4.2 Directional Derivatives, Gradients, and Tangent Planes

4.2 Directional Derivatives, Gradients, and Tangent Planes

Partial derivatives $f_x$ and $f_y$ measure rates of change parallel to coordinate axes. Directional derivatives generalize this concept to arbitrary trajectories in $\mathbb{R}^n$, while the gradient vector $\nabla f$ encapsulates all directional variation into a single vector field.

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Directional Derivatives from First Principles

Let $f: D \subset \mathbb{R}^n \to \mathbb{R}$ and let $\mathbf{u}$ be a unit vector ($|\mathbf{u}| = 1$). The directional derivative of $f$ at $\mathbf{x}_0$ along $\mathbf{u}$ is: $D_{\mathbf{u}} f(\mathbf{x}_0) = \lim_{h \to 0} \frac{f(\mathbf{x}_0 + h \mathbf{u}) - f(\mathbf{x}_0)}{h}$

• For $\mathbf{u} = \mathbf{i} = \langle 1, 0 \rangle$, $D_{\mathbf{i}} f = f_x$; for $\mathbf{u} = \mathbf{j} = \langle 0, 1 \rangle$, $D_{\mathbf{j}} f = f_y$.
• For an arbitrary non-zero vector $\mathbf{v}$, normalize first: $\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|}$.

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The Gradient Vector and Dot Product Formula

For a scalar field $f$, the gradient vector $\nabla f$ is: $\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle, \quad \nabla f(x, y, z) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle$

The Fundamental Directional Derivative Formula

If $f$ is differentiable at $\mathbf{x}_0$, the directional derivative along unit vector $\mathbf{u}$ is: $D_{\mathbf{u}} f(\mathbf{x}_0) = \nabla f(\mathbf{x}_0) \cdot \mathbf{u} = \|\nabla f(\mathbf{x}_0)\| \cos \theta$ where $\theta$ is the angle between $\nabla f(\mathbf{x}_0)$ and $\mathbf{u}$.

Requirement: Differentiability is mandatory. If $f$ is not differentiable, directional derivatives may exist along every ray without satisfying the dot product formula.

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Geometric Properties of the Gradient

Because $-1 \le \cos \theta \le 1$, the dot product formula yields three cardinal properties:

1. Steepest Ascent: Occurs when $\theta = 0$ ($\cos \theta = 1$).
   • Direction: $\mathbf{u} = \frac{\nabla f}{|\nabla f|}$ (along the gradient).
   • Maximum Value: $D_{\mathbf{u}} f = |\nabla f(\mathbf{x}_0)|$.
2. Steepest Descent: Occurs when $\theta = \pi$ ($\cos \theta = -1$).
   • Direction: $\mathbf{u} = -\frac{\nabla f}{|\nabla f|}$ (opposite to the gradient).
   • Minimum Value: $D_{\mathbf{u}} f = -|\nabla f(\mathbf{x}_0)|$.
3. Zero Change: Occurs when $\theta = \pi/2$ ($\cos \theta = 0$). Directions orthogonal to $\nabla f$ produce zero rate of change and lie tangent to level sets.

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Orthogonality to Level Curves and Surfaces

Let $F(x, y, z) = k$ define a smooth level surface $S$. Let $C$ be any curve on $S$ parameterized by $\mathbf{r}(t)$ with $\mathbf{r}(t_0) = P_0(x_0, y_0, z_0)$. Because $F(\mathbf{r}(t)) = k$ is constant: $\frac{d}{dt} F(\mathbf{r}(t)) = \nabla F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = 0$ At $t = t_0$, $\nabla F(P_0) \cdot \mathbf{r}'(t_0) = 0$. Since $\mathbf{r}'(t_0)$ is an arbitrary tangent vector to $S$ at $P_0$, the gradient $\nabla F(P_0)$ is perpendicular to every tangent vector to the level surface. Thus $\nabla F(P_0)$ is a normal vector $\mathbf{n}$ to the tangent plane.

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Tangent Planes and Normal Lines to Level Surfaces

For level surface $F(x, y, z) = k$ at $P_0(x_0, y_0, z_0)$ with $\nabla F(P_0) \neq \mathbf{0}$:

Cartesian Equation of the Tangent Plane

$F_x(P_0)(x - x_0) + F_y(P_0)(y - y_0) + F_z(P_0)(z - z_0) = 0$

Equations of the Normal Line

• Parametric Form: $\mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t \langle F_x(P_0), F_y(P_0), F_z(P_0) \rangle$
• Symmetric Form: $\frac{x - x_0}{F_x(P_0)} = \frac{y - y_0}{F_y(P_0)} = \frac{z - z_0}{F_z(P_0)}$ (for non-zero components)

Explicit Surfaces $z = f(x, y)$

Rewriting $z = f(x, y)$ as $F(x, y, z) = f(x, y) - z = 0$ gives normal $\nabla F = \langle f_x, f_y, -1 \rangle$, reproducing the explicit tangent plane formula $z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$.

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Total Differentials and Error Propagation

For differentiable $w = f(x, y, z)$, the total differential is: $dw = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz = \nabla f \cdot \langle dx, dy, dz \rangle$ For small increments $\Delta x, \Delta y, \Delta z$, the change $\Delta w \approx dw$:

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Step-by-Step Worked Problems

Problem 1: Directional Derivative along a Vector

Compute the directional derivative of $f(x, y) = x^2 e^{2y} - y^3$ at $P(2, 0)$ pointing toward $Q(5, 4)$.

Solution:

1. Gradient: $\nabla f = \langle 2x e^{2y}, 2x^2 e^{2y} - 3y^2 \rangle \implies \nabla f(2, 0) = \langle 4, 8 \rangle$.
2. Displacement: $\mathbf{v} = \vec{PQ} = \langle 3, 4 \rangle$.
3. Normalize: $|\mathbf{v}| = \sqrt{9 + 16} = 5 \implies \mathbf{u} = \langle 3/5, 4/5 \rangle$.
4. Dot product: $D_{\mathbf{u}} f(2, 0) = \langle 4, 8 \rangle \cdot \langle 3/5, 4/5 \rangle = \frac{12 + 32}{5} = \frac{44}{5}$.

Problem 2: Tangent Plane to an Ellipsoid

Find the tangent plane to $x^2 + 2y^2 + 3z^2 = 36$ at $P_0(1, -2, 3)$.

Solution:

1. Level surface $F(x, y, z) = x^2 + 2y^2 + 3z^2$. Verify $1 + 8 + 27 = 36$.
2. Gradient: $\nabla F = \langle 2x, 4y, 6z \rangle$.
3. At $P_0$: $\mathbf{n} = \nabla F(1, -2, 3) = \langle 2, -8, 18 \rangle = 2 \langle 1, -4, 9 \rangle$.
4. Tangent plane: $1(x - 1) - 4(y + 2) + 9(z - 3) = 0 \implies x - 4y + 9z = 36$.

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GRE Exam Traps & Pitfalls

Trap 1: Forgetting to Normalize the Direction Vector $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$ requires a unit vector. Computing $\nabla f \cdot \mathbf{v}$ for $|\mathbf{v}| \neq 1$ scales the derivative by $|\mathbf{v}|$.

Trap 2: Applying the Dot Product Formula Without Differentiability When $f$ is non-differentiable, directional derivatives along rays may exist while failing the dot product identity.

Trap 3: Sign Errors on Explicit Surfaces For $z = f(x, y)$, using $f(x, y) - z = 0$ gives normal $\langle f_x, f_y, -1 \rangle$. Do not flip individual signs.