Partial Derivatives & the Gradient

Why freezing works: setting $y = b$ turns $f$ into a function of $x$ alone, $g(x) = f(x, b)$, and its graph is the curve where the vertical plane $y = b$ slices the surface. The partial $f_x(a, b)$ is just $g'(a)$, the slope of that slice, so nothing new is being differentiated. For $f = x^3y^2$ the slice $y = 1$ is $z = x^3$, with slope $3$ at $x = 1$, and indeed $f_x(1, 1) = 3\cdot1^2\cdot1^2 = 3$. The other slice, $x = 1$, is $z = y^2$, with slope $2$ at $y = 1$, matching $f_y(1, 1) = 2\cdot1^3\cdot1 = 2$.

The directional derivative along a unit vector $\mathbf{u}$ is the dot product $D_{\mathbf u}f = \nabla f \cdot \mathbf u$ — so no direction can beat $|\nabla f|$, and the level curves are always perpendicular to the gradient.

Why the two partials are the two tilts: inside the plane, moving along $y = 2$ changes $z$ at the rate $f_x = 2$ per unit of $x$, and moving along $x = 1$ changes it at the rate $f_y = 4$; a non-vertical plane is fixed by one point and those two slopes. Multiplied out, the plane is $z = 2x + 4y - 5$. It touches the paraboloid at $(1, 2, 5)$ and stays below it everywhere, because the gap is $x^2 + y^2 - (2x + 4y - 5) = (x - 1)^2 + (y - 2)^2 \ge 0$.

Why it is accurate: the tangent plane matches the surface to first order, so the error is small compared with the step — for a smooth $f$, halving the step roughly quarters the error. That makes the differential the working tool for two jobs: estimating a value near a point where everything is easy, and tracking how uncertainty in the inputs spreads to the output. In both, you differentiate once and plug in the small changes.

The steps are $\Delta x = 0.02$ and $\Delta y = -0.03$, so $f \approx 5 + \tfrac35(0.02) + \tfrac45(-0.03) = 5 + 0.012 - 0.024 = 4.988$. The true value is $\sqrt{24.8813} \approx 4.98812$, so the estimate is off by about $0.0001$. The whole trick is the choice of base point: $(3, 4)$ is close to the target and makes $f$ and both partials exact.

The signs of the measurement errors are unknown, so take the worst case, where both terms push the same way: the relative error is at most $\tfrac{0.01}{1} + 2\cdot\tfrac{0.02}{2} = 1\% + 2\% = 3\%$. The measured value is $g = 4\pi^2\cdot\tfrac14 = \pi^2 \approx 9.87$, so $g = 9.87 \pm 0.30$ m/s$^2$. The period enters squared, so its error counts twice: to improve the experiment, time more swings rather than measure the string more carefully.

Why perpendicular: $f$ does not change along a level curve, so the directional derivative along the curve's tangent $\mathbf u$ is $0$, and $D_{\mathbf u}f = \nabla f\cdot\mathbf u = 0$ says exactly that $\nabla f\perp\mathbf u$. Why the spacing: to climb from the contour $c$ to the contour $c + \Delta c$ at the steepest rate $|\nabla f|$ takes a distance of about $\tfrac{\Delta c}{|\nabla f|}$, so a large gradient packs the contours tightly.

The tangent to the ellipse at $(2, 1)$ must be perpendicular to $(4, 8)$, and $(2, -1)$ is, since $8 - 8 = 0$. Implicit differentiation agrees: $2x + 8yy' = 0$ gives $y' = -\tfrac{x}{4y} = -\tfrac12$, the slope of $(2, -1)$. Walking straight toward the origin, $\mathbf u = -\tfrac{1}{\sqrt5}(2, 1)$, the rate of change is $\nabla f\cdot\mathbf u = -\tfrac{8 + 8}{\sqrt5} = -\tfrac{16}{\sqrt5} \approx -7.16$. The steepest descent rate is larger, $|\nabla f| = \sqrt{80} = 4\sqrt5 \approx 8.94$, in the direction $-(1, 2)$. On a bowl that is not round, the steepest way down does not point at the bottom.

The rate of change in the direction of $(2, -1, 2)$ needs the unit vector $\tfrac13(2, -1, 2)$: $D_{\mathbf u}f = \tfrac{12 - 3 + 4}{3} = \tfrac{13}{3}$. And since $\nabla f$ is perpendicular to the level surface $xyz = 6$, it is the normal of the tangent plane there: $6(x - 1) + 3(y - 2) + 2(z - 3) = 0$, that is, $6x + 3y + 2z = 18$.

Since $\cos\theta \le 1$ with equality only at $\theta = 0$, the slope is maximized precisely when $\mathbf u$ points along $\nabla f$, and the maximal slope is $|\nabla f|$. $\blacksquare$

Differentiability of $f$ at $(a,b)$ says exactly that $f(x,y)$ minus this plane is $o\!\left(\sqrt{(x-a)^2 + (y-b)^2}\right)$ — the plane matches the surface to first order, which is what makes it THE tangent plane and the basis of linear approximation. $\blacksquare$

That is why every one-variable rule (product, quotient, chain) transfers verbatim: during the computation, $y$ is a constant like any other. Two subtleties are worth knowing early. First, the mixed partials usually agree: $f_{xy} = f_{yx}$ for smooth functions (Clairaut's theorem), so the order of differentiation is a free choice. Second, having both partials is weaker than true differentiability — the honest statement is that when the partials exist and are continuous, the tangent-plane approximation $f(a+h, b+k) \approx f(a,b) + f_x h + f_y k$ has error shrinking faster than $\sqrt{h^2 + k^2}$, which is what every application (linearization, error propagation, Newton's method) actually uses. For $f = x^2y$ at $(1, 2)$ the slice is $g(x) = 2x^2$, and the difference quotient is $\dfrac{2(1 + h)^2 - 2}{h} = 4 + 2h$: slope $6$ for $h = 1$, $5$ for $h = \tfrac12$, and $4 = f_x(1, 2)$ in the limit.

Step 1 — partials: $f_x = 2xy$ (treat $y$ constant) and $f_y = x^2$ (treat $x$ constant).

Step 2 — evaluate: $\nabla f(1,2) = (2\cdot1\cdot2,\; 1^2) = (4, 1)$.

Step 3 — steepest ascent: the maximum slope at this point is $|\nabla f| = \sqrt{16 + 1} = \sqrt{17} \approx 4.12$, achieved walking in the direction of $(4,1)$.

Step 4 — slope in a GIVEN direction $\mathbf u = \left(\tfrac35, \tfrac45\right)$ (note $|\mathbf u| = 1$, required!): $D_{\mathbf u}f = \nabla f\cdot\mathbf u = 4\cdot\tfrac35 + 1\cdot\tfrac45 = \tfrac{12 + 4}{5} = \tfrac{16}{5} = 3.2$.

Step 5 — coherence check: $3.2 < 4.12$ ✓, as no direction can beat the gradient direction; and walking perpendicular to $(4,1)$, e.g. along $(1,-4)/\sqrt{17}$, gives slope $\tfrac{4 - 4}{\sqrt{17}} = 0$ — that is the direction the level curve runs.

Step 1 — partials: $f_x = 2x = 2$ and $f_y = 2y = 4$ at the point.

Step 2 — assemble the plane: $z = 5 + 2(x - 1) + 4(y - 2)$.

Step 3 — estimate: $z \approx 5 + 2(0.1) + 4(-0.1) = 5 + 0.2 - 0.4 = 4.8$.

Step 4 — compare with the truth: $f(1.1, 1.9) = 1.21 + 3.61 = 4.82$. The linear estimate missed by $0.02$ — and the miss is exactly the curvature the plane cannot see: $(0.1)^2 + (-0.1)^2 = 0.02$. For this surface the tangent-plane error is ALWAYS $h^2 + k^2$, a perfect illustration of "error shrinks like the square of the step."