4.6 Directional Derivatives and the Gradient

← Chapter 4 §4.6 Directional Derivatives and the Gradient 方向导数与梯度 · Chapter 4 · Differentiation of Functions of Several Variables

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Access for free at openstax.org · CC BY-NC-SA 4.0

Objective	Takeaways; say them first, then click
4.6.1 Determine the directional derivative in a given direction for a function of two variables.	slope depends on direction; tangent slope along a unit $\mathbf u$ ; $f_x\cos\theta+f_y\sin\theta$
4.6.2 Determine the gradient vector of a given real-valued function.	$\nabla f=f_x\,\mathbf i+f_y\,\mathbf j$ ; $D_{\mathbf u}f=\nabla f\cdot\mathbf u$ , $\mathbf u$ normalized
4.6.3 Explain the significance of the gradient vector with regard to direction of change along a surface.	steepest ascent along $\nabla f$ at rate $\\|\nabla f\\|$ ; descent along $-\nabla f$ ; crowded contours = steep
4.6.4 Use the gradient to find the tangent to a level curve of a given function.	$\nabla f$ is normal to the level curve; tangent $\langle -b,a\rangle$
4.6.5 Calculate directional derivatives and gradients in three dimensions.	$\langle f_x,f_y,f_z\rangle$ ; direction cosines; $D_{\mathbf u}f=\nabla f\cdot\mathbf u$

Mistake	Result	Fix
Dotting with $\langle-5,12\rangle$ itself	176 at $(-1,2)$ , not $\tfrac{176}{13}\approx13.54$	divide by the length first
Giving $\nabla f$ as the maximum rate	" $\langle-24,20\rangle$ " in Example 4.34	the rate is a number: $\\|\nabla f\\|=4\sqrt{61}$
$\theta=\arcsin\tfrac{5}{\sqrt{61}}$ alone	0.69 rad: the wrong quadrant	$\cos\theta<0$ , so $\theta=\pi-0.69\approx2.45$
Swapping components without a sign change	$\langle18,-9\rangle$ is not tangent in Example 4.35	$\langle-18,-9\rangle$ : its dot with $\nabla f$ is 0
Calling $\langle-9,18\rangle$ the tangent	a vector across the level curve	$\nabla f$ is the normal; the tangent is perpendicular to it
Reading ∇f as a vector on the surface	an arrow in space	∇f of $f(x,y)$ lives in the $xy$ -plane

#	task	answer
4.28	$3x^2y-4xy^3+3y^2-4x$ , $\theta=\tfrac\pi3$ , at $(3,4)$	$-94-\tfrac{525\sqrt3}{2}\approx-548.66$
4.29	$\nabla\dfrac{x^2-3y^2}{2x+y}$	$\dfrac{2x^2+2xy+6y^2}{(2x+y)^2}\mathbf i-\dfrac{x^2+12xy+3y^2}{(2x+y)^2}\mathbf j$
4.30	$4x-xy+2y^2$ at $(-2,3)$ : steepest way	$\nabla g=\mathbf i+14\mathbf j$ ; max $\sqrt{197}$ at $\theta\approx1.499$ rad
4.31	$x^2-2xy+5y^2+3x-2y+4$ at $(1,1)$	$\nabla f=3\mathbf i+6\mathbf j$ ; tangent $6\mathbf i-3\mathbf j$ (level 9*)
4.32	$\nabla\dfrac{x^2-3y^2+z^2}{2x+y-4z}$	$\dfrac{\langle 2x^2+2xy+6y^2-8xz-2z^2,\ -(x^2+12xy+3y^2-24yz+z^2),\ 4x^2-12y^2-4z^2+4xz+2yz\rangle}{(2x+y-4z)^2}$
4.33	$3x^2+xy-2y^2+4yz-z^2+2xz$ along $\langle-3,12,-4\rangle$ , at $(0,-2,5)$	$\tfrac{384}{13}$

#	task	answer	key step
261	$y^2\cos2x$ at $(\tfrac\pi3,2)$ , $\theta=\tfrac\pi4$	$-2\sqrt6-\sqrt2$	$\nabla f=\langle-4\sqrt3,-2\rangle$
263	$xy$ at $(0,-2)$	$-1$	$\nabla f=\langle-2,0\rangle$
265	$xyz$ at $(2,1,1)$ along $\langle2,1,-1\rangle$	$\tfrac{2}{\sqrt6}$	normalize
267	$x^2-y^2$ at $(1,0)$	$\sqrt3$	$\nabla f=\langle2,0\rangle$
269	$e^x\cos y$ at $(0,\tfrac\pi2)$ , $\mathbf u=\langle0,1\rangle$	$-1$	just $f_y$
271	$\ln(x^2+y^2)$ at $(1,2)$	$\tfrac{22}{25}$	$\nabla f=\langle\tfrac25,\tfrac45\rangle$
273	$y^2+xz$ at $(1,2,2)$ along $\langle2,-1,2\rangle$	$\tfrac23$	$\\|\mathbf v\\|=3$
275	$\tfrac{y}{x+2y}$ , $\theta=-\tfrac\pi4$	$-\dfrac{\sqrt2(x+y)}{2(x+2y)^2}$	quotient rule
277	$ye^x$ , $\theta=\tfrac\pi3$	$\tfrac{e^x(y+\sqrt3)}{2}$	—
279	$\ln(x+2y)$ , $\theta=\tfrac\pi3$	$\dfrac{1+2\sqrt3}{2(x+2y)}$	—
281	$xy+yz+xz$ at $(1,2,3)$	$\langle5,4,3\rangle$	$\langle y+z,x+z,x+y\rangle$
283	$4x^5y^2z^3$ at $(2,-1,1)$	$-320$	$\nabla f=\langle320,-256,384\rangle$
285	$\tfrac{y}{x+z}$ , $P(2,1,-1)$ toward $Q(-1,2,0)$	$\tfrac{3}{\sqrt{11}}$	$\overrightarrow{PQ}=\langle-3,1,1\rangle$
287	$\ln(5x+4y)$ at $(3,9)$ along $6\mathbf i+8\mathbf j$	$\tfrac{31}{255}$	$\mathbf u=\langle\tfrac35,\tfrac45\rangle$
289	level curve of $x^2+4y^2$ through $(-2,0)$	the ellipse $x^2+4y^2=4$ ; $\nabla f=-4\mathbf i$	field lab
291	$xe^y-\ln x$ at $(-3,0)$	$\tfrac43\mathbf i-3\mathbf j$	as printed; $\ln x$ needs $x>0$
293	$x\sqrt{y^2+z^2}$ at $(-2,-1,-1)$	$\sqrt2\,\mathbf i+\sqrt2\,\mathbf j+\sqrt2\,\mathbf k$	—
295	$e^{xy}$ at $(6,7)$ , fastest rate	$\approx1.6\times10^{19}$	$e^{42}\sqrt{85}$
297	$\ln(xy+yz+zx)$ at $(-9,-18,-27)$	$\tfrac{5\sqrt2}{99}$	$\\|\nabla f\\|$
299	$xe^{-y}$ at $(1,0)$	$\sqrt2$ along $\langle1,-1\rangle$	—
301	$\cos(3x+2y)$ at $(\tfrac\pi6,-\tfrac\pi8)$	$\sqrt{13/2}$ along $\langle-3,-2\rangle$	$3x+2y=\tfrac\pi4$
303	$xy+yz+xz=3$ at $(1,1,1)$	plane $x+y+z=3$ ; line $x-1=y-1=z-1$	$\nabla f=\langle2,2,2\rangle$
305	$xe^y\cos z-z=1$ at $(1,0,0)$	plane $x+y-z=1$ ; line $x-1=y=-z$	$\nabla f=\langle1,1,-1\rangle$
307	$V=5x^2-3xy+xyz$ at $(3,4,5)$	$\tfrac{32}{\sqrt3}$ ; $\langle38,6,12\rangle$ ; $2\sqrt{406}$	(a) along $\langle1,1,-1\rangle$
309	$\varphi=\sin\pi x\sin2\pi y$	$\langle\pi\cos\pi x\sin2\pi y,\ 2\pi\sin\pi x\cos2\pi y\rangle$	$\langle u,v\rangle=\nabla\varphi$

Directional Derivatives and the Gradient

On a surface the slope depends on which way you walk

The directional derivative is the tangent slope along a unit vector $\mathbf u$

From the limit: $D_{\mathbf u}f=\frac{2x+21y}{5}$ , so $D_{\mathbf u}f(-1,2)=8$

Split the step into east and north: $D_{\mathbf u}f=f_x\cos\theta+f_y\sin\theta$

The chain rule turns the limit into $f_x\cos\theta+f_y\sin\theta$

With partial derivatives, Example 4.31's answer 8 takes two lines

The gradient packs both partials, and $D_{\mathbf u}f=\nabla f\cdot\mathbf u$

Given a non-unit direction, divide by its length before taking the dot product

A gradient is the two partial derivatives as its $\mathbf i$ and $\mathbf j$ components

The slope along $\mathbf u$ is $\|\nabla f\|\cos\varphi$ : largest along $\nabla f$ , smallest against it

The steepest slope at $(-2,3)$ is $4\sqrt{61}\approx31.24$

Turn $\mathbf u$ : the slope follows a cosine and peaks along $\nabla f$

On a contour map the gradient points uphill and is longest where contours crowd

The climb is steepest when $\mathbf u$ lies over $\nabla f$

Always walking along $\nabla f$ traces a path of steepest ascent to the top

The gradient is perpendicular to the level curve through the point

Along a level curve $f$ is constant, so $\nabla f\cdot\mathbf r'(t)=0$

Reverse the gradient's components and negate one: that vector is tangent to the level curve

Everywhere on the map, $\nabla f$ crosses its level curve at a right angle

In three variables the gradient gains a $\mathbf k$ -component

Three partials, one per component: Example 4.36

Move the point over a level surface: $\nabla f$ always sticks straight out of it

In space the gradient is normal to the level surface, so it gives the tangent plane

In space the direction cosines make up $\mathbf u$ , and still $D_{\mathbf u}f=\nabla f\cdot\mathbf u$

Normalize, evaluate $\nabla f$ , take the dot product: $D_{\mathbf u}f(1,-2,3)=-\tfrac{25}{3}$

The gradient is the uphill arrow; every directional slope is its shadow

Common mistakes

Practice and answers

Directional Derivatives and the Gradient

On a surface the slope depends on which way you walk

The directional derivative is the tangent slope along a unit vector u\mathbf uu

From the limit: Duf=2x+21y5D_{\mathbf u}f=\frac{2x+21y}{5}Du​f=52x+21y​, so Duf(−1,2)=8D_{\mathbf u}f(-1,2)=8Du​f(−1,2)=8

Split the step into east and north: Duf=fxcos⁡θ+fysin⁡θD_{\mathbf u}f=f_x\cos\theta+f_y\sin\thetaDu​f=fx​cosθ+fy​sinθ

The chain rule turns the limit into fxcos⁡θ+fysin⁡θf_x\cos\theta+f_y\sin\thetafx​cosθ+fy​sinθ

With partial derivatives, Example 4.31's answer 8 takes two lines

The gradient packs both partials, and Duf=∇f⋅uD_{\mathbf u}f=\nabla f\cdot\mathbf uDu​f=∇f⋅u

Given a non-unit direction, divide by its length before taking the dot product

A gradient is the two partial derivatives as its i\mathbf ii and j\mathbf jj components

The slope along u\mathbf uu is ∥∇f∥cos⁡φ\|\nabla f\|\cos\varphi∥∇f∥cosφ: largest along ∇f\nabla f∇f, smallest against it

The steepest slope at (−2,3)(-2,3)(−2,3) is 461≈31.244\sqrt{61}\approx31.24461​≈31.24

Turn u\mathbf uu: the slope follows a cosine and peaks along ∇f\nabla f∇f