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1. 1.Sketch · 8 min

   Three pictures, each with a justification. A drawing with nothing written beside it earns nothing: every feature — an included or excluded boundary, the shape of a curve, the direction of an arrow — must be backed by an equation or an inequality.

   1. (a)Find the domain of $f(x,y)=\frac{\sqrt{y-x}}{x^2+y^2-1}$ and sketch it, making clear which boundary points belong to it and which do not.
   2. (b)Sketch the level curves of $g(x,y)=\frac{y}{x^2+1}$ for $k=-1,0,1,2$ on one set of axes and describe the shape of each. Then, reading only your contour map, say in which direction from the origin $g$ increases fastest, and why.
   3. (c)Describe the level surfaces $F(x,y,z)=k$ of $F(x,y,z)=z-\sqrt{x^2+y^2}$ for $k=-1,0,1$. What single surface are they all copies of?
   Show the solution to question 1

   (a) Two restrictions, then intersect. The square root needs $y-x\ge 0$, that is $y\ge x$; the denominator needs $x^2+y^2\ne 1$. So

   $$D=\{(x,y):y\ge x,x^2+y^2\ne 1\}.$$

   Draw the line $y=x$ solid, since points on it satisfy $y-x=0\ge 0$ and belong to $D$, and shade the closed half-plane above and to the left of it (test point $(0,1)$: $1-0\ge 0$). Then remove the unit circle. Only the part of the circle inside the shaded region matters: the arc running counter-clockwise from $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ through $(0,1)$ and $(-1,0)$ to $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$. Its two endpoints are where $y=x$ meets $x^2+y^2=1$, so $2x^2=1$; they lie on the circle and are therefore excluded, even though they sit on the included line. Draw the arc dashed and its endpoints as open circles. The rest of the circle lies below the line and was never in the domain.

   (b) Solve $g=k$ for $y$. $g(x,y)=k$ means $y=k\left(x^2+1\right)$. For $k=0$ this is the $x$-axis. For $k=1$ it is the parabola $y=x^2+1$, vertex $(0,1)$, opening upward; for $k=2$ it is $y=2x^2+2$, vertex $(0,2)$, opening upward and narrower; for $k=-1$ it is $y=-x^2-1$, vertex $(0,-1)$, opening downward. No two of them meet, as level curves for different values never can.

   Read the spacing. Along the $y$-axis the four curves cross at $y=-1,0,1,2$, one unit apart. Along the line $x=1$ they cross at $y=-2,0,2,4$, two units apart, and along $x=2$ at $y=-5,0,5,10$. The contours are packed tightest on the $y$-axis and spread out as $|x|$ grows, so the surface is steepest above the $y$-axis and flattens away from it.

   From the origin, which lies on the curve $k=0$, the nearest point of the curve $k=1$ is its vertex $(0,1)$, at distance $1$: any other point $\left(x,x^2+1\right)$ of that curve is at distance $\sqrt{x^2+{\left(x^2+1\right)}^2}>1$. So the shortest walk that gains one unit of $g$ is straight up the positive $y$-axis, and that is the direction of fastest increase. (The gradient $\nabla g(0,0)=⟨0,1⟩$ agrees.)

   (c) Solve for $z$. $F=k$ means $z=\sqrt{x^2+y^2}+k$. For $k=0$ this is $z=\sqrt{x^2+y^2}$: the upper half of the cone $z^2=x^2+y^2$, vertex at the origin, opening upward, with its sides at $45^{\circ }$ to the axis (in the plane $y=0$ it is $z=|x|$); its horizontal cross-section at height $z=c>0$ is a circle of radius $c$. For $k=1$ the same half-cone is shifted up one unit, vertex $(0,0,1)$; for $k=-1$ it is shifted down, vertex $(0,0,-1)$. Every level surface is a vertical translate of the single half-cone $z=\sqrt{x^2+y^2}$, and since $F$ is defined on all of ${\mathbb{R}}^3$ with range all of $\mathbb{R}$, every real $k$ produces one.

2. 2.Computation · 7 min

   Evaluate each limit or show that it does not exist. To show a limit exists, bound $|f(x,y)-L|$ by an expression that visibly tends to $0$ (polar coordinates are welcome); to show it fails, exhibit two paths and quote the two-path test.

   1. (a)$\underset{(x,y)\to (0,0)}{\lim }\frac{x^3-y^3}{x^2+y^2}$
   2. (b)$\underset{(x,y)\to (0,0)}{\lim }\frac{x{y}^3}{x^2+y^6}$
   3. (c)$\underset{(x,y)\to (1,2)}{\lim }\frac{x^{2}y-2x^2-y+2}{x-1}$
   Show the solution to question 2

   (a) Bound it in polar coordinates. Substitution gives $\frac{0}{0}$. With $x=r\cos \theta$, $y=r\sin \theta$,

   $$\frac{x^3-y^3}{x^2+y^2}=\frac{r^3\left({\cos }^3\theta -{\sin }^3\theta \right)}{r^2}=r\left({\cos }^3\theta -{\sin }^3\theta \right).$$

   Since $|{\cos }^3\theta -{\sin }^3\theta |\le |\cos \theta {|}^3+|\sin \theta {|}^3\le 2$ for every $\theta$,

   $$\left|\frac{x^3-y^3}{x^2+y^2}-0\right|\le 2r,$$

   which tends to $0$ as $r\to 0^{+}$ regardless of $\theta$. The limit is $0$; in $\epsilon$–$\delta$ terms, $\delta =\epsilon /2$ works. Without polar coordinates: $\frac{x^2}{x^2+y^2}\le 1$ gives $\frac{|x{|}^3}{x^2+y^2}\le |x|$, likewise for $y$, so $|f(x,y)|\le |x|+|y|\to 0$.

   (b) Every line agrees; a cubic curve does not. Along $y=mx$,

   $$f(x,mx)=\frac{x\cdot m^3{x}^3}{x^2+m^6{x}^6}=\frac{m^3{x}^2}{1+m^6{x}^4}⟶0,$$

   and along the $y$-axis $f(0,y)=0$. Every straight line reports $0$. But along the curve $x=y^3$,

   $$f\left(y^3,y\right)=\frac{y^3\cdot y^3}{y^6+y^6}=\frac{1}{2}(y\ne 0).$$

   Two paths into the origin, limits $0$ and $\frac{1}{2}$: by the two-path test the limit does not exist. Polar coordinates would not have rescued this one — after substitution the expression still depends on $\theta$ in a way no bound in $r$ alone controls, which is itself the warning sign.

   (c) Factor and cancel. At $(1,2)$ the numerator is $2-2-2+2=0$ and the denominator is $0$: the form is $\frac{0}{0}$. Group the numerator:

   $$x^{2}y-2x^2-y+2=x^2(y-2)-(y-2)=\left(x^2-1\right)(y-2)=(x-1)(x+1)(y-2).$$

   The function is undefined on the whole line $x=1$, but a limit only looks at points of the domain, and at those points $x-1\ne 0$ may be cancelled:

   $$\frac{x^{2}y-2x^2-y+2}{x-1}=(x+1)(y-2)(x\ne 1).$$

   The right-hand side is a polynomial, hence continuous, so its limit is its value: $(1+1)(2-2)=0$.

3. 3.Computation · 7 min

   Compute the following. In each part state which variable is being held constant, and name any theorem that lets you choose the order of differentiation.

   1. (a)$f_x$ and $f_y$ for $f(x,y)=x^2\cos (xy)$.
   2. (b)Show that $u(x,y)=\ln \left(x^2+y^2\right)$ satisfies Laplace's equation $u_{xx}+u_{yy}=0$ at every point other than the origin.
   3. (c)$g_{xyy}$ for $g(x,y)=x\sin y+y^3{e}^{2x}+\frac{x^5}{1+e^x}$, choosing the order of differentiation that makes the work shortest and justifying the choice.
   Show the solution to question 3

   (a) Product rule with the chain rule inside. Holding $y$ constant, $x^2\cos (xy)$ is a product of $x^2$ and $\cos (xy)$, and $\frac{\partial }{\partial x}(xy)=y$:

   $$f_x=2x\cos (xy)+x^2\cdot (-\sin (xy))\cdot y=2x\cos (xy)-x^{2}y\sin (xy).$$

   Holding $x$ constant, $x^2$ is a constant factor and $\frac{\partial }{\partial y}(xy)=x$:

   $$f_y=x^2\cdot (-\sin (xy))\cdot x=-x^3\sin (xy).$$

   (b) Differentiate twice in each variable. With $y$ held constant, $u_x=\frac{2x}{x^2+y^2}$, and by the quotient rule

   $$u_{xx}=\frac{2\left(x^2+y^2\right)-2x\cdot 2x}{{\left(x^2+y^2\right)}^2}=\frac{2y^2-2x^2}{{\left(x^2+y^2\right)}^2}.$$

   The formula for $u$ is unchanged when $x$ and $y$ are swapped, so swapping them in $u_{xx}$ gives $u_{yy}=\frac{2x^2-2y^2}{{\left(x^2+y^2\right)}^2}$ (a direct computation from $u_y=\frac{2y}{x^2+y^2}$ confirms it). Adding,

   $$u_{xx}+u_{yy}=\frac{2y^2-2x^2+2x^2-2y^2}{{\left(x^2+y^2\right)}^2}=0.$$

   Every step needed $x^2+y^2\ne 0$, which is exactly the punctured plane on which $u$ is defined. So $u$ is harmonic there.

   (c) Choose the order with Clairaut's Theorem. Every partial derivative of $g$, of every order, is built from polynomials, $\sin y$, $e^{2x}$ and the quotient $\frac{x^5}{1+e^x}$ whose denominator never vanishes, so they are all continuous on ${\mathbb{R}}^2$. Clairaut's Theorem, applied twice, gives $g_{xyy}=g_{yxy}=g_{yyx}$: only the number of differentiations in each variable matters. Differentiate in $y$ first, because the term $\frac{x^5}{1+e^x}$ contains no $y$ and vanishes at the first stroke, whereas differentiating it in $x$ first would cost a quotient rule for nothing:

   $$g_y=x\cos y+3y^2{e}^{2x},g_{yy}=-x\sin y+6y{e}^{2x},g_{yyx}=-\sin y+12y{e}^{2x}.$$

   Hence $g_{xyy}=-\sin y+12y{e}^{2x}$. Check by the other order on the two surviving terms: $g_x=\sin y+2y^3{e}^{2x}+(\text{a function of }x\text{ alone})$, $g_{xy}=\cos y+6y^2{e}^{2x}$, $g_{xyy}=-\sin y+12y{e}^{2x}$, the same.

4. 4.Computation · 6 min

   Let $f(x,y)=x^2{e}^y$.

   1. (a)Find the tangent plane to $z=f(x,y)$ at the point $(2,0,4)$. State the hypothesis on $f$ that the tangent plane formula requires, and check that the point lies on your plane.
   2. (b)Use the linearization of $f$ at $(2,0)$ to approximate $f(1.9,0.05)$.
   3. (c)The inputs are measured as $x=2$ and $y=0$ with possible errors of $0.01$ and $0.02$ respectively. Use the total differential $dz$ to estimate the maximum error in the computed value $f(2,0)=4$, and the maximum relative error.
   Show the solution to question 4

   (a) Two partials, one formula. $f_x=2x{e}^y$ and $f_y=x^2{e}^y$ are continuous on all of ${\mathbb{R}}^2$, which is the hypothesis the tangent plane theorem needs. At $(2,0)$: $f=4$, $f_x=4$, $f_y=4$. The tangent plane is

   $$z-4=4(x-2)+4(y-0),\text{that is,}z=4x+4y-4.$$

   Check: at $(2,0)$ the plane gives $8+0-4=4$, the height of the point.

   (b) Evaluate the plane, not the function. The linearization is $L(x,y)=4x+4y-4$, so

   $$f(1.9,0.05)\approx L(1.9,0.05)=7.6+0.2-4=\mathrm{3.8.}$$

   (For reference, the true value is $3.61e^{0.05}\approx 3.795$; the step from the base point is small and the estimate is good to two decimals.)

   (c) Differential, worst case. $dz=f_{x}dx+f_{y}dy=2x{e}^{y}dx+x^2{e}^{y}dy$, which at $(2,0)$ is $dz=4dx+4dy$. With $|dx|\le 0.01$ and $|dy|\le 0.02$, the worst case is both errors with the same sign:

   $$|dz|\le 4(0.01)+4(0.02)=0.04+0.08=\mathrm{0.12.}$$

   The maximum relative error is $\frac{0.12}{4}=0.03$, that is $3\%$. The same number comes from dividing through: $\frac{dz}{z}=\frac{2dx}{x}+dy=2(0.005)+0.02=0.03$. Note that an absolute error in $y$ becomes a relative error in $f$, because $y$ sits in the exponent.

5. 5.Proof · 10 min

   Work from the $\epsilon$–$\delta$ definition of ${\lim }_{(x,y)\to (a,b)}f(x,y)=L$. Throughout, $r=\sqrt{(x-a{)}^2+(y-b{)}^2}$ denotes the distance from $(x,y)$ to $(a,b)$.

   1. (a)Prove that $\underset{(x,y)\to (0,0)}{\lim }\frac{x^4-y^4}{x^2+y^2}=0$, giving an explicit $\delta$ in terms of $\epsilon$.
   2. (b)Let $f(x,y)=\frac{x^4-y^4}{x^2+y^2}$ for $(x,y)\ne (0,0)$ and $f(0,0)=c$. Find every value of $c$ for which $f$ is continuous on all of ${\mathbb{R}}^2$, justifying continuity away from the origin by naming the facts you use.
   3. (c)Prove that if ${\lim }_{(x,y)\to (a,b)}f(x,y)=L$, then for every real $m$ the one-variable limit $\underset{x\to a}{\lim }f(x,b+m(x-a))$ exists and equals $L$. Then explain in one sentence why this is what makes the two-path test work.
   Show the solution to question 5

   (a) Cancel first, then estimate. For $(x,y)\ne (0,0)$ the denominator is nonzero and

   $$\frac{x^4-y^4}{x^2+y^2}=\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{x^2+y^2}=x^2-y^2,$$

   and $|x^2-y^2|\le x^2+y^2=r^2$.

   Proof. Let $\epsilon >0$ and set $\delta =\sqrt{\epsilon }$. If $0<r<\delta$, then

   $$\left|\frac{x^4-y^4}{x^2+y^2}-0\right|=|x^2-y^2|\le r^2<{\delta }^2=\epsilon .$$

   Since $\epsilon$ was arbitrary, the limit is $0$. $■$

   (b) Only $c=0$. Away from the origin $f$ is a rational function whose denominator does not vanish, and rational functions are continuous wherever they are defined (limits of sums, products and quotients are the sums, products and quotients of the limits, with $\lim x=a$ and $\lim y=b$ from the definition). At the origin, continuity means ${\lim }_{(x,y)\to (0,0)}f(x,y)=f(0,0)$; by (a) the left side is $0$, so continuity holds exactly when $c=0$. A second way to see it: with $c=0$ the function is $f(x,y)=x^2-y^2$ at every point, the origin included, and that is a polynomial.

   (c) Proof. Assume, as the statement implicitly does, that $f$ is defined at the points of the line near $(a,b)$. Let $\epsilon >0$. By hypothesis there is $\delta >0$ such that $|f(x,y)-L|<\epsilon$ whenever $0<r<\delta$. A point of the line has the form $(x,b+m(x-a))$, and its distance from $(a,b)$ is

   $$r=\sqrt{(x-a{)}^2+m^2(x-a{)}^2}=|x-a|\sqrt{1+m^2}.$$

   Put ${\delta }^{\prime }=\frac{\delta }{\sqrt{1+m^2}}>0$. If $0<|x-a|<{\delta }^{\prime }$, then $0<r=|x-a|\sqrt{1+m^2}<\delta$, so $\left|f\right(x,b+m(x-a))-L|<\epsilon$. That is precisely the $\epsilon$–$\delta$ statement of ${\lim }_{x\to a}f(x,b+m(x-a))=L$, with ${\delta }^{\prime }$ as the tolerance. $■$

   The two-path test is the contrapositive: if the limit existed, every line (and, by the same argument with the distance along any curve into $(a,b)$, every path) would have to report the same value $L$, so two paths reporting different values rule the limit out. Note the direction of the implication — the result never says that lines agreeing makes the limit exist.

6. 6.Critique · 9 min

   Let $f(x,y)=\frac{x^{2}y}{x^4+y^2}$ for $(x,y)\ne (0,0)$ and $f(0,0)=0$. A student writes:

   Step 1. $f(x,0)=0$ for every $x$, so $f_x(0,0)=\underset{h\to 0}{\lim }\frac{f(h,0)-f(0,0)}{h}=0$; likewise $f(0,y)=0$ for every $y$, so $f_y(0,0)=0$.

   Step 2. Both partial derivatives exist at the origin, so $f$ is differentiable at $(0,0)$.

   Step 3. Hence the tangent plane at the origin is $z=0$, and the linear approximation gives $f(x,y)\approx 0$ for all $(x,y)$ near $(0,0)$.

   Step 4. Since differentiability implies continuity, $f$ is continuous at $(0,0)$.

   1. (a)Identify the first step that is wrong. Name the theorem the student believes they are using, state its actual hypothesis, and explain why that hypothesis has not been established.
   2. (b)Prove that $f$ is not continuous at $(0,0)$, and deduce — naming the result you use — that $f$ is not differentiable there. Which of Steps 1–4 make true claims, whatever their justification?
   3. (c)Show directly from the definition of differentiability that no functions ${\epsilon }_1,{\epsilon }_2\to 0$ can make $\Delta z=f_x(0,0)\Delta x+f_y(0,0)\Delta y+{\epsilon }_1\Delta x+{\epsilon }_2\Delta y$ hold along the path $(\Delta x,\Delta y)=\left(t,t^2\right)$.
   Show the solution to question 6

   (a) Step 2 is the first error. Step 1 is correct: $f(h,0)=\frac{0}{h^4}=0$ and $f(0,k)=\frac{0}{k^2}=0$, so both difference quotients are identically $0$. Step 2 appeals to the theorem continuous partials imply differentiability: if $f_x$ and $f_y$ exist near $(a,b)$ and are continuous at $(a,b)$, then $f$ is differentiable at $(a,b)$. The hypothesis is continuity of the partial derivatives, not their existence, and Step 1 established only two values. Existence of $f_x(a,b)$ and $f_y(a,b)$ says nothing about differentiability, because the partials look along two lines only while differentiability is a statement about every direction at once.

   The hypothesis in fact fails here. For $(x,y)\ne (0,0)$ the quotient rule gives

   $$f_x=\frac{2xy\left(x^4+y^2\right)-x^{2}y\cdot 4x^3}{{\left(x^4+y^2\right)}^2}=\frac{2x{y}^3-2x^{5}y}{{\left(x^4+y^2\right)}^2},$$

   and along $y=2x^2$ this is $\frac{16x^7-4x^7}{25x^8}=\frac{12}{25x}$, which is unbounded as $x\to 0$. So $f_x$ is not continuous at the origin, and the theorem is silent. (Part (b) gives a shorter reason: were the partials continuous, $f$ would be differentiable, hence continuous, and it is not.)

   (b) Two paths. Along the $x$-axis, $f(x,0)=0\to 0$. Along the parabola $y=x^2$,

   $$f\left(x,x^2\right)=\frac{x^2\cdot x^2}{x^4+x^4}=\frac{1}{2}(x\ne 0),$$

   so $f\to \frac{1}{2}$ on that path. By the two-path test ${\lim }_{(x,y)\to (0,0)}f(x,y)$ does not exist, and continuity at the origin — which requires that limit to exist and equal $f(0,0)=0$ — fails. The corollary differentiability implies continuity, read in contrapositive, now says: not continuous, therefore not differentiable at $(0,0)$.

   So Step 2's claim is false, Step 4's claim is false, and Step 3's claim is false too: $f$ equals $\frac{1}{2}$ at points $\left(t,t^2\right)$ arbitrarily close to the origin, so “$f(x,y)\approx 0$ near $(0,0)$” is wrong — the plane $z=0$ misses the surface by $\frac{1}{2}$ no matter how close you zoom. Only Step 1 makes a true claim.

   (c) From the definition. With $(\Delta x,\Delta y)=\left(t,t^2\right)$ and $t\ne 0$,

   $$\Delta z=f\left(t,t^2\right)-f(0,0)=\frac{1}{2}-0=\frac{1}{2},$$

   so the required identity would read

   $$\frac{1}{2}=0\cdot t+0\cdot t^2+{\epsilon }_{1}t+{\epsilon }_2{t}^2.$$

   If ${\epsilon }_1,{\epsilon }_2\to 0$ as $(\Delta x,\Delta y)\to (0,0)$, then as $t\to 0$ the right-hand side tends to $0\cdot 0+0\cdot 0=0$, while the left-hand side is the constant $\frac{1}{2}$. Contradiction; no such ${\epsilon }_1,{\epsilon }_2$ exist, and $f$ is not differentiable at the origin. The error in the tangent-plane approximation along this path is $\frac{1}{2}$ — not small compared with the distance moved, not even small — which is exactly what differentiability forbids.

7. 7.Computation · 9 min

   Draw the tree diagram for each part before you differentiate, and write one term per root-to-leaf path.

   1. (a)Let $z=f(x,y)$ be differentiable with $f_x(2,3)=-1$ and $f_y(2,3)=4$, and let $x=t^2+1$, $y=3e^{t-1}$. Find $\frac{dz}{dt}$ at $t=1$.
   2. (b)Let $w=f(u,v)$ have continuous partial derivatives, where $u=x^2-y^2$ and $v=2xy$. Show that $${\left(\frac{\partial w}{\partial x}\right)}^2+{\left(\frac{\partial w}{\partial y}\right)}^2=4\left(x^2+y^2\right)\left[{\left(\frac{\partial f}{\partial u}\right)}^2+{\left(\frac{\partial f}{\partial v}\right)}^2\right].$$
   3. (c)The equation $z^3+xz-y^2=0$ defines $z$ implicitly as a function of $x$ and $y$ near the point $(3,2,1)$. Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ there, saying why the implicit differentiation formula applies, and write down the tangent plane to the surface at that point.
   Show the solution to question 7

   (a) Case 1 of the chain rule. The tree has $z$ over $x$ and $y$, each over $t$: two paths, two terms,

   $$\frac{dz}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}.$$

   At $t=1$: $x=1+1=2$ and $y=3e^0=3$, so the partials are to be evaluated at $(2,3)$, which is where they were given. Also $\frac{dx}{dt}=2t=2$ and $\frac{dy}{dt}=3e^{t-1}=3$. Hence

   $$\frac{dz}{dt}{|}_{t=1}=(-1)(2)+(4)(3)=10.$$

   (b) Case 2, then square and add. The four inner partials are $u_x=2x$, $u_y=-2y$, $v_x=2y$, $v_y=2x$. Case 2 gives

   $$w_x=f_u\cdot 2x+f_v\cdot 2y,w_y=f_u\cdot (-2y)+f_v\cdot 2x.$$

   Squaring,

   $$w_x^2=4x^2{f}_u^2+8xy{f}_u{f}_v+4y^2{f}_v^2,w_y^2=4y^2{f}_u^2-8xy{f}_u{f}_v+4x^2{f}_v^2.$$

   The cross terms cancel, and collecting the rest gives

   $$w_x^2+w_y^2=4\left(x^2+y^2\right)f_u^2+4\left(x^2+y^2\right)f_v^2=4\left(x^2+y^2\right)\left(f_u^2+f_v^2\right).$$

   (c) Implicit differentiation formula. Let $F(x,y,z)=z^3+xz-y^2$. Then $F(3,2,1)=1+3-4=0$, so the point is on the surface; $F$ is a polynomial, so its partials are continuous; and

   $$F_x=z,F_y=-2y,F_z=3z^2+x,$$

   which at $(3,2,1)$ are $1$, $-4$ and $6$. Since $F_z=6\ne 0$, the formulas apply:

   $$\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{1}{6},\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=\frac{4}{6}=\frac{2}{3}.$$

   The tangent plane to $z=z(x,y)$ at $(3,2,1)$ is

   $$z-1=-\frac{1}{6}(x-3)+\frac{2}{3}(y-2).$$

   Multiplying by $6$: $6z-6=-x+3+4y-8$, so $x-4y+6z=1$. Check: $3-8+6=1$, and the coefficients $⟨1,-4,6⟩$ are exactly $\nabla F(3,2,1)$, as they must be.

8. 8.Computation · 9 min

   Every direction below must be a unit vector before it is dotted with a gradient; say so each time.

   1. (a)Let $f(x,y,z)=x{e}^y-y{z}^2$ and $P=(1,0,2)$. Find the rate of change of $f$ at $P$ in the direction from $P$ toward $Q=(3,-1,4)$, the direction in which $f$ increases fastest at $P$, and that maximum rate.
   2. (b)A differentiable function $g(x,y)$ has, at a point $P$, directional derivative $3$ in the direction $\mathbf{u}=⟨\frac{3}{5},\frac{4}{5}⟩$ and directional derivative $-1$ in the direction $\mathbf{v}=⟨\frac{4}{5},-\frac{3}{5}⟩$. Find $\nabla g(P)$, the maximum rate of change of $g$ at $P$, and the two unit directions in which the rate of change is $0$.
   3. (c)Find the tangent plane and the normal line to the surface $xyz=6$ at $(1,2,3)$, naming the theorem that supplies the normal vector.
   Show the solution to question 8

   (a) Gradient, then a unit vector. $\nabla f=⟨e^y,x{e}^y-z^2,-2yz⟩$, so $\nabla f(P)=⟨1,1-4,0⟩=⟨1,-3,0⟩$. The vector from $P$ to $Q$ is $⟨2,-1,2⟩$, of length $\sqrt{4+1+4}=3$, so the unit direction is $\mathbf{u}=⟨\frac{2}{3},-\frac{1}{3},\frac{2}{3}⟩$ and

   $$D_{\mathbf{u}}f(P)=⟨1,-3,0⟩\cdot ⟨\frac{2}{3},-\frac{1}{3},\frac{2}{3}⟩=\frac{2}{3}+1+0=\frac{5}{3}.$$

   By the theorem on maximising the directional derivative, $f$ increases fastest in the direction of $\nabla f(P)=⟨1,-3,0⟩$ — as a unit vector, $\frac{1}{\sqrt{10}}⟨1,-3,0⟩$ — and the maximum rate is $|\nabla f(P)|=\sqrt{10}$. Sanity check: $\frac{5}{3}<\sqrt{10}$, as every directional derivative must be.

   (b) Two known rates pin down the gradient. Both $\mathbf{u}$ and $\mathbf{v}$ are unit vectors ($\frac{9}{25}+\frac{16}{25}=1$) and they are perpendicular ($\frac{12}{25}-\frac{12}{25}=0$), so they form an orthonormal basis of the plane. Write $\nabla g(P)=\alpha \mathbf{u}+\beta \mathbf{v}$; then $D_{\mathbf{u}}g=\nabla g\cdot \mathbf{u}=\alpha$ and $D_{\mathbf{v}}g=\beta$, so $\alpha =3$, $\beta =-1$ and

   $$\nabla g(P)=3⟨\frac{3}{5},\frac{4}{5}⟩-⟨\frac{4}{5},-\frac{3}{5}⟩=⟨\frac{9}{5}-\frac{4}{5},\frac{12}{5}+\frac{3}{5}⟩=⟨1,3⟩.$$

   (Alternatively, solve the linear system $\frac{3}{5}a+\frac{4}{5}b=3$, $\frac{4}{5}a-\frac{3}{5}b=-1$ for $\nabla g=⟨a,b⟩$: multiplying by $3$ and $4$ and adding gives $5a=5$, then $b=3$.) Check: $⟨1,3⟩\cdot \mathbf{u}=\frac{3}{5}+\frac{12}{5}=3$ and $⟨1,3⟩\cdot \mathbf{v}=\frac{4}{5}-\frac{9}{5}=-1$.

   The maximum rate of change is $|\nabla g(P)|=\sqrt{10}$, in the direction $\frac{1}{\sqrt{10}}⟨1,3⟩$. The rate is $0$ exactly when $\mathbf{w}\perp \nabla g(P)$, i.e. along the level curve: $\mathbf{w}=\pm \frac{1}{\sqrt{10}}⟨3,-1⟩$, since $⟨1,3⟩\cdot ⟨3,-1⟩=0$.

   (c) The gradient is normal to level sets. The surface is the level set $F=6$ of $F(x,y,z)=xyz$, and by the theorem that the gradient is normal to level sets, $\nabla F(1,2,3)$ is a normal vector, provided it is nonzero:

   $$\nabla F=⟨yz,xz,xy⟩,\nabla F(1,2,3)=⟨6,3,2⟩\ne 0.$$

   Tangent plane: $6(x-1)+3(y-2)+2(z-3)=0$, that is $6x+3y+2z=18$; check $6+6+6=18$. Normal line, through $(1,2,3)$ in the direction $⟨6,3,2⟩$:

   $$\frac{x-1}{6}=\frac{y-2}{3}=\frac{z-3}{2},\text{or}x=1+6t,y=2+3t,z=3+2t.$$

9. 9.Computation · 9 min

   Let $f(x,y)=x^{2}y-2xy+y^2-3y$.

   1. (a)Find all critical points of $f$. Show your case analysis — a factored equation has more than one solution branch, and each must be followed to the end.
   2. (b)Classify each critical point with the Second Derivatives Test, stating the value of $D$ at each one and the hypothesis the test requires.
   3. (c)Is the local minimum value you found the absolute minimum of $f$ on ${\mathbb{R}}^2$? Prove your answer.
   Show the solution to question 9

   (a) Factor $f_x$, then split into cases. $f$ is a polynomial, so both partials exist everywhere and the critical points are exactly the simultaneous zeros of

   $$f_x=2xy-2y=2y(x-1),f_y=x^2-2x+2y-3.$$

   $f_x=0$ forces $y=0$ or $x=1$.

   Case $y=0$. Then $f_y=x^2-2x-3=(x-3)(x+1)=0$, so $x=3$ or $x=-1$: the points $(3,0)$ and $(-1,0)$.

   Case $x=1$. Then $f_y=1-2+2y-3=2y-4=0$, so $y=2$: the point $(1,2)$.

   Three critical points: $(3,0)$, $(-1,0)$, $(1,2)$.

   (b) Second Derivatives Test. $f_{xx}=2y$, $f_{xy}=2x-2$, $f_{yy}=2$, all polynomials, hence continuous on every disc — the hypothesis of the test. So

   $$D=f_{xx}{f}_{yy}-f_{xy}^2=4y-4(x-1{)}^2.$$

   At $(3,0)$: $D=0-16=-16<0$, a saddle point. At $(-1,0)$: $D=0-16=-16<0$, a saddle point. At $(1,2)$: $D=8-0=8>0$ and $f_{xx}=4>0$, a local minimum, with value

   $$f(1,2)=2-4+4-6=-4.$$

   (c) No — $f$ is unbounded below. Restrict $f$ to the line $y=-1$:

   $$f(x,-1)=-x^2+2x+1+3=-x^2+2x+4⟶-\infty \text{as }x\to \infty ;$$

   for instance $f(10,-1)=-100+20+4=-76<-4$. So no value is an absolute minimum, and $-4$ is a local minimum only. (Likewise $f(0,y)=y^2-3y\to \infty$, so there is no absolute maximum either.) There is no contradiction with the test: critical points are where a local extremum can be, and on an unbounded domain nothing — no Extreme Value Theorem — promises that an absolute extremum exists at all.

10. 10.Computation · 11 min

    Find the absolute maximum and minimum values of $f(x,y)=x^2-y^2+2y$ on the closed disc $x^2+y^2\le 4$, and the points where they occur. Before you begin, say why you know they exist.

    1. (a)Find and classify the critical points of $f$ in the open disc, and explain why the value at a critical point classified as a saddle can be neither the absolute maximum nor the absolute minimum.
    2. (b)Reduce $f$ on the boundary circle to a function of one variable on a closed interval — by parametrising, or by substituting $x^2=4-y^2$ — and find its extreme values, endpoints included.
    3. (c)Combine to state the absolute extrema. Then explain what changes if the region is the open disc $x^2+y^2<4$.
    Show the solution to question 10

    Existence. $f$ is a polynomial, hence continuous, and the closed disc is closed (it contains its boundary circle) and bounded (it lies inside the disc of radius $2$). By the Extreme Value Theorem for two variables, $f$ attains an absolute maximum and an absolute minimum on it.

    (a) Interior. $f_x=2x$ and $f_y=-2y+2$ vanish together only at $(0,1)$, which is inside the disc since $0+1<4$; there $f(0,1)=-1+2=1$. The second partials $f_{xx}=2$, $f_{yy}=-2$, $f_{xy}=0$ are continuous and give $D=(2)(-2)-0=-4<0$: a saddle point. An absolute extremum attained at an interior point is in particular a local extremum there, and a saddle point is neither a local maximum nor a local minimum; so the value $1$ is out, and both absolute extrema must lie on the boundary circle.

    (b) Boundary. On the circle $x^2=4-y^2$ with $-2\le y\le 2$, and because $f$ depends on $x$ only through $x^2$, the substitution is legitimate:

    $$h(y)=\left(4-y^2\right)-y^2+2y=-2y^2+2y+4,-2\le y\le 2.$$

    Closed-interval method: $h^{\prime }(y)=-4y+2=0$ at $y=\frac{1}{2}$, and

    $$h\left(\frac{1}{2}\right)=-\frac{1}{2}+1+4=\frac{9}{2},h(2)=-8+4+4=0,h(-2)=-8-4+4=-8.$$

    Translate back to points: $y=\frac{1}{2}$ gives $x^2=4-\frac{1}{4}=\frac{15}{4}$, so $x=\pm \frac{\sqrt{15}}{2}$; $y=2$ gives $(0,2)$; $y=-2$ gives $(0,-2)$. (Parametrising instead, $x=2\cos t$, $y=2\sin t$ gives $g(t)=4-8{\sin }^{2}t+4\sin t$, the same quadratic in $s=\sin t$, with its maximum at $\sin t=\frac{1}{4}$, i.e. $y=\frac{1}{2}$ — the substitution route simply avoids an ugly angle.)

    Check one directly: $f\left(\frac{\sqrt{15}}{2},\frac{1}{2}\right)=\frac{15}{4}-\frac{1}{4}+1=\frac{14}{4}+1=\frac{9}{2}$.

    (c) Compare. The candidates are $1$ (interior saddle, already excluded), $\frac{9}{2}$, $0$ and $-8$. The absolute maximum is $\frac{9}{2}$, at the two points $\left(\pm \frac{\sqrt{15}}{2},\frac{1}{2}\right)$; the absolute minimum is $-8$, at $(0,-2)$.

    On the open disc the set is not closed, so the Extreme Value Theorem no longer applies, and in fact both extrema disappear: the only interior critical point is a saddle, so $f$ has no local extremum anywhere in the open disc, hence no absolute one. The values $\frac{9}{2}$ and $-8$ are approached — $f(0,y)=-y^2+2y\to -8$ as $y\to -2$ from inside — but never attained. The open disc has a supremum and an infimum of $f$, not a maximum and a minimum.

11. 11.Computation · 11 min

    Use Lagrange multipliers. In each part say why the extreme values exist before you look for them, and treat every case that arises when you eliminate $\lambda$ — a division by a quantity that might be $0$ is a case split, not a step.

    1. (a)Find the maximum and minimum values of $f(x,y)=x^{2}y$ on the ellipse $x^2+2y^2=6$, and the points where they occur.
    2. (b)Find the highest and lowest points on the curve in which the cylinder $x^2+y^2=1$ meets the plane $x+y+z=1$ — that is, the extreme values of $z$ on that curve.
    Show the solution to question 11

    (a) Existence, gradient check, then cases. The ellipse is closed and bounded and $f$ is continuous, so by the Extreme Value Theorem both extremes exist. With $g(x,y)=x^2+2y^2$, $\nabla g=⟨2x,4y⟩$ vanishes only at the origin, which is not on the ellipse, so the method applies everywhere on it. Since $\nabla f=⟨2xy,x^2⟩$, the equations $\nabla f=\lambda \nabla g$ and $g=6$ are

    $$2xy=2\lambda x(1),x^2=4\lambda y(2),x^2+2y^2=6(3).$$

    Equation (1) says $x(y-\lambda )=0$, so either $x=0$ or $\lambda =y$.

    Case $x=0$. Then (3) gives $2y^2=6$, $y=\pm \sqrt{3}$, and (2) reads $0=4\lambda y$, so $\lambda =0$, consistent. Candidates $(0,\pm \sqrt{3})$, where $f=0$.

    Case $x\ne 0$, so $\lambda =y$. Then (2) gives $x^2=4y^2$, and (3) becomes $4y^2+2y^2=6$, so $y=\pm 1$ and $x=\pm 2$: four candidates $(\pm 2,1)$ and $(\pm 2,-1)$, with

    $$f(\pm 2,1)=4,f(\pm 2,-1)=-4.$$

    Comparing $0,4,-4$: the maximum is $4$ at $(2,1)$ and $(-2,1)$, the minimum is $-4$ at $(2,-1)$ and $(-2,-1)$. Dividing (1) through by $x$ would have silently discarded the first case; here those points happen not to be extremes, but nothing guaranteed that in advance.

    (b) Two constraints. The curve is the intersection of the cylinder with a non-vertical plane, an ellipse: it is bounded, since $x^2+y^2=1$ bounds $x$ and $y$ and then $z=1-x-y$ is bounded too, and closed, so $z$ attains a maximum and a minimum on it. Take $f(x,y,z)=z$, $g=x^2+y^2$, $h=x+y+z$:

    $$\nabla f=⟨0,0,1⟩,\nabla g=⟨2x,2y,0⟩,\nabla h=⟨1,1,1⟩.$$

    These last two are never parallel on the curve (a multiple of $\nabla h$ has nonzero third component, while $\nabla g$ has third component $0$ and is itself nonzero on the cylinder), so the two-constraint condition $\nabla f=\lambda \nabla g+\mu \nabla h$ holds at each extremum:

    $$0=2\lambda x+\mu ,0=2\lambda y+\mu ,1=\mu .$$

    So $\mu =1$ and $2\lambda x=-1=2\lambda y$. This forces $\lambda \ne 0$, and then $x=y$. The cylinder gives $2x^2=1$, $x=y=\pm \frac{1}{\sqrt{2}}$, and the plane gives $z=1-x-y=1\mp \sqrt{2}$. The two candidates are

    $$\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1-\sqrt{2}\right)\text{and}\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},1+\sqrt{2}\right),$$

    so the lowest point has $z=1-\sqrt{2}$ and the highest has $z=1+\sqrt{2}$. Check by parametrising: $x=\cos t$, $y=\sin t$ gives $z=1-\cos t-\sin t=1-\sqrt{2}\sin \left(t+\frac{\pi }{4}\right)$, whose range is exactly $\left[1-\sqrt{2},1+\sqrt{2}\right]$.

12. 12.Proof · 12 min

    Let $k>0$ and let $T$ be the triangle

    $$T=\{(x,y,z):x+y+z=k,x\ge 0,y\ge 0,z\ge 0\}.$$

    This question uses Lagrange multipliers to prove an inequality, so the logic — why a maximum exists, where it can be, and why the candidate found is the maximum — matters as much as the algebra.

    1. (a)Explain why $f(x,y,z)=xyz$ attains a maximum on $T$, and why that maximum is attained at a point with $x$, $y$ and $z$ all strictly positive.
    2. (b)Use the method of Lagrange multipliers to show that the maximum is ${\left(\frac{k}{3}\right)}^3$, attained only at $x=y=z=\frac{k}{3}$.
    3. (c)Deduce the arithmetic–geometric mean inequality $\sqrt[3]{abc}\le \frac{a+b+c}{3}$ for all $a,b,c\ge 0$, with equality if and only if $a=b=c$.
    4. (d)Explain why the same argument proves nothing if the sign conditions are dropped and $T$ is replaced by the whole plane $x+y+z=k$.
    Show the solution to question 12

    (a) Extreme Value Theorem, then push off the edges. $T$ is bounded: each coordinate is at least $0$ and, since the other two are at least $0$ and the three sum to $k$, at most $k$; so $T$ lies in a ball of radius $k\sqrt{3}$. $T$ is closed: it is cut out by an equation and non-strict inequalities in continuous functions, so any limit of points of $T$ satisfies the same conditions and is in $T$. And $f$ is a polynomial, hence continuous. By the Extreme Value Theorem, $f$ attains an absolute maximum $M$ at some point of $T$.

    On the edges of $T$ — the points where at least one coordinate is $0$ — the product $xyz$ is $0$. But the point $\left(\frac{k}{3},\frac{k}{3},\frac{k}{3}\right)$ belongs to $T$ and gives $f=\frac{k^3}{27}>0$. Hence $M\ge \frac{k^3}{27}>0$, and a maximiser cannot lie on an edge: at the maximiser, $x,y,z$ are all strictly positive.

    (b) Lagrange at an interior point. Near a point of $T$ with all coordinates positive, the constraint set is the smooth level surface $g(x,y,z)=x+y+z=k$, with $\nabla g=⟨1,1,1⟩\ne 0$. So at the maximiser $\nabla f=\lambda \nabla g$, that is,

    $$yz=\lambda ,xz=\lambda ,xy=\lambda ,x+y+z=k.$$

    From $yz=xz$ and $z>0$ we get $x=y$; from $xz=xy$ and $x>0$ we get $z=y$. (These divisions are legal precisely because (a) put us in the interior.) So $x=y=z$, and the constraint gives $3x=k$, $x=\frac{k}{3}$. There is exactly one candidate. Since a maximiser exists and must be a candidate, the maximum is

    $$M=f\left(\frac{k}{3},\frac{k}{3},\frac{k}{3}\right)={\left(\frac{k}{3}\right)}^3,$$

    and it is attained only there: any other maximiser would be an interior point solving the same system, hence the same point. $■$

    (c) The inequality. Let $a,b,c\ge 0$. If $a+b+c=0$ then $a=b=c=0$, both sides are $0$, and equality holds with $a=b=c$. Otherwise put $k=a+b+c>0$; then $(a,b,c)\in T$, so by (b)

    $$abc\le {\left(\frac{k}{3}\right)}^3={\left(\frac{a+b+c}{3}\right)}^3,$$

    and taking cube roots, which is an increasing operation on $[0,\infty )$, gives $\sqrt[3]{abc}\le \frac{a+b+c}{3}$. For equality: if $\sqrt[3]{abc}=\frac{a+b+c}{3}$ then $abc={\left(\frac{k}{3}\right)}^3$, so $(a,b,c)$ is a maximiser of $f$ on $T$, and by (b) $a=b=c=\frac{k}{3}$. Conversely, if $a=b=c$ both sides equal $a$. $■$

    (d) Without the sign conditions there is no maximum to find. On the whole plane $x+y+z=k$ the function is unbounded in both directions: along $(t,t,k-2t)$,

    $$f=t^2(k-2t)⟶-\infty \text{ as }t\to +\infty ,f⟶+\infty \text{ as }t\to -\infty .$$

    The plane is unbounded, so the Extreme Value Theorem does not apply, and indeed neither a maximum nor a minimum exists. The Lagrange system is unchanged and still produces the candidate $\left(\frac{k}{3},\frac{k}{3},\frac{k}{3}\right)$ — a local maximum on the plane, but not an absolute one, since $f$ exceeds $\frac{k^3}{27}$ far out. The method produces candidates only; the existence argument in (a) is what turned the candidate into a theorem.