Exercises — Directional derivative — definition, formula

What we check: the formula demands a unit vector — a direction arrow of length exactly $1$. $\Vert \mathbf{v}\Vert =\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\ne 1.$ So $\mathbf{v}$ is not valid as-is. Normalise (divide by its length): $\mathbf{u}=\frac{\mathbf{v}}{\Vert \mathbf{v}\Vert }=\left(\frac{3}{5},\frac{4}{5}\right).$ Check: $\sqrt{(3/5{)}^2+(4/5{)}^2}=\sqrt{9/25+16/25}=\sqrt{25/25}=1.$ ✓

What a gradient is: the list of partial derivatives — one slope per axis (see Gradient vector). $f_x=\frac{\partial }{\partial x}(x^{3}y-y^2)=3x^{2}y,f_y=\frac{\partial }{\partial y}(x^{3}y-y^2)=x^3-2y.$ So $\nabla f=(3x^{2}y,x^3-2y)$. At $(1,-1)$: $\nabla f(1,-1)=(3(1)(-1),1-2(-1))=(-3,3).$

False. $D_{\mathbf{u}}f=\nabla f\cdot \mathbf{u}$ is a scalar (a dot product of two vectors is a single number — see Dot product and cosine of angle). The vector pointing most uphill is $\nabla f$ itself, and the steepest slope is the number $\Vert \nabla f\Vert$.

G — Gradient. $f_x=2x+3y$, $f_y=3x+2y$, so $\nabla f=(2x+3y,3x+2y)$. $\nabla f(1,1)=(2+3,3+2)=(5,5).$ U — Unit direction. Direction vector $=(4,5)-(1,1)=(3,4)$, length $\sqrt{3^2+4^2}=5$, so $\mathbf{u}=(3/5,4/5)$. N — Number (dot product). $D_{\mathbf{u}}f=(5,5)\cdot \left(\frac{3}{5},\frac{4}{5}\right)=\frac{15+20}{5}=\frac{35}{5}=7.$

Why an angle gives a unit vector for free: $(\cos \alpha ,\sin \alpha )$ always has length $1$, since ${\cos }^2\alpha +{\sin }^2\alpha =1$. So $\mathbf{u}=(\cos 60^{\circ },\sin 60^{\circ })=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$. Gradient. $f_x=e^x\cos y$, $f_y=-e^x\sin y$. At $(0,0)$: $e^0=1$, $\cos 0=1$, $\sin 0=0$: $\nabla f(0,0)=(1,0).$ Dot product. $D_{\mathbf{u}}f=(1,0)\cdot \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\frac{1}{2}.$ The $y$-rate is $0$ here, so only the horizontal lean of the walk matters.

Normalise: $\Vert (1,2,2)\Vert =\sqrt{1+4+4}=\sqrt{9}=3$, so $\mathbf{u}=\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)$. Gradient: $\nabla f=(yz,xz,xy)$. At $(1,2,3)$: $\nabla f=(2\cdot 3,1\cdot 3,1\cdot 2)=(6,3,2)$. Dot product: $D_{\mathbf{u}}f=(6,3,2)\cdot \left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)=\frac{6}{3}+\frac{6}{3}+\frac{4}{3}=2+2+\frac{4}{3}=\frac{16}{3}\approx \mathrm{5.33.}$

Gradient: $\nabla f=(2x,2y)=(6,8)$, with $\Vert \nabla f\Vert =\sqrt{6^2+8^2}=\sqrt{100}=10$. (a) Steepest ascent is along $\nabla f$: $\mathbf{u}=\frac{1}{10}(6,8)=(0.6,0.8)$, rate $=\Vert \nabla f\Vert =10$. (b) Steepest descent is opposite: rate $=-\Vert \nabla f\Vert =-10$.

(c) Zero rate happens when $\mathbf{u}\perp \nabla f$ — walking along a level curve, where $f$ is constant. We need a direction perpendicular to $(6,8)$. How to build a perpendicular systematically (the "$90^{\circ }$ turn" rule): two vectors are perpendicular exactly when their dot product is $0$. If $\nabla f=(p,q)$, try $(-q,p)$: $(p,q)\cdot (-q,p)=-pq+qp=0.✓$ So swap the two components and negate one of them — that always turns a vector by a right angle. This is why the rule $(p,q)\mapsto (-q,p)$ works; it isn't magic, it's forced by the dot product being zero. With $\nabla f=(6,8)$: perpendicular $(-8,6)$, length $\sqrt{64+36}=10$, so ${\mathbf{u}}_1=(-0.8,0.6).$ The second choice (edge case). A right-angle turn can go clockwise or counter-clockwise, so there are two opposite unit directions with zero slope. The other one is just $-{\mathbf{u}}_1$ (equivalently the rule $(p,q)\mapsto (q,-p)$): ${\mathbf{u}}_2=(0.8,-0.6).$ Both check out: $(6,8)\cdot (-0.8,0.6)=-4.8+4.8=0$ and $(6,8)\cdot (0.8,-0.6)=4.8-4.8=0.$ ✓ Geometrically these are the two tangent directions to the level circle at $\mathbf{a}$ — same line, opposite arrows.

The figure below shows all four vectors at once: the yellow gradient (ascent), the red opposite (descent), and the two green tangents (zero slope) sitting on the level circle.

Reading the figure. The blue circle is the level curve $x^2+y^2=25$. The yellow arrow $\nabla f=(6,8)$ points straight out of the circle — steepest ascent, rate $10$. The red arrow is its exact reverse — steepest descent, rate $-10$. The two green arrows lie along the circle (tangent), $90^{\circ }$ from yellow; walking either way keeps you on the same contour, so the slope is $0$. Notice green is perpendicular to yellow — that is the geometric meaning of "swap-and-negate."

Gradient: $\nabla f=(2x,-2y)=(4,-2)$. Along $\mathbf{u}$: $(4,-2)\cdot \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=\frac{4-2}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\approx 1.414>0$ — the surface rises this way. Along $\mathbf{w}$: $(4,-2)\cdot \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)=\frac{4+2}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}\approx 4.243>0$ — rises faster. Interpretation: both are positive, but $\mathbf{w}$ leans more toward $\nabla f=(4,-2)$ (which points down-and-right), so it climbs harder. The sign tells you up/down; the size tells you how steep.

Why cosine enters: the dot product satisfies $\nabla f\cdot \mathbf{u}=\Vert \nabla f\Vert \Vert \mathbf{u}\Vert \cos \theta$, and $\Vert \mathbf{u}\Vert =1$ (see Dot product and cosine of angle). So $D_{\mathbf{u}}f=\Vert \nabla f\Vert \cos \theta$ — a knob turned by the angle only. Here $\Vert \nabla f\Vert =\sqrt{3^2+4^2}=5$, so $\cos \theta =\frac{D_{\mathbf{u}}f}{\Vert \nabla f\Vert }=\frac{2.5}{5}=0.5\Rightarrow \theta =60^{\circ }.$

Gradient: $\nabla f=(2x+y,x)=(4,1)$, so $\Vert \nabla f\Vert =\sqrt{16+1}=\sqrt{17}$. Half of steepest: we need $D_{\mathbf{u}}f=\frac{1}{2}\Vert \nabla f\Vert$. Since $D_{\mathbf{u}}f=\Vert \nabla f\Vert \cos \theta$: $\Vert \nabla f\Vert \cos \theta =\frac{1}{2}\Vert \nabla f\Vert \Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =60^{\circ }.$ Building a heading $60^{\circ }$ from the gradient (why the rotation works). Any unit vector in the plane can be written as a mix of two perpendicular unit vectors: $\hat{\mathbf{n}}$ pointing along the gradient, and ${\hat{\mathbf{n}}}^{\perp }$ pointing $90^{\circ }$ across it. If we take $\mathbf{u}=\cos \theta \hat{\mathbf{n}}+\sin \theta {\hat{\mathbf{n}}}^{\perp },$ then $\mathbf{u}$ is automatically a unit vector (since $\hat{\mathbf{n}}\perp {\hat{\mathbf{n}}}^{\perp }$ gives $\Vert \mathbf{u}{\Vert }^2={\cos }^2\theta +{\sin }^2\theta =1$), and it makes angle $\theta$ with $\hat{\mathbf{n}}$ because $\mathbf{u}\cdot \hat{\mathbf{n}}=\cos \theta$. That is exactly a rotation by $\theta$ — it's how the rotation matrix is built, one perpendicular axis at a time. Here $\hat{\mathbf{n}}=\frac{1}{\sqrt{17}}(4,1)$, and using the swap-and-negate rule from Problem 3.1, ${\hat{\mathbf{n}}}^{\perp }=\frac{1}{\sqrt{17}}(-1,4)$. With $\theta =60^{\circ }$: $\mathbf{u}=\frac{1}{2}\hat{\mathbf{n}}+\frac{\sqrt{3}}{2}{\hat{\mathbf{n}}}^{\perp }.$ Numerically $\hat{\mathbf{n}}\approx (0.970,0.243)$, ${\hat{\mathbf{n}}}^{\perp }\approx (-0.243,0.970)$, so $\mathbf{u}\approx 0.5(0.970,0.243)+0.866(-0.243,0.970)\approx (0.274,0.962).$ Check the slope: $(4,1)\cdot (0.274,0.962)=1.096+0.962=2.058\approx \frac{\sqrt{17}}{2}=\mathrm{2.06.}$ ✓ (Choosing $-\sin \theta$ instead gives the other valid heading — rotating the other way.)

Key link: the level curve through $(3,4)$ is the circle $x^2+y^2=25$. The gradient $\nabla f=(6,8)$ is perpendicular to that circle. A tangent direction is therefore perpendicular to $(6,8)$. Perp unit vector (swap-and-negate, then normalise): $(-8,6)/10=(-0.8,0.6)$. $D_{\mathbf{u}}f=(6,8)\cdot (-0.8,0.6)=-4.8+4.8=0.$ Why zero makes sense: along a level curve $f$ is constant, so its rate of change is $0$ — no matter which of the two tangent directions you pick.

Idea: the directional derivative along an axis equals that partial derivative (dotting the gradient with $(1,0)$ picks out $f_x$). $D_{(1,0)}f=f_x=2x=10\Rightarrow x=5,D_{(0,1)}f=f_y=2y=6\Rightarrow y=3.$ So $\mathbf{a}=(5,3)$, and $\nabla f=(10,6)$, giving $\Vert \nabla f\Vert =\sqrt{10^2+6^2}=\sqrt{136}=2\sqrt{34}\approx \mathrm{11.66.}$

$\nabla f=(2x,2y)=(0,0)$ at the origin. Then for any unit $\mathbf{u}$: $D_{\mathbf{u}}f=(0,0)\cdot (u_1,u_2)=0.$ Geometry: the origin is the bottom of the bowl (a minimum). Every direction is momentarily flat — the surface is level to first order in all directions. This is the degenerate case where "steepest ascent" has no preferred direction because there is no ascent at all instantaneously.

Direct limit. Take unit $\mathbf{u}=(a,b)$, so $a^2+b^2=1$. Along the ray $(ha,hb)$: $\frac{f(ha,hb)-f(0,0)}{h}=\frac{1}{h}\cdot \frac{(ha{)}^2(hb)}{(ha{)}^2+(hb{)}^2}=\frac{1}{h}\cdot \frac{h^3{a}^{2}b}{h^2(a^2+b^2)}=\frac{a^{2}b}{a^2+b^2}=a^{2}b,$ using $a^2+b^2=1$. So $D_{\mathbf{u}}f(0,0)=a^{2}b$ — exists for every direction. The partials. Put $\mathbf{u}=(1,0)$: $f_x(0,0)=1^2\cdot 0=0$. Put $\mathbf{u}=(0,1)$: $f_y(0,0)=0^2\cdot 1=0$. So $\nabla f(0,0)=(0,0)$, and the formula would predict $\nabla f\cdot \mathbf{u}=(0,0)\cdot (a,b)=0\text{for all directions.}$ But the true value is $a^{2}b$, which is nonzero for e.g. $\mathbf{u}=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$: there $a^{2}b=\frac{1}{2}\cdot \frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\approx 0.354\ne 0$. Moral: the dot-product shortcut requires $f$ to be differentiable at the point (see Tangent planes and differentiability). Here $f$ is not, so the formula fails even though all directional derivatives exist.

Tool: the cosine form of the dot product (see Dot product and cosine of angle): $D_{\mathbf{u}}f=\nabla f\cdot \mathbf{u}=\Vert \nabla f\Vert \Vert \mathbf{u}\Vert \cos \theta =\Vert \nabla f\Vert \cos \theta ,$ since $\Vert \mathbf{u}\Vert =1$. Here $\theta$ is the angle between $\mathbf{u}$ and $\nabla f$. Why this settles it: $\Vert \nabla f\Vert$ is a fixed positive number, so $D_{\mathbf{u}}f$ is largest exactly when $\cos \theta$ is largest. And $\cos \theta \le 1$ with equality iff $\theta =0$, i.e. $\mathbf{u}$ points the same way as $\nabla f$. That direction is $\mathbf{u}=\nabla f/\Vert \nabla f\Vert$, and then $D_{\mathbf{u}}f=\Vert \nabla f\Vert \cdot 1=\Vert \nabla f\Vert .■$ Similarly the minimum is $\theta =\pi$, giving $-\Vert \nabla f\Vert$ (steepest descent).