MATH 213 — Exam II Review

Exam II Review — Partial Derivatives and Double Integrals
MATH 213  Calculus III  Stewart 9e, §14.3–14.7 and §15.1–15.2

How to use this note. This is a consolidation, not new material. The whole of Chapter 14 is one idea repeated: freeze every variable but one, and you are back in Calculus I. Everything — partial derivatives, the chain rule, directional derivatives, the second derivative test — is bookkeeping on top of that. Chapter 15 then reverses the move: a double integral is two ordinary integrals, done one variable at a time. Each section below gives what you must know cold, the move that uses it, and the error that costs the most points.

What is covered.

§	Topic	What you must be able to do
14.3	Partial derivatives	Compute $f_x, f_y$fx​,fy​; second and mixed partials; Clairaut
14.4	Tangent planes	Write the plane; use it as a linear approximation
14.5	Chain rule	$dz/dt$dz/dt along a path; implicit differentiation
14.6	Gradient	Directional derivative; direction and rate of fastest increase
14.7	Optimization	Critical points; the $D$D-test; extrema on a closed bounded region
15.1–15.2	Double integrals	Over rectangles and over general regions; choosing the order

Not on this exam: §14.1–14.2 (limits and continuity — those were Exam 1) and §15.3 (polar coordinates — we cover that today, and it belongs to Exam 3). No problem on Wednesday needs polar.

Partial derivatives (§14.3)

Know cold. $f_{x}$fx​ means: treat $y$y as a constant and differentiate in $x$x. Notation you must read fluently:

$f_{xy} = \frac{\partial^{2}f}{\partial y\,\partial x} = \frac{\partial}{\partial y}\left(f_{x}\right).$fxy​=∂y∂x∂2f​=∂y∂​(fx​).

Clairaut: if the mixed partials are continuous, $f_{xy} = f_{yx}$fxy​=fyx​.

The move. Cover the other variable with your finger. If you see $x^{2}y^{3}$x2y3 and are computing $f_{x}$fx​, you are differentiating $(\text{constant})\cdot x^{2}$(constant)⋅x2.

Trap. Mixed versus pure. $f_{xy}$fxy​ differentiates once in each variable; $f_{xx}$fxx​ differentiates twice in $x$x. A problem that says “verify $f_{xy}=f_{yx}$fxy​=fyx​” and one that says “verify $f_{xx}+f_{yy}=0$fxx​+fyy​=0” are asking for different computations, and confusing them costs the whole part.

Trap. “Verify” means compute both sides separately and compare. Citing Clairaut to avoid computing the second one answers a different question.

Example 1 (Reading the subscripts). $f(x,y) = x^{3}y^{2} - y^{4}$f(x,y)=x3y2−y4. Then $f_{x} = 3x^{2}y^{2}$fx​=3x2y2 and $f_{y} = 2x^{3}y - 4y^{3}$fy​=2x3y−4y3, so

$f_{xy} = 6x^{2}y = f_{yx}, \qquad\text{while}\qquad f_{xx} = 6xy^{2}, \quad f_{yy} = 2x^{3}-12y^{2}.$fxy​=6x2y=fyx​,whilefxx​=6xy2,fyy​=2x3−12y2.

Four different second partials, two of which happen to coincide.

Tangent planes and linear approximation (§14.4)

Know cold. For $z=f(x,y)$z=f(x,y) at $(a,b)$(a,b):

$z = f(a,b) + f_{x}(a,b)\,(x-a) + f_{y}(a,b)\,(y-b).$z=f(a,b)+fx​(a,b)(x−a)+fy​(a,b)(y−b).

That same expression, called $L(x,y)$L(x,y), is the linearization: for $(x,y)$(x,y) near $(a,b)$(a,b), $f(x,y) \approx L(x,y)$f(x,y)≈L(x,y).

For a surface given implicitly as $F(x,y,z) = k$F(x,y,z)=k, the vector $\nabla F$∇F is normal to the surface, and the tangent plane at $P_{0}$P0​ is $F_{x}(x-x_{0})+F_{y}(y-y_{0})+F_{z}(z-z_{0}) = 0$Fx​(x−x0​)+Fy​(y−y0​)+Fz​(z−z0​)=0.

Trap. Three evaluations, all at the old point $(a,b)$(a,b): $f(a,b)$f(a,b), $f_{x}(a,b)$fx​(a,b), $f_{y}(a,b)$fy​(a,b). The new point enters only as $(x,y)$(x,y) at the very end. Plugging the new point into $f$f defeats the purpose — that is the exact value you were asked to approximate.

Trap. A surface given as $F(x,y,z)=k$F(x,y,z)=k is not of the form $z=f(x,y)$z=f(x,y). Use the gradient, or solve for $z$z first — but do not mix the two formulas.

Example 2 (Plane, then estimate). $f(x,y)=\sqrt{x^{2}+y^{2}}$f(x,y)=x2+y2​ at $(3,4)$(3,4): $f(3,4)=5$f(3,4)=5, $f_{x}=x/\sqrt{x^{2}+y^{2}} = 3/5$fx​=x/x2+y2​=3/5, and $f_{y}=4/5$fy​=4/5. So $L(x,y) = 5 + \tfrac35(x-3)+\tfrac45(y-4)$L(x,y)=5+53​(x−3)+54​(y−4), and $f(3.02,3.99) \approx 5 + \tfrac35(0.02)+\tfrac45(-0.01) = 5.004$f(3.02,3.99)≈5+53​(0.02)+54​(−0.01)=5.004.

The chain rule and the gradient (§14.5–14.6)

Know cold. Along a path $x=x(t)$x=x(t), $y=y(t)$y=y(t):

$\frac{dz}{dt} = f_{x}\frac{dx}{dt} + f_{y}\frac{dy}{dt} = \nabla f \cdot \mathbf{r}'(t).$dtdz​=fx​dtdx​+fy​dtdy​=∇f⋅r′(t).

In a direction given by a unit vector $\mathbf{u}$u:

$D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u}.$Du​f=∇f⋅u.

$f$f increases fastest in the direction of $\nabla f$∇f, at rate $\left\lvert \nabla f \right\rvert$∣∇f∣; fastest decrease is $-\nabla f$−∇f at rate $-\left\lvert \nabla f \right\rvert$−∣∇f∣; and $D_{\mathbf{u}}f = 0$Du​f=0 exactly along the level curve.

Trap. $\mathbf{u}$u must be a unit vector. Given a direction like $\langle 3,4\rangle$⟨3,4⟩, divide by $5$5 first. Forgetting to normalise scales your whole answer.

Trap. The direction of fastest increase is $\nabla f$∇f at that point — a vector of numbers, not a formula in $x$x and $y$y. Evaluate before you report.

Example 3 (The two questions are different). $f(x,y)=x^{2}y$f(x,y)=x2y at $(1,2)$(1,2): $\nabla f = \langle 2xy,\ x^{2}\rangle = \langle 4,1\rangle$∇f=⟨2xy, x2⟩=⟨4,1⟩. In the direction $\langle 3,4 \rangle$⟨3,4⟩: $\mathbf{u} = \langle \tfrac35,\tfrac45\rangle$u=⟨53​,54​⟩ and $D_{\mathbf{u}}f = \tfrac{12}{5}+\tfrac45 = \tfrac{16}{5}$Du​f=512​+54​=516​. The maximum rate in any direction is $\left\lvert \langle 4,1\rangle \right\rvert = \sqrt{17} \approx 4.12$∣⟨4,1⟩∣=17​≈4.12, which is larger — as it must be.

Optimization (§14.7)

Know cold. Critical points solve $f_{x}=0$fx​=0 and $f_{y}=0$fy​=0. With $D = f_{xx}f_{yy}-\left(f_{xy}\right)^{2}$D=fxx​fyy​−(fxy​)2 evaluated at the critical point:

$D>0$D>0 and $f_{xx}>0$fxx​>0	local minimum
$D>0$D>0 and $f_{xx}<0$fxx​<0	local maximum
$D<0$D<0	saddle point
$D=0$D=0	no information — argue directly

On a closed and bounded region, the Extreme Value Theorem guarantees both absolute extrema exist. Find them by comparing: (i) values at interior critical points, (ii) values on the boundary. Then simply pick the largest and smallest. You do not classify anything.

Trap. $D=0$D=0 does not mean “saddle.” It means the test failed and you must reason about $f$f directly.

Trap. On a closed region the extrema are often on the boundary, where $\nabla f \neq 0$∇f=0. A solution that only hunts for critical points will miss them entirely.

Trap. To handle a boundary, reduce it to one variable — substitute the constraint, or parametrise it — and then it is a Calculus I max/min problem on a closed interval, endpoints included.

Example 4 (Boundary matters). $f(x,y)=x+y$f(x,y)=x+y on the closed disk $x^{2}+y^{2}\le 1$x2+y2≤1 has $\nabla f = \langle 1,1\rangle$∇f=⟨1,1⟩, never zero: no interior critical point at all. Yet extrema exist. On the boundary, substituting $y = \pm\sqrt{1-x^{2}}$y=±1−x2​ (or $x=\cos t$x=cost, $y=\sin t$y=sint, giving $f=\cos t+\sin t$f=cost+sint) yields maximum $\sqrt2$2​ and minimum $-\sqrt2$−2​.

Double integrals (§15.1–15.2)

Know cold. Over a rectangle, Fubini says the order does not matter. Over a general region you must first decide the type:

$\text{Type I:}\quad a\le x\le b,\ g_{1}(x)\le y\le g_{2}(x) \qquad\Longrightarrow\qquad \int_{a}^{b}\!\!\int_{g_{1}(x)}^{g_{2}(x)} f\,dy\,dx,$Type I:a≤x≤b, g1​(x)≤y≤g2​(x)⟹∫ab​∫g1​(x)g2​(x)​fdydx, $\text{Type II:}\quad c\le y\le d,\ h_{1}(y)\le x\le h_{2}(y) \qquad\Longrightarrow\qquad \int_{c}^{d}\!\!\int_{h_{1}(y)}^{h_{2}(y)} f\,dx\,dy.$Type II:c≤y≤d, h1​(y)≤x≤h2​(y)⟹∫cd​∫h1​(y)h2​(y)​fdxdy.

The move. Draw the region first — always. Then slice it: vertical strips give Type I, horizontal strips give Type II. The outer limits are always constants; the inner limits may involve the outer variable, never the reverse.

Trap. Inner limits containing the inner variable. $\int_0^1\!\int_0^y f\,dy\,dx$∫01​∫0y​fdydx is meaningless — the $y$y in the limit has nowhere to come from. If you write this, your region description is wrong.

Trap. Reversing the order is not swapping the limits. You must redescribe the region from the picture. $\int_0^1\!\int_x^1 \to \int_0^1\!\int_0^y$∫01​∫x1​→∫01​∫0y​, not $\int_x^1\!\int_0^1$∫x1​∫01​.

Trap. Sometimes only one order is possible, because the other produces an integral with no elementary antiderivative — $e^{y^{2}}$ey2, $\sin(y^{2})$sin(y2), $\frac{\sin y}{y}$ysiny​ are the usual suspects. If the inner integral looks impossible, that is a signal, not a dead end: reverse.

Example 5 (The sketch decides). $\displaystyle\int_{0}^{4}\!\!\int_{\sqrt{x}}^{2} \sin\!\left(y^{3}\right) dy\,dx$∫04​∫x​2​sin(y3)dydx cannot be started as written. The region is $0\le x\le 4$0≤x≤4, $\sqrt x \le y \le 2$x​≤y≤2 — i.e. the area between $y=\sqrt x$y=x​ and $y=2$y=2. Redescribing with horizontal strips: $0\le y\le 2$0≤y≤2 and $0 \le x \le y^{2}$0≤x≤y2. Then

$\int_{0}^{2}\!\!\int_{0}^{y^{2}} \sin\!\left(y^{3}\right) dx\,dy = \int_{0}^{2} y^{2}\sin\!\left(y^{3}\right)dy = \frac{1-\cos 8}{3}.$∫02​∫0y2​sin(y3)dxdy=∫02​y2sin(y3)dy=31−cos8​.

Practice problems

Deliberately not shaped like the exam — the point is to stretch the same skills over different ground. Work them without notes first.

A. Say it exactly

1. State the tangent-plane formula, the chain rule for $dz/dt$dz/dt, the formula for $D_{\mathbf{u}}f$Du​f, and the second derivative test — all four from memory.

2. Explain in one sentence each: why $\mathbf{u}$u must be a unit vector; why $D=0$D=0 gives no information; why the outer limits of a double integral must be constants.

B. Find the error

1. “$f(x,y)=x^{2}y^{3}$f(x,y)=x2y3, so $f_{xx} = \frac{\partial}{\partial y}\left(2xy^{3}\right) = 6xy^{2}$fxx​=∂y∂​(2xy3)=6xy2.”

2. “$D_{\mathbf{u}}f$Du​f at $(1,1)$(1,1) in the direction $\langle 1,1 \rangle$⟨1,1⟩ equals $\nabla f(1,1)\cdot\langle 1,1\rangle$∇f(1,1)⋅⟨1,1⟩.”

3. “$D = 0$D=0 at the critical point, so it is a saddle point.”

4. “Reversing the order: $\int_{0}^{2}\!\int_{0}^{x} f\,dy\,dx = \int_{0}^{x}\!\int_{0}^{2} f\,dx\,dy$∫02​∫0x​fdydx=∫0x​∫02​fdxdy.”

C. Partial derivatives

1. For $f(x,y) = x^{2}\ln\!\left(x+y\right)$f(x,y)=x2ln(x+y), find $f_{x}$fx​, $f_{y}$fy​, $f_{xy}$fxy​ and $f_{yx}$fyx​.

2. Show that $u(x,y) = \ln\!\sqrt{x^{2}+y^{2}}$u(x,y)=lnx2+y2​ satisfies $u_{xx}+u_{yy}=0$uxx​+uyy​=0.

3. For $f(x,y)=xe^{y}+y e^{x}$f(x,y)=xey+yex, verify Clairaut by computing both mixed partials separately.

D. Planes, chains, gradients

1. Find the tangent plane to $z = x^{2}+4y^{2}$z=x2+4y2 at $(1,1,5)$(1,1,5), and use it to estimate the value at $(1.02, 0.97)$(1.02,0.97).

2. Find the tangent plane to the sphere $x^{2}+y^{2}+z^{2}=14$x2+y2+z2=14 at $(1,2,3)$(1,2,3).

3. If $z = x^{2}y$z=x2y with $x = 2t+1$x=2t+1 and $y = t^{3}$y=t3, find $dz/dt$dz/dt at $t=1$t=1 two ways: by the chain rule, and by substituting first.

4. For $f(x,y)=x e^{y}$f(x,y)=xey at $(2,0)$(2,0), find $D_{\mathbf{u}}f$Du​f in the direction $\langle -3, 4 \rangle$⟨−3,4⟩, and the maximum rate of increase.

E. Optimization

1. Find and classify all critical points of $f(x,y) = x^{2}+y^{2}-4x-6y+13$f(x,y)=x2+y2−4x−6y+13.

2. Find and classify all critical points of $f(x,y) = xy - x^{2} - y^{2} - 2x - 2y$f(x,y)=xy−x2−y2−2x−2y.

3. Find the absolute extrema of $f(x,y)=x^{2}+2y^{2}$f(x,y)=x2+2y2 on the closed disk $x^{2}+y^{2}\le 4$x2+y2≤4.

4. Find the absolute extrema of $f(x,y)=x+2y$f(x,y)=x+2y on the closed triangle with vertices $(0,0)$(0,0), $(2,0)$(2,0), $(0,3)$(0,3).  (Three boundary segments — do each separately, endpoints included.)

F. Double integrals

1. Evaluate $\displaystyle\int_{1}^{3}\!\!\int_{0}^{2}\left(x+y^{2}\right)dy\,dx$∫13​∫02​(x+y2)dydx, then again in the other order, and confirm the two agree.

2. Sketch the region and evaluate $\displaystyle\iint_{D} 2xy \,dA$∬D​2xydA where $D$D is bounded by $y=x^{2}$y=x2 and $y=2x$y=2x.

3. Sketch the region, then reverse the order of integration in $\displaystyle\int_{0}^{1}\!\!\int_{y}^{1} f(x,y)\,dx\,dy$∫01​∫y1​f(x,y)dxdy. (Do not evaluate.)

4. Evaluate $\displaystyle\int_{0}^{1}\!\!\int_{3y}^{3} e^{x^{2}}\,dx\,dy$∫01​∫3y3​ex2dxdy.  (Only one order is possible — decide which, and why, before computing.)