Answer: No, it is invalid; using a non-unit vector scales the result by $|\mathbf{w}|$, so the answer would be $|\mathbf{w}|$ times too large.

The Unit Vector Requirement: The directional derivative formula is $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$ where $|\mathbf{u}| = 1$. If $\mathbf{w}$ is used instead: $\nabla f \cdot \mathbf{w} = \nabla f \cdot (|\mathbf{w}| \hat{\mathbf{w}}) = |\mathbf{w}| (\nabla f \cdot \hat{\mathbf{w}}) = |\mathbf{w}| D_{\hat{\mathbf{w}}} f$. Effect for $\mathbf{w} = \langle 2, 2 \rangle$: $|\mathbf{w}| = \sqrt{4+4} = 2\sqrt{2}$. So the student's result is $2\sqrt{2}$ times the true directional derivative. Why distractors fail: Option A is incorrect because direction does matter for the *value* of the directional derivative, but normalization is needed to measure rate of change *per unit distance*. Option C is false; $\mathbf{w}$ already lies in the $xy$-plane. Option D is the student's erroneous claim and is mathematically wrong.

Answer: The claim is false; $D_{\mathbf{u}} f$ varies from $-5$ to $5$, and it equals zero in the direction $\langle 4, 3 \rangle / 5$.

Directional Derivative Varies by Direction: $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} = \langle 3, -4 \rangle \cdot \mathbf{u} = 5\cos\theta$, which ranges from $-5$ to $5$ depending on the angle $\theta$ between $\mathbf{u}$ and $\nabla f$. Finding the Zero Direction: $D_{\mathbf{u}} f = 0$ when $\nabla f \cdot \mathbf{u} = 0$, i.e., when $\mathbf{u} \perp \langle 3, -4 \rangle$. A perpendicular unit vector is $\langle 4, 3 \rangle / 5$ (since $3 \cdot 4 + (-4) \cdot 3 = 0$). Why distractors fail: Option A incorrectly states the average directional derivative equals $|\nabla f|$ (it actually equals zero by symmetry). Option C is wrong; $D_{\mathbf{u}} f$ can be negative. Option D confuses the maximum value with all values.

Answer: The level curve $f = e$ is the unit circle, and $\nabla f(1, 0) = \langle 2e, 0 \rangle$ is its outward normal at $(1, 0)$.

Identify the Level Curve: $f(x,y) = e^{x^2 + y^2} = e^1 = e$ implies $x^2 + y^2 = 1$, the unit circle. Compute the Gradient: $f_x = 2x e^{x^2+y^2}$, $f_y = 2y e^{x^2+y^2}$. At $(1,0)$: $f_x = 2e^1 = 2e$, $f_y = 0$. So $\nabla f(1,0) = \langle 2e, 0 \rangle$. Why option B is correct: The level curve is indeed the unit circle, and $\nabla f = \langle 2e, 0 \rangle$ points outward (radially) at $(1,0)$, serving as the outward normal. Why distractors fail: Option A gets the shape and gradient magnitude wrong. Option C correctly identifies the shape but misses the factor $e$ in the gradient. Option D is wrong on both the shape and the gradient.

Answer: Move in a unit direction $\mathbf{u}$ such that $\nabla h \cdot \mathbf{u} = |\nabla h| / \sqrt{2}$, i.e., at a $45°$ angle to $\nabla h$.

Setup the Equation: We want $D_{\mathbf{u}} h = |\nabla h| / \sqrt{2}$. Since $D_{\mathbf{u}} h = |\nabla h| \cos\theta$, we need $\cos\theta = 1/\sqrt{2}$, so $\theta = 45°$. Compute $\nabla h$ at $(2,1)$: $h_x = -2x = -4$, $h_y = -6y = -6$. $\nabla h = \langle -4, -6 \rangle$, $|\nabla h| = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$. Finding the Direction: Any unit vector making a $45°$ angle with $\nabla h$ satisfies the requirement. The robot has multiple valid directions (two, symmetric about $\nabla h$). Why distractors fail: Option A modifies the gradient's magnitude but not direction — this is not a unit vector in general. Option C moves perpendicular to $\nabla h$, giving $D_{\mathbf{u}} h = 0$, not $|\nabla h|/\sqrt{2}$. Option D is a 3D construct not applicable here.

Answer: The direction of $-\nabla f$

Steepest Descent: $D_{\mathbf{u}} f = |\nabla f|\cos\theta$ is minimized when $\cos\theta = -1$, i.e., $\theta = 180°$. This means $\mathbf{u}$ points opposite to $\nabla f$, giving the direction $-\nabla f / |\nabla f|$. Why option B is correct: Moving in the direction $-\nabla f$ decreases $f$ most rapidly, giving a directional derivative of $-|\nabla f|$, the most negative possible value. Why distractors fail: Option A is the direction of steepest ascent, not descent. Option C gives zero rate of change (along level curves). Option D is nonsensical since $|\nabla f|$ is a scalar, not a direction.

Answer: Student A, because the directional derivative requires a unit vector.

The Unit Vector Requirement: The directional derivative $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$ is only defined when $\mathbf{u}$ is a unit vector ($|\mathbf{u}| = 1$). This ensures the result measures the rate of change per unit distance. Why Student A is correct: $|\langle 3/5, 4/5 \rangle| = \sqrt{9/25 + 16/25} = 1$. So Student A correctly gets $D_{\mathbf{u}} f = 9/5 + 16/5 = 25/5 = 5 = |\nabla f|$, which is the maximum directional derivative. Why distractors fail: Option A validates Student B's error. Using a non-unit vector inflates the result by $|\mathbf{v}|$. Option C is incorrect because the two computations measure different things. Option D is wrong; the direction vector need not be the gradient.

Answer: $\langle 2, 4 \rangle$

Compute the Partial Derivatives: $f_x = 2x$ and $f_y = 2y$, so $\nabla f = \langle 2x, 2y \rangle$. Evaluate at $(1, 2)$: $\nabla f(1, 2) = \langle 2(1), 2(2) \rangle = \langle 2, 4 \rangle$. Why distractors fail: Option A forgets to multiply by 2 in both components. Option C computes $\langle 2x, 2x \rangle$ using $x = 1$ for both, ignoring that $y = 2$. Option D correctly computes $f_y = 4$ but takes $f_x = 1$ (the value of $x$) instead of $2x = 2$.

Answer: $15$

Compute the Gradient: $f_x = y^2$, $f_y = 2xy$. At $(2, 3)$: $f_x = 9$, $f_y = 12$. So $\nabla f(2,3) = \langle 9, 12 \rangle$. Compute the Magnitude: $|\nabla f| = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$. Why option C is correct: The maximum rate of change equals $|\nabla f| = 15$, achieved in the direction of the gradient. Why distractors fail: Option A ($9$) is just the $x$-component of the gradient ($f_x$). Option B ($12$) is just the $y$-component ($f_y$). Option D ($\sqrt{45} \approx 6.7$) could arise from incorrectly computing $f_x = 3$ and $f_y = 6$ (dividing each partial by some factor).

Answer: $f$, because $\nabla f(1,1) = \langle 3, 2 \rangle$ and $|\nabla f| = \sqrt{13} > \sqrt{10} = |\nabla g|$.

Compute $\nabla f$ at $(1,1)$: $f_x = 3x^2y^2 = 3$, $f_y = 2x^3y = 2$. So $\nabla f(1,1) = \langle 3, 2 \rangle$ and $|\nabla f| = \sqrt{9+4} = \sqrt{13}$. Compute $\nabla g$ at $(1,1)$: $g_x = y^3 = 1$, $g_y = 3xy^2 = 3$. So $\nabla g(1,1) = \langle 1, 3 \rangle$ and $|\nabla g| = \sqrt{1+9} = \sqrt{10}$. Compare: $\sqrt{13} > \sqrt{10}$, so $f$ has the larger maximum rate of change. Why distractors fail: Option B gives an incorrect gradient for $g$ and wrong magnitude. Option C is false; the functions are not symmetric (they have different partial derivative structures at $(1,1)$). Option D is confused reasoning — function values don't determine gradient magnitude.

Answer: The strategy is locally valid — it decreases $f$ most rapidly from $P$ — but may converge to a local minimum rather than the global minimum.

Gradient Descent is Locally Valid: Moving in the direction $-\nabla f$ reduces $f$ at the maximum possible rate locally. This is the basis of the gradient descent algorithm. Key Limitation: Local vs. Global: Gradient descent follows the steepest downward path locally but may get trapped in a local minimum if $f$ has multiple critical points. There is no guarantee of reaching the global minimum. Why distractors fail: Option A overstates the guarantee — for non-convex functions, gradient descent may converge to a local, not global, minimum. Option C is a common misconception; the direction of descent is $-\nabla f / |\nabla f|$ (normalized), but the direction $-\nabla f$ is still valid for descent. Option D is incorrect — the presence of critical points is precisely when the method might stall, not a condition for its validity.

Answer: $\langle 3, 4 \rangle$

Requirements: We need $|\nabla f(1,0)| = 5$ (since the maximum directional derivative equals the gradient magnitude) and $\nabla f(1,0)$ pointing in the direction of $\langle 3, 4 \rangle$. Compute the Required Gradient: The unit vector in the direction $\langle 3, 4 \rangle$ is $\langle 3/5, 4/5 \rangle$. Scaling by $5$: $\nabla f(1,0) = 5 \cdot \langle 3/5, 4/5 \rangle = \langle 3, 4 \rangle$. Check: $|\langle 3, 4 \rangle| = \sqrt{9 + 16} = 5$. ✓ Why option A is correct: $\langle 3, 4 \rangle$ has magnitude $5$ and points in the direction $\langle 3, 4 \rangle$, satisfying both conditions. Why distractors fail: Option B ($\langle 3/5, 4/5 \rangle$) is the correct direction but has magnitude $1$, not $5$. Option C ($\langle 15, 20 \rangle$) points in the right direction but has magnitude $25$, not $5$. Option D ($\langle 4, 3 \rangle$) has the correct magnitude $5$ but points in the wrong direction.

Answer: $f(x, y) = 2x - 3y$

Check Each Option: For option A: $f_x = 2$, $f_y = -3$, so $\nabla f = \langle 2, -3 \rangle$ everywhere, including at $(0,0)$. Why option A is correct: The linear function $f(x,y) = 2x - 3y$ has constant gradient equal to $\langle 2, -3 \rangle$, which matches the requirement. Why distractors fail: Option B gives $\nabla f = \langle -3, 2 \rangle$, swapping the components. Option C gives $\nabla f = \langle 2x, -2y \rangle = \langle 0, 0 \rangle$ at $(0,0)$. Option D gives $\nabla f = \langle 4x, -6y \rangle = \langle 0, 0 \rangle$ at $(0,0)$.

Answer: $\mathbf{u}$ is in the direction of $\nabla f$ and $\mathbf{v} = -\mathbf{u}$.

Interpreting the Given Information: The maximum directional derivative is $|\nabla f| = 3$. Since $D_{\mathbf{u}} f = 3 = |\nabla f|$, $\mathbf{u}$ must equal $\nabla f / |\nabla f|$, i.e., the unit vector in the direction of $\nabla f$. Interpreting $D_{\mathbf{v}} f = -3$: The minimum directional derivative is $-|\nabla f| = -3$, achieved when $\mathbf{v} = -\nabla f / |\nabla f| = -\mathbf{u}$. Why distractors fail: Option B claims perpendicularity, but $\mathbf{u}$ and $\mathbf{v}$ are antiparallel, not perpendicular. Option C is false since $|\nabla f| = 3 \neq 0$. Option D gives $3 + (-3) = 0 \neq |\nabla f| = 3$.

Answer: $\nabla f = \langle 2, 2, -2 \rangle$; the level surface $f = 1$ is a hyperboloid, and $\nabla f$ is its normal at $(1,1,1)$.

Compute the Gradient: $f_x = 2x = 2$, $f_y = 2y = 2$, $f_z = -2z = -2$ at $(1,1,1)$. So $\nabla f = \langle 2, 2, -2 \rangle$. Identify the Level Surface: $x^2 + y^2 - z^2 = 1$ is a hyperboloid of one sheet — a standard quadric surface. Gradient as Normal: $\nabla f$ is always normal (perpendicular) to the level surface, not tangent. Why distractors fail: Option B has the wrong sign for $f_z$. Option C uses the wrong gradient (halved values). Option D correctly gives the gradient but makes the false claim that it is tangent to the surface.

Answer: In any direction perpendicular to $\nabla f$

Zero Directional Derivative: $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} = |\nabla f|\cos\theta$. This equals zero when $\cos\theta = 0$, i.e., when $\theta = 90°$, meaning $\mathbf{u}$ is perpendicular to $\nabla f$. Why option C is correct: Any unit vector $\mathbf{u}$ orthogonal to $\nabla f$ makes $D_{\mathbf{u}} f = 0$. This corresponds to moving along a level curve. Why distractors fail: Option A gives the maximum, not zero, directional derivative. Option B gives the most negative (minimum) directional derivative. Option D is not a geometric/directional criterion and is not mathematically meaningful in this context.

Answer: $1$

Compute the Gradient: $f_x = \frac{2x}{x^2+y^2}$, $f_y = \frac{2y}{x^2+y^2}$. At $(1,0)$: $f_x = 2$, $f_y = 0$. So $\nabla f = \langle 2, 0 \rangle$. Find the Unit Direction Vector: The direction at angle $\pi/3$: $\mathbf{u} = \langle \cos(\pi/3), \sin(\pi/3) \rangle = \langle 1/2, \sqrt{3}/2 \rangle$. Compute the Directional Derivative: $D_{\mathbf{u}} f = \langle 2, 0 \rangle \cdot \langle 1/2, \sqrt{3}/2 \rangle = 2 \cdot (1/2) + 0 = 1$. Why distractors fail: Option B ($\sqrt{3}$) would arise from using $\sin(\pi/3)$ instead of $\cos(\pi/3)$ in the projection. Option C ($1/2$) could come from forgetting the factor of 2 in $f_x$. Option D ($2$) is the maximum directional derivative (along the $x$-axis direction), not in the direction $\pi/3$.

Answer: $\nabla f = \langle f_x, f_y, f_z \rangle$

Gradient in Three Dimensions: For $f(x,y,z)$, the gradient extends naturally: $\nabla f = \langle \partial f/\partial x, \partial f/\partial y, \partial f/\partial z \rangle = \langle f_x, f_y, f_z \rangle$. Why option C is correct: Each component of the gradient is the first-order partial derivative with respect to the corresponding variable. Why distractors fail: Option A is the 2D gradient, missing the $z$-component. Option B is a scalar product, not a vector. Option D uses second-order partials, which are not the gradient.

Answer: $\mathbf{u}$ is not in the direction of $\nabla \phi$, and the angle between $\mathbf{u}$ and $\nabla \phi$ is $\arccos(5/7)$.

Use the Angle Formula: $D_{\mathbf{u}} \phi = |\nabla \phi| \cos\theta = 7\cos\theta = 5$, so $\cos\theta = 5/7$, giving $\theta = \arccos(5/7) \approx 44.4°$. Why option A is correct: Since $5 < 7 = |\nabla \phi|$, $\mathbf{u}$ is not aligned with $\nabla \phi$. The angle $\arccos(5/7)$ is well-defined and between $0°$ and $90°$. Why distractors fail: Option B is incorrect; the maximum value $|\nabla \phi| = 7$ is achieved only when $\mathbf{u}$ is in the gradient direction — not here since $D_{\mathbf{u}} \phi = 5 \neq 7$. Option C is false; any value in $[-7, 7]$ is achievable. Option D inverts the inequality — $|\nabla \phi|$ is the maximum, so $|\nabla \phi| \geq D_{\mathbf{u}} \phi$.

Answer: The claim is incorrect; $\nabla f$ points in the direction of steepest local increase, not necessarily toward the global maximum.

Local vs. Global Direction: The gradient $\nabla f$ at a point $P$ indicates the direction of steepest increase of $f$ in an infinitesimal neighborhood of $P$. It is a local quantity and does not encode global information about where the maximum of $f$ lies. Why option B is correct: Following the gradient at each step (gradient ascent) may lead toward a local maximum, not necessarily the global one. For example, if $f$ has multiple peaks, $\nabla f$ at a point leads to the nearest local maximum. Why distractors fail: Option A overstates what the gradient tells us — it says nothing about global extrema directly. Option C adds a condition that doesn't fix the fundamental misunderstanding. Option D confuses the gradient with an integral relationship that doesn't exist in this context.

Answer: It equals $|\nabla f|$

Maximum Rate of Change Principle: Since $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} = |\nabla f||\mathbf{u}|\cos\theta = |\nabla f|\cos\theta$ (because $\mathbf{u}$ is a unit vector), the maximum occurs when $\cos\theta = 1$, i.e., $\theta = 0$. This gives a maximum value of $|\nabla f|$. Why option B is correct: The maximum directional derivative is the magnitude of the gradient, $|\nabla f|$, achieved when $\mathbf{u}$ points in the same direction as $\nabla f$. Why distractors fail: Option A gives $|\nabla f|^2$ (the dot product of the gradient with itself), which is larger. Option C also gives $|\nabla f|^2$. Option D is incorrect because the maximum is determined by the gradient magnitude, not the normalization of $f$.

Answer: In any direction perpendicular to $\nabla h = \langle -4, -12 \rangle$

Contour Direction: To remain at constant elevation means moving along a level curve of $h$. The gradient is perpendicular to level curves, so the hiker must move in a direction perpendicular to $\nabla h$. Compute the Gradient at $(1,2)$: $h_x = -4x = -4$, $h_y = -6y = -12$. So $\nabla h = \langle -4, -12 \rangle$. A direction perpendicular to this is, for example, $\langle -12, 4 \rangle$ or $\langle 12, -4 \rangle$. Why distractors fail: Option A is the direction of steepest ascent — the hiker would gain elevation. Option C is the direction of steepest descent. Option D has no clear geometric interpretation for remaining on a contour.

Answer: $\nabla f$ is perpendicular to the level curve at that point.

Gradient Perpendicularity Principle: A level curve is defined by $f(x,y) = c$. Along this curve, $f$ does not change. Since the directional derivative in the tangent direction to the level curve is zero, and $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} = 0$ implies $\nabla f \perp \mathbf{u}$, the gradient must be perpendicular to the level curve. Why option C is correct: The gradient points in the direction of steepest increase, which is always orthogonal to the curves of constant function value. Why distractors fail: Options A and B say the gradient is tangent or parallel to the level curve, which contradicts the perpendicularity result. Option D is incorrect; $\nabla f$ can be nonzero at points on a level curve — it just points perpendicular to it.

Answer: $\langle -2, -8 \rangle$

Find the Gradient: $T_x = -2x$, $T_y = -8y$. At $(1,1)$: $\nabla T = \langle -2, -8 \rangle$. Direction of Steepest Ascent: The steepest increase in $T$ is in the direction of $\nabla T = \langle -2, -8 \rangle$. This points toward the origin, where $T$ is maximized. Why option A is correct: The gradient $\langle -2, -8 \rangle$ is the direction of steepest ascent. A heat-seeking particle moves toward increasing temperature, so it follows $\nabla T$. Why distractors fail: Option B gives the opposite direction (steepest descent — temperature decreases most rapidly). Option C is proportional to the gradient but has incorrect components ($T_y = -8y = -8$, not $-4$). Option D has the wrong sign on the first component.

Answer: Because along a level curve, $f$ is constant, so its directional derivative in the tangential direction is zero, implying $\nabla f$ is orthogonal to that direction.

Formal Reasoning: Let $\mathbf{r}(t)$ be a parametric curve lying on the level curve $f = c$. Then $f(\mathbf{r}(t)) = c$ for all $t$. Differentiating: $\nabla f \cdot \mathbf{r}'(t) = 0$. Since $\mathbf{r}'(t)$ is tangent to the level curve, $\nabla f$ must be orthogonal to all tangent vectors. Why option C is correct: This is the rigorous chain-rule argument. It applies in all dimensions and for any shape of level curve. Why distractors fail: Option A is only true for $f = x^2 + y^2$ and is not general. Option B is incorrect; perpendicularity to level curves is a theorem, not a convention. Option D is false — the partial derivatives need not sum to zero on level curves.

Answer: $\langle 4, 8 \rangle$

Gradient is Normal to Level Curve: $f_x = 2x$, $f_y = 8y$. At $(2, 1)$: $\nabla f = \langle 4, 8 \rangle$. This is the normal to the level curve. Verify the Point is on the Level Curve: $f(2,1) = 4 + 4 = 8$. Confirmed. Why distractors fail: Option A swaps $f_x$ and $f_y$ values. Option B uses $(f_x, f_y) = (2, 4)$ — these are the values of $x$ and $y$ doubled, not the correct partial derivative evaluations ($f_y = 8y = 8$, not $4$). Option D is a simplified version using $x=1$ and $y=2$, swapped.

Answer: $|\nabla f(A)| > |\nabla f(B)|$ because the level curves are denser at $A$, indicating a steeper rate of change.

Interpreting Contour Maps: Closely spaced level curves indicate that $f$ changes rapidly over short distances — a large magnitude gradient. Widely spaced level curves indicate a gentle slope — a small gradient magnitude. Why option B is correct: Near $A$, the same unit change in $f$ (e.g., from 2 to 3) occurs over a much shorter spatial distance than near $B$. This means $|\nabla f(A)|$ is larger. Why distractors fail: Option A reverses the correct interpretation. Option C is incorrect; density of level curves encodes gradient magnitude. Option D is false; contour maps are a standard tool for visualizing gradient information.

Answer: $\nabla f$ is perpendicular to the level surface at each point.

Gradient Perpendicular to Level Surfaces: A level surface is defined by $f(x,y,z) = c$. For any curve $\mathbf{r}(t)$ lying on this surface, $f(\mathbf{r}(t)) = c$, so $\nabla f \cdot \mathbf{r}'(t) = 0$ by the chain rule. Since this holds for all tangent vectors to the surface, $\nabla f$ is perpendicular (normal) to the surface. Why option B is correct: The gradient serves as the normal vector to a level surface, pointing in the direction of steepest increase away from the surface. Why distractors fail: Option A contradicts the perpendicularity result. Option C is false; $\nabla f$ is defined at all points where $f$ is differentiable. Option D is false; $\nabla f = \mathbf{0}$ only at critical points, not necessarily on all level surfaces.

Answer: $D_{\mathbf{u}} f = 0$ for all unit vectors $\mathbf{u}$.

Zero Gradient Implies Zero Directional Derivatives: By the formula $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$, if $\nabla f = \mathbf{0}$, then $D_{\mathbf{u}} f = \mathbf{0} \cdot \mathbf{u} = 0$ for every direction $\mathbf{u}$. Why option C is correct: A zero gradient means the function has no first-order change in any direction — the point is a critical point. Why distractors fail: Option A is false; the directional derivative is well-defined and equals zero. Option B is arbitrary and has no mathematical justification. Option D is incorrect; $|\mathbf{u}| = 1$ for unit vectors, but the directional derivative is still zero.