4.4 Tangent Planes and Linear Approximations

← Chapter 4 §4.4 Tangent Planes and Linear Approximations 切平面与线性近似 · Chapter 4 · Differentiation of Functions of Several Variables

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Objective	Takeaways; say them first, then click
4.4.1 Determine the equation of a plane tangent to a given surface at a point.	all tangent lines lie in one plane; $z=f+f_x\,(x-x_0)+f_y\,(y-y_0)$ : evaluate at the point, then substitute
4.4.2 Use the tangent plane to approximate a function of two variables at a point.	$f(x,y)\approx L(x,y)$ near an easy point
4.4.3 Explain when a function of two variables is differentiable.	partials alone are not enough; $E/\text{dist}\to0$ ; continuous partials ⇒ differentiable ⇒ continuous
4.4.4 Use the total differential to approximate the change in a function of two variables.	$\Delta z\approx dz=f_x\,dx+f_y\,dy$

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	$dV/V=2\,dr/r+dh/h=8\%+5\%=13\%$

#	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 $(1,-1)$ , $dx=0.03$ , $dy=-0.02$	$dz=0.18$ ; $\ \Delta z=0.180682$

#	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	$z=0$

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 $\langle f_x,f_y,-1\rangle$ 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 $e^{-0.5}\approx0.6065$

Zoomed in, a smooth surface is its tangent plane

The plane's error grows like the distance squared

Both partial derivatives can exist with no tangent plane

Partials exist, yet the slope depends on the direction

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

$f=2x^2-4y$ 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

Continuous partials ⇒ differentiable ⇒ continuous

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

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

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

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

$dz$ 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 $(1,-1)$ : $dz=0.18$ , $\Delta z=0.180682$

Halve the step: $dz$ halves, $\Delta z-dz$ drops to a quarter

Zoomed in, a differentiable surface is its tangent plane

Common mistakes

Practice and answers

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

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