Tangent planes and linear approximations to surfaces: Exercises

Level 1 — Recognition

Can you read the formula and pull the right numbers out of it?

L1.1

Given $f (x, y) = 3 x - 2 y + 7$ . Write down $f (2, 1)$ , $f_{x} (2, 1)$ , $f_{y} (2, 1)$ , and the normal vector $n$ .

Recall Solution L1.1

$f (2, 1) = 3 (2) - 2 (1) + 7 = 6 - 2 + 7 = 11$ .
$f_{x} = 3$ (coefficient of $x$ ), so $f_{x} (2, 1) = 3$ .
$f_{y} = - 2$ (coefficient of $y$ ), so $f_{y} (2, 1) = - 2$ .
Normal $n = ⟨ f_{x}, f_{y}, - 1 ⟩ = ⟨ 3, - 2, - 1 ⟩$ .

What we did: since $f$ is already linear, its partials are just the constant coefficients. What it looks like: the surface is its own tangent plane — a flat sheet, tilted $3$ in $x$ and $- 2$ in $y$ .

L1.2

A surface has $f (a, b) = 4$ , $f_{x} (a, b) = - 1$ , $f_{y} (a, b) = 5$ at $(a, b) = (0, 3)$ . Write the tangent plane equation.

Recall Solution L1.2

Plug straight into $L$ : $z = 4 + (- 1) (x - 0) + 5 (y - 3) = 4 - x + 5 y - 15 = - x + 5 y - 11.$ Why the $- 3$ appears: the y-step is $(y - b) = (y - 3)$ , not $y$ . The base point is $b = 3$ .

Level 2 — Application

Compute the plane and use it to approximate a value.

L2.1

Find the tangent plane to $f (x, y) = x^{2} y$ at $(2, 1)$ .

Recall Solution L2.1

Step 1 (height): $f (2, 1) = 2^{2} \cdot 1 = 4$ . Step 2 (slopes): $f_{x} = 2 x y \Rightarrow f_{x} (2, 1) = 2 \cdot 2 \cdot 1 = 4$ ; $f_{y} = x^{2} \Rightarrow f_{y} (2, 1) = 4$ . Step 3 (assemble): $z = 4 + 4 (x - 2) + 4 (y - 1) = 4 x + 4 y - 8.$

L2.2

Use L2.1's plane to estimate $f (1.9, 1.05) = (1.9)^{2} (1.05)$ .

Recall Solution L2.2

Steps from base: $d x = 1.9 - 2 = - 0.1$ , $d y = 1.05 - 1 = 0.05$ . $L = 4 + 4 (- 0.1) + 4 (0.05) = 4 - 0.4 + 0.2 = 3.8.$ True value $= 3.61 \times 1.05 = 3.7905$ . Error $\approx 0.0095$ — small, because both steps are small. ✓

L2.3

Find the tangent plane to $f (x, y) = e^{x} cos y$ at $(0, 0)$ and use it to approximate $e^{0.1} cos (0.05)$ .

Recall Solution L2.3

Step 1: $f (0, 0) = e^{0} cos 0 = 1$ . Step 2: $f_{x} = e^{x} cos y \Rightarrow f_{x} (0, 0) = 1$ . $f_{y} = - e^{x} sin y \Rightarrow f_{y} (0, 0) = - 1 \cdot 0 = 0$ . Step 3: $z = 1 + 1 (x - 0) + 0 (y - 0) = 1 + x$ . Approximate: $d x = 0.1$ , $d y = 0.05$ : $L = 1 + 0.1 = 1.1$ . True value $= e^{0.1} cos (0.05) \approx 1.10517 \times 0.99875 \approx 1.1038$ . Error $\approx 0.0038$ . ✓ Why $f_{y}$ vanished: $cos y$ is flat at its peak $y = 0$ — moving in $y$ doesn't change height to first order, so the plane is level front-to-back there.

Level 3 — Analysis

Interpret geometry, use differentials, handle chained functions.

L3.1

For $f (x, y) = x^{2} + y^{2}$ at $(3, 4)$ , find the tangent plane, then verify the normal vector $n = ⟨ f_{x}, f_{y}, - 1 ⟩$ is genuinely perpendicular to both cross-section tangent directions $⟨ 1, 0, f_{x} ⟩$ and $⟨ 0, 1, f_{y} ⟩$ .

Recall Solution L3.1

Slopes: $f_{x} = \frac{x}{x ^{2} + y ^{2}} = \frac{3}{5}$ , $f_{y} = \frac{4}{5}$ (see parent Example 2). Plane: $z = 5 + \frac{3}{5} (x - 3) + \frac{4}{5} (y - 4) .$ Normal $n = ⟨ \frac{3}{5}, \frac{4}{5}, - 1 ⟩$ . Perpendicularity check (dot product = 0 means right angle): $n \cdot ⟨ 1, 0, \frac{3}{5} ⟩ = \frac{3}{5} + 0 + (- 1) \frac{3}{5} = 0. ✓$ $n \cdot ⟨ 0, 1, \frac{4}{5} ⟩ = 0 + \frac{4}{5} + (- 1) \frac{4}{5} = 0. ✓$ What it means: the normal is at a right angle to both tangent lines, so it's perpendicular to the whole plane they span. That's exactly why The gradient vector view works.

L3.2

A cylinder has radius $r = 5$ cm and height $h = 12$ cm, each measurable to within $0.05$ cm. Volume $V = π r^{2} h$ . Estimate the maximum error $d V$ .

Recall Solution L3.2

Partials: $V_{r} = 2 π r h$ , $V_{h} = π r^{2}$ . At $(5, 12)$ : $V_{r} = 2 π (5) (12) = 120 π$ , $V_{h} = π (25) = 25 π$ . Differential (worst case, both errors positive): $d V = ∣ V_{r} ∣ d r + ∣ V_{h} ∣ d h = 120 π (0.05) + 25 π (0.05) = 6 π + 1.25 π = 7.25 π \approx 22.78 cm^{3} .$ Why absolute values: to bound the error we assume the two contributions add constructively rather than cancel.

L3.3

The plane $z = 2 x + 4 y - 5$ is tangent to $f = x^{2} + y^{2}$ at $(1, 2)$ (parent Example 1). Without recomputing partials, explain why the linear estimate of $f (1.5, 2.5)$ is worse than that of $f (1.1, 1.9)$ , and quantify roughly.

Recall Solution L3.3

Error $E = f - L$ for this paraboloid equals $(x - 1)^{2} + (y - 2)^{2}$ exactly (algebra below), i.e. the squared distance from the base point.

At $(1.1, 1.9)$ : distance $^{2} = 0. 1^{2} + 0. 1^{2} = 0.02$ .
At $(1.5, 2.5)$ : distance $^{2} = 0. 5^{2} + 0. 5^{2} = 0.50$ . Ratio $= 0.50/0.02 = 25$ : the far point's error is 25× larger. Algebra: $f - L = (x^{2} + y^{2}) - (5 + 2 (x - 1) + 4 (y - 2)) = x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = (x - 1)^{2} + (y - 2)^{2}$ . ✓ This is the concrete face of "error grows like the square of the distance."

Level 4 — Synthesis

Combine ideas: solve for unknowns, work backwards, cross to related surfaces.

L4.1

The tangent plane to some $f (x, y)$ at $(1, - 1)$ is $z = 2 + 3 (x - 1) - 4 (y + 1)$ . Estimate $f (1.02, - 0.97)$ and state $f (1, - 1)$ , $f_{x} (1, - 1)$ , $f_{y} (1, - 1)$ .

Recall Solution L4.1

Read the plane like the formula $L = f (a, b) + f_{x} (x - a) + f_{y} (y - b)$ with $(a, b) = (1, - 1)$ : $f (1, - 1) = 2$ , $f_{x} (1, - 1) = 3$ , $f_{y} (1, - 1) = - 4$ (note $y - b = y - (- 1) = y + 1$ ). Steps: $d x = 0.02$ , $d y = - 0.97 - (- 1) = 0.03$ . $f (1.02, - 0.97) \approx 2 + 3 (0.02) - 4 (0.03) = 2 + 0.06 - 0.12 = 1.94.$

L4.2

Find the point on the paraboloid $z = x^{2} + y^{2}$ where the tangent plane is parallel to the plane $z = 6 x - 4 y + 1$ .

Recall Solution L4.2

Idea: two non-vertical planes are parallel when their $x$ - and $y$ -slopes match. The target plane has $x$ -slope $6$ , $y$ -slope $- 4$ . Set the surface slopes equal: $f_{x} = 2 x = 6 \Rightarrow x = 3, f_{y} = 2 y = - 4 \Rightarrow y = - 2.$ Height: $z = 3^{2} + (- 2)^{2} = 9 + 4 = 13$ . Point: $(3, - 2, 13)$ . What it looks like: the paraboloid tilts more steeply the farther you go from the axis; the unique spot with the demanded tilt is $(3, - 2)$ .

L4.3

Show that at every point of the sphere piece $f (x, y) = 9 - x^{2} - y^{2}$ (upper hemisphere, radius 3), the surface normal $n = ⟨ f_{x}, f_{y}, - 1 ⟩$ points along the radius direction. Verify at $(1, 2)$ .

Recall Solution L4.3

Partials (chain rule on the square root): $f_{x} = \frac{- x}{9 - x ^{2} - y ^{2}}, f_{y} = \frac{- y}{9 - x ^{2} - y ^{2}} .$ At $(1, 2)$ : $9 - 1 - 4 = 4 = 2$ , so $z = 2$ and $f_{x} = - \frac{1}{2}, f_{y} = - 1$ . $n = ⟨ - \frac{1}{2}, - 1, - 1 ⟩$ . Multiply by $- 2$ : $⟨ 1, 2, 2 ⟩$ — exactly the position vector $(x, y, z) = (1, 2, 2)$ from the centre. Why: on a sphere the outward radius is the normal, so this confirms the tangent-plane machinery reproduces classical geometry. See The gradient vector.

Level 5 — Mastery

Prove, generalise, and confront degenerate cases.

L5.1

Consider $f (x, y) = ⎩ ⎨ ⎧ \frac{x y}{x ^{2} + y ^{2}}, 0, (x, y) \neq = (0, 0) (x, y) = (0, 0) .$ (a) Show both partials $f_{x} (0, 0)$ and $f_{y} (0, 0)$ exist and equal $0$ , so the candidate tangent plane is $z = 0$ . (b) Show $f$ is not differentiable at $(0, 0)$ , so no true tangent plane exists — by testing the diagonal $y = x$ .

Recall Solution L5.1

(a) Partials at the origin — use the limit definition along each axis. Along $y = 0$ : $f (x, 0) = \frac{x \cdot 0}{x ^{2} + 0} = 0$ for all $x$ . So $f_{x} (0, 0) = lim_{h \to 0} \frac{f ( h , 0 ) - f ( 0 , 0 )}{h} = lim_{h \to 0} \frac{0 - 0}{h} = 0.$ By symmetry $f_{y} (0, 0) = 0$ . Candidate plane: $L = 0 + 0 \cdot x + 0 \cdot y = 0$ .

(b) Differentiability fails. The precise test (parent note): does $\frac{f ( x , y ) - L ( x , y )}{x ^{2} + y ^{2}} \to 0 ?$ Approach along $y = x$ : $f (x, x) = \frac{x \cdot x}{x ^{2} + x ^{2}} = \frac{x ^{2}}{2 x ^{2}} = \frac{1}{2}$ (constant!). And $L = 0$ . Distance $= x^{2} + x^{2} = ∣ x ∣ 2$ . So the ratio is $\frac{\frac{1}{2} - 0}{∣ x ∣ 2} = \frac{1}{2 2 ∣ x ∣} \to \infty \neq = 0.$ The error does not shrink faster than distance — it blows up. So no tangent plane exists even though both partials are $0$ . What it looks like (Figure): the surface is level along both axes but jumps to height $\frac{1}{2}$ along the diagonal — a twisted saddle the flat plane can never hug.

Figure — Tangent planes and linear approximations to surfaces

L5.2

Prove that for any function of the form $f (x, y) = g (x) + h (y)$ (separable), the tangent plane at $(a, b)$ is just the sum of the two single-variable tangent lines. Verify with $f = sin x + y^{3}$ at $(0, 1)$ .

Recall Solution L5.2

Proof. $f_{x} = g^{'} (x)$ (the $h (y)$ term is constant in $x$ ), similarly $f_{y} = h^{'} (y)$ . So $L = g (a) + h (b) + g^{'} (a) (x - a) + h^{'} (b) (y - b) .$ Group by variable: $L = tangent line to g [g (a) + g^{'} (a) (x - a)] + tangent line to h [h (b) + h^{'} (b) (y - b)] .$ This links directly to Tangent line and linear approximation (single variable): the 2D plane decouples into two independent 1D approximations. Verify $f = sin x + y^{3}$ at $(0, 1)$ : $f (0, 1) = 0 + 1 = 1$ . $f_{x} = cos x \Rightarrow 1$ ; $f_{y} = 3 y^{2} \Rightarrow 3$ . $L = 1 + 1 (x - 0) + 3 (y - 1) = 1 + x + 3 y - 3 = x + 3 y - 2.$ The $g$ -part $0 + 1 \cdot x$ is the tangent line to $sin x$ at $0$ ; the $h$ -part $1 + 3 (y - 1)$ is the tangent line to $y^{3}$ at $1$ . ✓

L5.3

The surface $z = f (x, y)$ has tangent plane $z = 5 + 2 (x - 1) - 3 (y - 2)$ at $(1, 2)$ . A bug walks from $(1, 2)$ in the direction of the unit vector $u = ⟨ \frac{3}{5}, \frac{4}{5} ⟩$ . Using only the tangent plane, find the bug's rate of height gain (the directional derivative) — and show it equals $\nabla f \cdot u$ .

Recall Solution L5.3

From the plane: $f_{x} (1, 2) = 2$ , $f_{y} (1, 2) = - 3$ , so $\nabla f = ⟨ 2, - 3 ⟩$ . On the tangent plane the height change over a small step $(d x, d y)$ is $d z = f_{x} d x + f_{y} d y$ . Moving distance $t$ along $u$ gives $d x = \frac{3}{5} t$ , $d y = \frac{4}{5} t$ , so $d z = 2 \cdot \frac{3}{5} t + (- 3) \cdot \frac{4}{5} t = (\frac{6}{5} - \frac{12}{5}) t = - \frac{6}{5} t .$ Rate of height gain $= d z / t = - \frac{6}{5} = - 1.2$ (the bug goes downhill). Check against the dot product (this is the definition, see Directional derivatives): $\nabla f \cdot u = ⟨ 2, - 3 ⟩ \cdot ⟨ \frac{3}{5}, \frac{4}{5} ⟩ = \frac{6}{5} - \frac{12}{5} = - \frac{6}{5} . ✓$ Why they agree: the directional derivative is nothing but the slope of the flat tangent plane read along $u$ — the plane already contains every directional slope.

Recall One-line self-check

The tangent plane equation ::: $z = f (a, b) + f_{x} (a, b) (x - a) + f_{y} (a, b) (y - b)$ "Parallel to plane P" means match ::: the $x$ - and $y$ -slopes only (not the constant) Error of the linear approximation grows like ::: the square of the distance from $(a, b)$ Directional derivative from the plane ::: $\nabla f \cdot u = f_{x} u_{1} + f_{y} u_{2}$

Connections

Partial derivatives — every $f_{x}, f_{y}$ above.
The gradient vector — normals in L3.1, L4.3, L5.3.
Differentiability of multivariable functions — the L5.1 failure.
Tangent line and linear approximation (single variable) — the L5.2 decoupling.
Directional derivatives — L5.3.
The chain rule (multivariable) — used in the $\cdot$ partials.