Multivariable-Calculus · Unit 8 · Video 1 · Interactive Practice

Change x and y Together: Why the Effects Add, and the Tangent Plane Behind It

IKey Formulas

Formula	Name	What it says
$\Delta z \approx f_x\,\Delta x + f_y\,\Delta y$	Approximation formula	Both inputs move: the two effects add (small steps)
$y = y_0,\ z = z_0 + a(x - x_0)$ and $x = x_0,\ z = z_0 + b(y - y_0)$	The two tangent lines	$a = f_x(x_0, y_0)$ , $b = f_y(x_0, y_0)$
$z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$	Tangent plane	The plane containing both tangent lines
$\Delta z = f_x\,\Delta x + f_y\,\Delta y + (\text{higher-order terms})$	Exact change	The extra terms involve second and higher derivatives

Key Insight: On the tangent plane $\Delta z = a\,\Delta x + b\,\Delta y$ holds exactly; the approximation formula says the graph of $f$ stays close to that plane near $(x_0, y_0)$ .

IIVisualization 1 — The Two Effects Add

For $f = x^3 y + y^2$ at $(1, 1)$ , where $f_x = f_y = 3$ : step $\Delta x$ , then $\Delta y$ , in either order.

Δx 0.01 Δy 0.02

IIIVisualization 2 — Two Tangent Lines Make the Tangent Plane

The tangent lines of the slices $y = y_0$ and $x = x_0$ both lie in one plane: the tangent plane.

Base point Step 1 View angle 35°

Step 1 — Slice $y = y_0$ , slope $a = f_x(x_0, y_0)$

y = y_0, \qquad z = z_0 + a(x - x_0)

💡 The same plane comes from vectors: $\langle 1, 0, a\rangle \times \langle 0, 1, b\rangle = \langle -a, -b, 1\rangle$ is a normal vector, giving $-a(x - x_0) - b(y - y_0) + (z - z_0) = 0$ .

IVVisualization 3 — The Graph Hugs Its Tangent Plane

Along $\Delta x = h$ , $\Delta y = 2h$ from $(1, 1)$ , the error of the formula shrinks like $h^2$ , faster than the step.

Step h 0.500

VQuiz Questions

Problem 1 · Predict the Change

Given: $f(x, y) = x^2 y + 3y$ at the point $(2, 1)$ . Estimate the change $\Delta z$ when $x$ increases by $0.1$ and $y$ decreases by $0.05$ .

✅ Correct!

f_x = 4

and

f_y = 7

(2, 1)

, so

\Delta z \approx 4(0.1) + 7(-0.05) = 0.4 - 0.35 = 0.05

❌ Check the sign of $\Delta y$ .

y

decreases, so

\Delta y = -0.05

and the

y

-effect

7(-0.05) = -0.35

is negative.

❌ The pairing is swapped.

f_x

multiplies

\Delta x

and

f_y

multiplies

\Delta y

4(0.1) + 7(-0.05)

, not

7(0.1) + 4(-0.05)

❌ Check $f_y$ . Holding

x

fixed,

\frac{\partial}{\partial y}(3y) = 3

, so

f_y = x^2 + 3 = 7

, not

x^2 = 4

❌ Not quite. Compute

f_x

and

f_y

(2, 1)

, then use

\Delta z \approx f_x\,\Delta x + f_y\,\Delta y

Show solution

The partial derivatives are

f_x = 2xy, \qquad f_y = x^2 + 3,

so at $(2, 1)$ : $f_x(2, 1) = 2 \cdot 2 \cdot 1 = 4$ and $f_y(2, 1) = 4 + 3 = 7$ . With $\Delta x = 0.1$ and $\Delta y = -0.05$ :

\Delta z \approx 4(0.1) + 7(-0.05) = 0.4 - 0.35 = 0.05.

The two effects partly cancel: increasing $x$ raises $f$ by about $0.4$ , decreasing $y$ lowers it by about $0.35$ .

Problem 2 · Write the Tangent Plane

Given: $f(x, y) = x^2 + x y^3$ . Find the tangent plane to the graph $z = f(x, y)$ at the point above $(1, 2)$ .

✅ Correct!

z_0 = 9

f_x(1, 2) = 10

f_y(1, 2) = 12

, and each slope multiplies the change from the point.

❌ The plane misses the point. At

(1, 2)

this gives

z = 0

, but the plane must pass through

(1, 2, 9)

: add

z_0 = f(1, 2) = 9

❌ The slopes are swapped.

f_x(1, 2) = 10

goes with

(x - 1)

and

f_y(1, 2) = 12

with

(y - 2)

❌ Measure from the point. The slopes multiply

x - x_0

and

y - y_0

; at

(1, 2)

this plane gives

z = 43

, not

9

❌ Not quite. Use

z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)

Show solution

Value at the point: $z_0 = f(1, 2) = 1 + 1 \cdot 8 = 9$ .

Partials: $f_x = 2x + y^3$ , $f_y = 3xy^2$ , so

f_x(1, 2) = 2 + 8 = 10, \qquad f_y(1, 2) = 3 \cdot 1 \cdot 4 = 12.

Tangent plane:

z = 9 + 10(x - 1) + 12(y - 2).

Check: setting $y = 2$ leaves the tangent line $z = 9 + 10(x - 1)$ of the slice $y = 2$ , whose slope is $f_x(1, 2) = 10$ .

Problem 3 · Estimate a Value

Given: $f(x, y) = x^3 y + y^2$ , with $f(1, 1) = 2$ and $f_x(1, 1) = f_y(1, 1) = 3$ . Estimate $f(0.98,\, 1.03)$ with the approximation formula.

✅ Correct!

\Delta z \approx 3(-0.02) + 3(0.03) = 0.03

, so

f \approx 2 + 0.03 = 2.03

(the exact value is

2.0303\ldots

❌ That is $\Delta z$ , not $f$ . The formula gives the change; add it to the starting value

f(1, 1) = 2

❌ Check the sign of $\Delta x$ .

x

goes from

1

0.98

, so

\Delta x = -0.02

and its effect

3(-0.02) = -0.06

is negative.

❌ The signs are reversed.

\Delta x = 0.98 - 1 = -0.02

and

\Delta y = 1.03 - 1 = +0.03

, so

\Delta z \approx -0.06 + 0.09 = +0.03

❌ Not quite. Find

\Delta x

and

\Delta y

from

(1, 1)

, estimate

\Delta z

, then add

f(1, 1)

Show solution

From the base point $(1, 1)$ : $\Delta x = 0.98 - 1 = -0.02$ and $\Delta y = 1.03 - 1 = 0.03$ .

\Delta z \approx f_x\,\Delta x + f_y\,\Delta y = 3(-0.02) + 3(0.03) = -0.06 + 0.09 = 0.03

f(0.98, 1.03) \approx f(1, 1) + \Delta z = 2 + 0.03 = 2.03

The exact value is $0.98^3 \cdot 1.03 + 1.03^2 = 0.96942776 + 1.0609 = 2.03032776$ , off by about $0.0003$ : the second-order terms.

Problem 4 · Read a Tangent Plane

Given: the tangent plane to $z = f(x, y)$ at the point above $(2, -1)$ is $z = 5 - 4(x - 2) + 3(y + 1)$ . Find $f_x(2, -1)$ and estimate $f(2.1,\, -1.2)$ .

The partial derivative in x at (2, −1)

The estimate at (2.1, −1.2)

✅ Correct! The coefficients of

(x - 2)

and

(y + 1)

are

f_x = -4

and

f_y = 3

, and

f(2.1, -1.2) \approx 5 - 0.4 - 0.6 = 4

❌ $5$ is the height $z_0 = f(2, -1)$ .

f_x

is the coefficient of

(x - 2)

❌ Keep the sign. The term is

-4(x - 2)

, so the coefficient of

(x - 2)

f_x(2, -1) = -4

, not

4

❌ $3$ is $f_y(2, -1)$ . It multiplies

(y + 1) = y - y_0

;

f_x

is the coefficient of

(x - 2)

❌ Check $f_x$ . In

z = z_0 + a(x - x_0) + b(y - y_0)

f_x

a

, the coefficient of

(x - 2)

, sign included.

❌ Check $\Delta y$ .

y

goes from

-1

-1.2

, so

\Delta y = y + 1 = -0.2

, not

+0.2

❌ That is $\Delta z$ . The plane value includes the height

z_0 = 5

5 + (-1) = 4

❌ Check the estimate. Substitute

x - 2 = 0.1

and

y + 1 = -0.2

into the plane, keeping the sign of

-4

Show solution

Compare with $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ at $(x_0, y_0) = (2, -1)$ , where $y - y_0 = y + 1$ :

z_0 = f(2, -1) = 5, \qquad f_x(2, -1) = -4, \qquad f_y(2, -1) = 3.

For $(2.1, -1.2)$ : $\Delta x = x - 2 = 0.1$ and $\Delta y = y + 1 = -0.2$ , so

f(2.1, -1.2) \approx 5 + (-4)(0.1) + 3(-0.2) = 5 - 0.4 - 0.6 = 4.

Near the point, the graph of $f$ is close to its tangent plane, so the plane's height is the estimate.

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