Calculus Volume 3 · Chapter 4 · Section 4.4

Tangent Planes and Linear Approximations

A tangent line, one dimension up: near a point, a smooth surface is a plane, and the plane is easy to compute with.

Idea 1 · two slopes make a plane $z=f+f_x\,(x-x_0)+f_y\,(y-y_0)$ The slope in

x

and the slope in

y

fix it. Idea 2 · zoom in and it is flatDifferentiable = locally a plane.The gap shrinks faster than the distance. Idea 3 · the plane predicts $\Delta z\approx dz=f_x\,dx+f_y\,dy$ Small changes add, weighted by the slopes.

Colour key surface $z=f(x,y)$ tangent plane slices, $dz$ the point, $\Delta z$ key idea · 3D figures turn when you drag them · page links open the textbook

L1Many tangent lines, one plane
L1The plane from two slopes
L1Zoom in: the surface turns flat
L2Examples 4.21–4.25
L2Linear approximation
L2Lab 2 · how good is the plane?
L2Differentiable
L2Lab 3 · partials, but no plane
L2Differentials
L2Lab 4 · Δz against dz
L2Common mistakes
L2Practice and answers
L3Where the formula comes from
L3Why the error vanishes
L3Why partials are not enough
L3Three variables, and what comes next

Level 1 · see itWhat a tangent plane isbefore any formula

Many tangent lines, one plane

On a curve there is one tangent line. On a surface, every curve through a point has its own tangent line there (p347).

If the surface is smooth, all those lines lie in one plane: the tangent plane.

The figure: $z=\sqrt{41-4x^2-y^2}$ at $(2,3,4)$ and tangent lines in several directions. Turn it until you look along the plane edge-on: every line lies in it.

Five directions, five tangent lines, one plane. Drag to turn.

The plane from two slopes

Two lines fix a plane. Take the two easiest: the slice $y=y_0$ (slope $f_x$ ) and the slice $x=x_0$ (slope $f_y$ ).

Tangent plane · (4.24) · p348

\begin{aligned} z=f(x_0,y_0)&+f_x(x_0,y_0)\,(x-x_0)\\ &+f_y(x_0,y_0)\,(y-y_0) \end{aligned}

Normal vector $\langle f_x,\ f_y,\ -1\rangle$ .

A tangent line, one dimension up: a height plus a slope per direction.

slice y = y₀ slice x = x₀ plane

Example 4.23 at

(2,3)

: slopes

f_x=-2

f_y=-\tfrac34

Zoom in: the surface turns flat

Lab 1. The window around the point shrinks; the figure is rescaled equally in $x$ , $y$ and $z$ , so nothing is stretched.

surface

x₀0 y₀0 window ±1

Try: play the zoom — at ±0.05 the surface and the plane are one sheet. On the saddle the surface crosses its plane, yet it still flattens. Differentiable means: zoomed in far enough, the graph is its tangent plane.

Near the point, the surface is the plane.

Level 2 · compute itPlanes, approximations, differentialsand three labs

Examples 4.21–4.25

Example	given	result
4.21 · p349	$f=2x^2-3xy+8y^2+2x-4y+4$ at $(2,-1)$	$f=34,\ f_x=13,\ f_y=-26$ : $\ z=13x-26y-18$
4.22 · p350	$f=\sin 2x\cos 3y$ at $(\tfrac\pi3,\tfrac\pi4)$	$f=-\tfrac{\sqrt6}{4},\ f_x=\tfrac{\sqrt2}{2},\ f_y=-\tfrac{3\sqrt6}{4}$
4.23 · p352	$f=\sqrt{41-4x^2-y^2}$ , estimate $f(2.1,\,2.9)$ from $(2,3)$	$L=\tfrac{41}{4}-2x-\tfrac34y$ ; $\ L=3.875$ , true $3.8665$
4.24 · p353	$f=2x^2-4y$ differentiable at $(2,-3)$ ?	$E=2(x-2)^2$ , $\ E/\text{dist}\to0$ : yes
4.25 · p357	$f=3x^2-2xy+y^2$ at $(2,-3)$ , $dx=0.1$ , $dy=-0.05$	$dz=18(0.1)-10(-0.05)=2.3$ ; $\ \Delta z=2.3425$

Every example is the same three numbers: $f$ , $f_x$ , $f_y$ at the point.

Linear approximation

Definition · (4.25) · p351

f(x,y)\approx L(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

The tangent plane, used as a formula.

Example 4.23

$L(2.1,\,2.9)=10.25-4.2-2.175=3.875$

$f(2.1,\,2.9)=\sqrt{14.95}\approx3.8665$ : about $0.2\%$ off.

The plane is cheap to evaluate and close to the surface near the point — that is the whole trade.

Lab 2 · how good is the plane?

$f=\sqrt{41-4x^2-y^2}$ , plane at $(2,3)$ . Move the point you estimate.

x2.1 y2.9

Curves of equal gap

|L-f|=0.01,\ 0.05,\ 0.2,\ 0.5

around

(2,3)

. Dashed: the edge of the domain,

4x^2+y^2=41

Try: step away along a line — double the distance, about four times the gap. Near the dashed edge the surface turns vertical and the plane fails fast. The gap grows like distance squared, so close in it is tiny.

Differentiable

Definition · (4.26) · p353

f(x,y)=L(x,y)+E(x,y),\qquad \lim_{(x,y)\to(x_0,y_0)}\frac{E(x,y)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0

The gap must vanish faster than the distance.

Theorem	says	so
4.6 · p355	differentiable ⇒ continuous	a jump rules out a tangent plane
4.7 · p355	$f_x$ , $f_y$ continuous near the point ⇒ differentiable	the everyday test: polynomials, $e^{\cdots}$ , $\sin$ , away from trouble
warning · p350	$f_x$ , $f_y$ exist ⇏ differentiable	$\dfrac{xy}{\sqrt{x^2+y^2}}$ at the origin (Lab 3)

In one variable, "the derivative exists" was enough. In two, two slopes along the axes say nothing about the other directions.

Lab 3 · partials, but no plane

$f=\dfrac{xy}{\sqrt{x^2+y^2}}$ , $f(0,0)=0$ . Flat along both axes, so $f_x(0,0)=f_y(0,0)=0$ and the candidate plane is $z=0$ .

direction θ45°

candidate plane z = 0

The graph is made of straight lines through the origin: a creased cone, not a plane.

Try: θ = 0° or 90° — the slice lies in $z=0$ , as the partials promised. θ = 45° — the slice rises with slope $\tfrac{\sqrt2}{2}\approx0.71$ on both sides: a crease. $E/\text{dist}$ stays at $\tfrac12\sin2\theta$ however close you get, so it does not go to 0.

Differentials

Total differential · (4.27) · p356

dz=f_x(x_0,y_0)\,dx+f_y(x_0,y_0)\,dy,\qquad \Delta z\approx dz

$dx=\Delta x$ , $dy=\Delta y$ . $\Delta z$ climbs the surface; $dz$ climbs the plane.

Example 4.25 · error propagation

$dz=18(0.1)-10(-0.05)=2.3$

exact: $\Delta z=2.3425$

Measuring errors in $x$ and $y$ propagate to $f$ the same way (p357).

A small change in $z$ is a weighted sum of the small changes you make, weighted by the slopes.

A slice through the step:

dz

rises along the plane,

\Delta z

along the surface.

Lab 4 · Δz against dz

Example 4.25: $f=3x^2-2xy+y^2$ at $(2,-3)$ .

dx0.1 dy−0.05

Heights over the moved point: green to the plane (

dz

), red to the surface (

\Delta z

). Heights are drawn at 1/20 of the horizontal scale.

Try: halve both steps — $dz$ halves, the difference $\Delta z-dz$ drops to a quarter. The first-order change is the plane; what is left over is second order.

Common mistakes

Mistake	Result	Fix
Partials left as functions: $z=f+(4x-3y+2)(x-2)+\cdots$	a curved surface, not a plane	evaluate $f_x$ , $f_y$ at the point first: 13 and −26
Forgetting $f(x_0,y_0)$	a plane through the wrong height	check: $z=f$ at $(x_0,y_0)$
$(x+x_0)$ for $(x-x_0)$ with a negative coordinate	Ex 4.21: $y-1$ instead of $y+1$	$y-(-1)=y+1$
Normal $\langle f_x,f_y,1\rangle$ for $z=f(x,y)$	a plane tilted the other way	$\langle f_x,f_y,-1\rangle$ , or the gradient of $f(x,y)-z$
"The partials exist, so it is differentiable"	wrong for $xy/\sqrt{x^2+y^2}$ at the origin	continuous partials (Theorem 4.7), or the limit of $E/\text{dist}$
Using $dz$ as the exact change	Ex 4.25: 2.3 against 2.3425	$\Delta z=f(x_0+dx,y_0+dy)-f(x_0,y_0)$ ; $dz$ only estimates it
Percentage errors added without weights	Ex 201: 9% for the cylinder	$dV/V=2\,dr/r+dh/h=8\%+5\%=13\%$

Practice and answers

Exercises from p358; the book's key: p862, p868.

Checkpoints 4.19–4.22

#	task	answer
4.19	tangent plane to $x^3-x^2y+y^2-2x+3y-2$ at $(-1,3)$	$f=14,\ f_x=7,\ f_y=8$ : $\ z=7x+8y-3$
4.20	estimate $f(4.1,\,0.9)$ for $e^{5-2x+3y}$ from $(4,1)$	$L=6-2x+3y=0.5$ ; true $e^{-0.5}\approx0.6065$
4.21	$3x-4y^2$ differentiable at $(-1,2)$	$E=-4(y-2)^2$ , $\ \|E\|/\text{dist}\le4\,\text{dist}\to0$
4.22	$dz$ and $\Delta z$ for $4y^2+x^2y-2xy$ at $(1,-1)$ , $dx=0.03$ , $dy=-0.02$	$dz=0.18$ ; $\ \Delta z=0.180682$

Exercises

#	task	answer
163	unit normal to $z=x^3$ at $(2,-1,8)$	$\tfrac{1}{\sqrt{145}}(12\mathbf i-\mathbf k)$
165	$x^2+xy+y^2=3$ at $(-1,-1)$ : normal, tangent	$\mathbf i+\mathbf j$ ; $\ \mathbf i-\mathbf j$
167	$xy^2-2x^2+y+5x=6$ at $(4,2)$	$7\mathbf i-17\mathbf j$ ; $\ 17\mathbf i+7\mathbf j$
169	$ze^{x^2-y^2}-3=0$ at $(2,2,3)$ : normal	$-12\mathbf i+12\mathbf j-\mathbf k$
171	tangent plane, $z=-9x^2-3y^2$ , $(2,1,-39)$	$-36x-6y-z=-39$
173	$z=\ln(10x^2+2y^2+1)$ , $(0,0,0)$	$z=0$
175	$xy+yz+zx=11$ , $(1,2,3)$	$5x+4y+3z-22=0$
177	$x^3+y^3=3xyz$ , $(1,2,\tfrac32)$	$4x-5y+4z=0$
179	$z=\sin x+\sin y+\sin(x+y)$ , $(0,0,0)$	$2x+2y-z=0$
181	$z=x^2-2xy+y^2$ , $(1,2,1)$	$-2(x-1)+2(y-2)-(z-1)=0$
183	normal line, $z=5x^2-2y^2$ , $(2,1,18)$	$x=20t+2,\ y=-4t+1,\ z=-t+18$
185	normal line, $z=\ln(3x^2+7y^2+1)$ , $(0,0,0)$	$x=0,\ y=0,\ z=t$
187	normal line, $z=x^2-2xy+y^2$ , $(1,2,1)$	$x-1=2t,\ y-2=-2t,\ z-1=t$
189	segment $BC$ in the figure (p359)	$dz=f_x\,dx+f_y\,dy$
191	$e^{xy}x$ differentiable at $(1,0)$	$e^{xy}x\approx x+y$ , error → 0
193	$x^2+3y$ differentiable everywhere	$\Delta z=2x\Delta x+3\Delta y+(\Delta x)^2$
195	$z=xe^y$ , $P(1,2)$ to $Q(1.05,\,2.1)$	$\Delta z\approx1.185422$ , $\ dz\approx1.108$
197	aluminium in a can, $d=8$ , $h=12$ , 0.04 cm thick	$\approx16\ \text{cm}^3$
199	$x^2+3xy-y^2$ , $(2,3)\to(2.05,\,2.96)$ (p360)	$\Delta z=0.6449$ , $\ dz=0.65$
201	cylinder, 4% in $r$ , 5% in $h$ : error in $V$	13%
203	$R_1=7$ , $R_2=6$ in parallel, ±0.05 Ω each	$\approx0.025\ \Omega$
205	pendulum, 0.5% in $L$ , 0.1% in $g$ : error in $T$	0.3%
207	$L$ for $x\sqrt y$ at $(1,4)$ (p361)	$2x+\tfrac14y-1$
209	$L$ for $\arctan(x+2y)$ at $(1,0)$	$\tfrac12x+y+\tfrac\pi4-\tfrac12$
211	$L$ for $\sqrt{x^2+y^2+z^2}$ at $(3,2,6)$	$\tfrac37x+\tfrac27y+\tfrac67z$
213	tangent plane, $z=\ln(10x^2+2y^2+1)$ at the origin	$z=0$

Level 3 · why it worksWhere the formulas come fromshort arguments

Where the formula comes from

Two tangent lines, crossed (p348):

\mathbf a=\mathbf j+f_y\,\mathbf k,\quad \mathbf b=\mathbf i+f_x\,\mathbf k,\qquad \mathbf a\times\mathbf b=f_x\,\mathbf i+f_y\,\mathbf j-\mathbf k

A plane through $(x_0,y_0,f(x_0,y_0))$ with that normal:

f_x\,(x-x_0)+f_y\,(y-y_0)-\big(z-f(x_0,y_0)\big)=0

Solve for $z$ and you have (4.24). The cross product of §2.4 turns two slopes into one normal.

Why the error vanishes

Example 4.24, $f=2x^2-4y$ at $(2,-3)$ :

E=f-L=2x^2-4y-(8x-4y-8)=2(x-2)^2

0\le\frac{E}{\sqrt{(x-2)^2+(y+3)^2}}\le\frac{2\big((x-2)^2+(y+3)^2\big)}{\sqrt{(x-2)^2+(y+3)^2}}=2\sqrt{(x-2)^2+(y+3)^2}\to0

A polynomial's error is made of squares and products of the steps: second order, so divided by the distance it still vanishes.

Why partials are not enough

For $f=\dfrac{xy}{\sqrt{x^2+y^2}}$ the candidate plane is $z=0$ , so $E=f$ and

\frac{E}{\sqrt{x^2+y^2}}=\frac{xy}{x^2+y^2}=\tfrac12\sin2\theta\quad\text{on the ray at angle }\theta

Different on different rays, so no limit (p354). Theorem 4.7 is not contradicted, because the partials are not continuous there:

f_x=\frac{y^3}{(x^2+y^2)^{3/2}},\qquad f_x(ky,\,y)=\frac{y^3}{|y|^3\,(k^2+1)^{3/2}}=\pm\frac{1}{(k^2+1)^{3/2}}

Along $x=ky$ the value depends on $k$ and on the sign of $y$ , so $f_x$ has no limit at the origin (p356).

Three variables, and what comes next

Three variables · (4.28) · p358

$f(x,y,z)=f(P)+f_x\,\Delta x+f_y\,\Delta y+f_z\,\Delta z+E$ , with $E/\text{dist}\to0$ . Same theorems.

§4.5: the chain rule is $dz=f_x\,dx+f_y\,dy$ divided by $dt$ . §4.6: $\langle f_x,f_y\rangle$ becomes the gradient, and the tangent plane to $F(x,y,z)=c$ has normal $\nabla F$ (Exercises 169–177).

Everything in this chapter is the plane in disguise.

Objective	You can now
4.4.1 tangent plane	$z=f+f_x\,(x-x_0)+f_y\,(y-y_0)$ : evaluate at the point, then substitute
4.4.2 approximate	$f(x,y)\approx L(x,y)$ near an easy point
4.4.3 differentiable	$E/\text{dist}\to0$ ; test with continuous partials; a crease fails
4.4.4 total differential	$\Delta z\approx dz=f_x\,dx+f_y\,dy$

Many tangent lines, one plane

The plane from two slopes

Zoom in: the surface turns flat

Examples 4.21–4.25

Linear approximation

Lab 2 · how good is the plane?

Differentiable

Lab 3 · partials, but no plane

Differentials

Lab 4 · Δz against dz

Common mistakes

Practice and answers

Where the formula comes from

Why the error vanishes

Why partials are not enough

Three variables, and what comes next

Tangent Planes and Linear Approximations

On a smooth surface, all tangent lines at a point lie in one plane

Two trace slopes fix the tangent plane

The normal ⟨fx,fy,−1⟩\langle f_x,f_y,-1\rangle⟨fx​,fy​,−1⟩ is the cross product of the two trace directions

Evaluate fff, fxf_xfx​, fyf_yfy​ at the point, then substitute: z=13x−26y−18z=13x-26y-18z=13x−26y−18

Checkpoint 4.19: the tangent plane at (−1,3)(-1,3)(−1,3) is z=7x+8y−3z=7x+8y-3z=7x+8y−3

With π\piπ in the point, leave the tangent plane in point-slope form

Both partial derivatives can exist with no tangent plane

Near the point, fff is almost its tangent plane: f(x,y)≈L(x,y)f(x,y)\approx L(x,y)f(x,y)≈L(x,y)

L(2.1,2.9)=3.875L(2.1,2.9)=3.875L(2.1,2.9)=3.875 against 14.95≈3.8665\sqrt{14.95}\approx3.866514.95​≈3.8665: about 0.2% off

A step that moves the exponent by −0.5 is not small: L=0.5L=0.5L=0.5 against e−0.5≈0.6065e^{-0.5}\approx0.6065e−0.5≈0.6065

Differentiable: the gap to the tangent plane shrinks faster than the distance

f=2x2−4yf=2x^2-4yf=2x2−4y is differentiable at (2,−3)(2,-3)(2,−3): E=2(x−2)2≤2 dist2E=2(x-2)^2\le2\,\text{dist}^2E=2(x−2)2≤2dist2

3x−4y23x-4y^23x−4y2 is differentiable at (−1,2)(-1,2)(−1,2): −4 dist≤E/dist≤0-4\,\text{dist}\le E/\text{dist}\le0−4dist≤E/dist≤0

The creased surface is not differentiable at the origin: E/distE/\text{dist}E/dist depends on the path

Continuous partials ⇒ differentiable ⇒ continuous

Differentiable ⇒ continuous, because every term of f−f(x0,y0)f-f(x_0,y_0)f−f(x0​,y0​) goes to 0

Continuous partials ⇒ differentiable: the Mean Value Theorem, once in each direction

The crease's fxf_xfx​ has no limit at the origin, so Theorem 4.7 cannot apply

dzdzdz is the rise along the tangent plane; Δz\Delta zΔz is the rise along the surface

dz=2.3dz=2.3dz=2.3 against Δz=2.3425\Delta z=2.3425Δz=2.3425: the differential is the linear part of the change

For 4y2+x2y−2xy4y^2+x^2y-2xy4y2+x2y−2xy at (1,−1)(1,-1)(1,−1): dz=0.18dz=0.18dz=0.18, Δz=0.180682\Delta z=0.180682Δz=0.180682

Three variables: one more slope, the same definition, the same theorems

Zoomed in, a differentiable surface is its tangent plane

The normal $\langle f_x,f_y,-1\rangle$ is the cross product of the two trace directions

Evaluate $f$ , $f_x$ , $f_y$ at the point, then substitute: $z=13x-26y-18$

Checkpoint 4.19: the tangent plane at $(-1,3)$ is $z=7x+8y-3$

With $\pi$ in the point, leave the tangent plane in point-slope form

Near the point, $f$ is almost its tangent plane: $f(x,y)\approx L(x,y)$

$L(2.1,2.9)=3.875$ against $\sqrt{14.95}\approx3.8665$ : about 0.2% off

A step that moves the exponent by −0.5 is not small: $L=0.5$ against $e^{-0.5}\approx0.6065$

$f=2x^2-4y$ is differentiable at $(2,-3)$ : $E=2(x-2)^2\le2\,\text{dist}^2$

$3x-4y^2$ is differentiable at $(-1,2)$ : $-4\,\text{dist}\le E/\text{dist}\le0$

The creased surface is not differentiable at the origin: $E/\text{dist}$ depends on the path

Differentiable ⇒ continuous, because every term of $f-f(x_0,y_0)$ goes to 0

The crease's $f_x$ has no limit at the origin, so Theorem 4.7 cannot apply

$dz$ is the rise along the tangent plane; $\Delta z$ is the rise along the surface

$dz=2.3$ against $\Delta z=2.3425$ : the differential is the linear part of the change

For $4y^2+x^2y-2xy$ at $(1,-1)$ : $dz=0.18$ , $\Delta z=0.180682$