-based optimization.

📖 Explanation: This scenario-based application computes displacement $\mathbf{d} = \mathbf{v} \Delta t = \langle 50/\sqrt{2}, 50/\sqrt{2} \rangle \times 0.1$ hr ≈ $\langle 3.54, 3.54 \rangle$ km, then $\Delta P \approx \nabla P \cdot \mathbf{d} = (-4)(3.54) + (3)(3.54) = -3.54$ kPa. The linear approximation $\Delta P \approx \nabla P \cdot \Delta \mathbf{r}$ assumes differentiability and negligible higher-order terms over short distances, a key modeling assumption in applied calculus.

📖 Explanation: This Olympiad-style conceptual challenge distinguishes directional derivatives from differentiability. For $f = r$, $D_{\mathbf{u}}f(0,0) = \lim_{h \to 0} |h|/h = 1$ for any unit $\mathbf{u}$, so all directional derivatives exist. However, $f$ isn't differentiable at origin because the linear approximation fails (the cone lacks a tangent plane), proving that existence of all directional derivatives doesn't guarantee gradient existence or differentiability—a subtle but crucial distinction.

📖 Explanation: Despite textual inconsistency in option A, this question tests the key concept that directional derivatives in an orthonormal basis completely determine the gradient vector via $\nabla f = \sum (D_{\mathbf{e}_i}f) \mathbf{e}_i$. With orthonormal $\mathbf{u}, \mathbf{v}$, $\| \nabla f \|^2 = 6^2 + 2^2 = 40$, so magnitude is $\sqrt{40}$. This demonstrates that two properly chosen directional derivatives suffice, countering the misconception that infinite directions are needed, and highlights the power of orthonormal decompositions in gradient reconstruction.

📖 Explanation: This error analysis clarifies that Lagrange multiplier conditions precisely enforce $\nabla f \parallel \nabla g$, and since $\nabla g$ is normal to the constraint curve, $\nabla f$ must also be normal, making its tangential component zero. Thus $D_{\mathbf{t}}f = 0$ is necessary at constrained critical points, not evidence against optimality. Expecting nonzero tangential derivative misunderstands that constrained extrema occur where the function is stationary along the constraint, analogous to unconstrained critical points having zero gradient.

📖 Explanation: This mixed-concepts question recognizes that $\nabla f = \langle y,x \rangle = \nabla(xy)$, implying $f(x,y) = xy + C$. On the constraint $xy=1$, $f$ is constant (=1+C), so its directional derivative along any tangent to the level curve must be zero. This elegantly connects gradient fields, level sets, and directional derivatives without computation, testing deep conceptual linkage between conservative fields and constancy on level surfaces.

📖 Explanation: This direct-recall-with-misconception question targets the pervasive error of omitting normalization. The directional derivative $D_{\mathbf{u}}f$ is defined only for unit vectors $\mathbf{u}$; using non-unit $\mathbf{v}$ computes $\nabla f \cdot \mathbf{v} = \| \mathbf{v} \| D_{\mathbf{u}}f$, which scales the true rate by vector magnitude. Here, $\| \mathbf{v} \| = 5$, so 16 is five times too large. Emphasizing this definition prevents confusion between directional derivatives and general dot products with gradients.

📖 Explanation: This graph-based conceptual question links contour density to gradient magnitude. Closer contour spacing indicates rapid elevation change over short horizontal distance, corresponding to larger $\| \nabla h \|$. Since uphill directional derivative equals $\| \nabla h \|$ (as gradient points uphill), it must be greater at B. This reinforces visual interpretation of gradients and counters the misconception that identical gradient direction implies identical magnitude, emphasizing that contour spacing encodes rate information independent of direction.

📖 Explanation: This application combines radial symmetry, trigonometric dependence, and geometric interpretation. At (a,0), $\nabla f = \langle -2a e^{-a^2}, 0 \rangle$, so $D_{\mathbf{u}_\theta}f = \nabla f \cdot \langle \cos \theta, \sin \theta \rangle = -2a e^{-a^2} \cos \theta$. This cosine dependence confirms maximum descent toward origin (θ=π), zero change tangentially (θ=π/2), and illustrates how directional derivatives encode angular variation even for radially symmetric functions, linking algebraic computation to geometric intuition about level curve tangency.

📖 Explanation: This conceptual understanding question tests the fundamental inequality $|D_{\mathbf{u}}f| \leq \| \nabla f \|$, which holds for any unit vector $\mathbf{u}$ regardless of orthogonality or spanning properties. Since $D_{\mathbf{v}}f = 4$ was observed, it must be that $\| \nabla f \| \geq 4$; the gradient magnitude is the supremum of directional derivatives. The distractors exploit misconceptions about orthogonality requirements or spanning conditions, but the inequality is universal, making B correct and reinforcing that observed directional derivatives provide lower bounds on gradient magnitude.