7 concepts19 worked examples28 exercises4 exam-level6 figures
How often this comes up in exams: 79 questions, 36/39 of the years
all years (1988-2026)Midterm 2 46 · Final 24 · Midterm 1 9
last ten yearsFinal 10 · Midterm 2 7 · Midterm 1 4
also touched by22 questions that are filed under another section but still use this one
The questions in this section name themselves clearly in the archive.
Counted from the department's own exam archive: 106 exams, 599 question items. On an independent sample of 60 questions read by hand, the topic labels were 83% right, so treat these counts as a direction, not a certainty.
What are you here for?
10 Directional derivatives, the gradient vector, and maximum and minimum values
Start with this
One question before you read anything. Getting it wrong is the point: it shows you what this section is for.
§10.0 — partials assemble into a gradient
A warm-up on the one tool everything here is built from: the partial derivatives you met earlier this term.
Find(a) Find $f_x$ and $f_y$, then evaluate both at $(2,1)$.
Given
$f(x,y)=x^{2}y-y^{3}$
the point $(2,1)$
Hint 1/4
You are being asked for two ordinary derivatives: one treating $y$ as a constant, one treating $x$ as a constant.
Units check: both are pure rates of change of $f$ per unit change in one variable, as expected for first partials.
You are on a hillside whose height above sea level is a function of your map position. Step due east and the ground rises at one rate; step due north and it rises at another; step northeast and you get a third number that is not the average of the first two. Somewhere there is a single compass bearing that is the steepest way up — and the contour lines printed on the map already know which way it points, because they always cross it at a right angle.
By the end you can take a rate of change in any direction, name the one direction of fastest increase and its size, and locate and classify every peak, valley, and pass of a surface.
In 60 seconds
Package both partial derivatives into one vector, the gradient: dot it with a unit direction to get any rate of change, and set it to zero to find the candidates for peaks and valleys.
a critical point must be classified as min, max, or saddle
Three most common mistakes
Using the given direction vector without turning it into a first. $D_{\mathbf u}f$ is a rate per unit distance, so a direction $\langle 3,4\rangle$ must be divided by its length $5$; skipping this scales every answer.
Reading $\nabla f=\mathbf 0$ as an extremum. It only makes the point a candidate — it can be a saddle. The Second Derivatives Test is what decides.
Forgetting to square $f_{xy}$ in the discriminant, or concluding min-versus-max from $f_{xx}$ before checking that $D>0$.
The course lists a weekly quiz (10%) and homework (10%); both draw straight from the routines on this page. The reliable earners are a directional-derivative computation and a full classify-the-critical-points question — practise each to a finished number, not just a setup.
How much time do you have?
10 minutes
You walk in able to compute a directional derivative, name the steepest direction and its size, and set a gradient to zero to find critical points.
In 60 seconds · The gradient packages both partials · Steepest ascent and the level-curve angle · Critical points · Formula card
45 minutes
You add the two questions that carry the most marks: writing a tangent plane to an implicit surface, and running the Second Derivatives Test all the way to a verdict.
everything in the 10-minute path · The gradient is normal to a level surface · The Second Derivatives Test · Worked exam-style classification · Practice B · Practice C
full read
You cover absolute extrema on a closed region (interior plus boundary), the interleaved links back to the chain rule and tangent planes, and every trap in the mistake ledger.
all concepts in order · Absolute maxima and minima on a closed region · The four-rung ladder · Practice A through D · Mistake ledger · Self-audit
By the end of this section
Compute the directional derivative $D_{\mathbf u}f$ of a function of two variables in a given direction, remembering to use a unit vector.
Assemble the gradient vector $\nabla f$ and use the form $D_{\mathbf u}f=\nabla f\cdot\mathbf u$ to read off any directional derivative from one vector.
Identify the direction of fastest increase and its magnitude $|\nabla f|$, and explain why the rate along a level curve is zero.
Write the tangent plane and to a level surface $F(x,y,z)=k$ using $\nabla F$ as the normal vector.
Locate all critical points of a function of two variables by solving $f_x=0$ and $f_y=0$ together.
Classify each critical point as a local minimum, , or using the Second Derivatives Test.
Determine the and minimum of a continuous function on a closed, bounded region by comparing interior and boundary values.
Optimise a quantity that carries one side condition by spending the constraint to reach two free variables, then confirming the verdict with the Second Derivatives Test.
Syllabus coverage
covered
14.6 — covered
Directional derivatives and the gradient vector
direction of fastest change
gradient normal to level curves and surfaces
tangent planes to level surfaces
Spread over four concepts: the directional derivative itself, the gradient as its package, the steepest-ascent geometry, and the jump to three variables, where one gradient produces both the tangent plane and the normal line to a level surface.
covered
14.7 — covered
Local maxima, minima, and saddle points
critical points
the Second Derivatives Test
absolute extrema on a closed bounded region
Four strands: finding critical points, classifying them with the discriminant, the closed-region procedure that mixes interior critical points with boundary work, and the applied optimisation that spends a constraint before the test is run.
Recall first
Partial derivatives
$f_x$ is the derivative of $f$ with $y$ held constant, $f_y$ with $x$ held constant. They are the rates of change of $f$ along the $x$- and $y$-directions only.
Every object in this section — the gradient, the critical-point equations, the discriminant — is built directly out of partial derivatives.
Dot product and length
$\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2=|\mathbf a|\,|\mathbf b|\cos\theta$, and a unit vector is any $\mathbf u$ with $|\mathbf u|=1$, obtained as $\mathbf v/|\mathbf v|$.
The directional derivative is a dot product, and the $\cos\theta$ form is what makes the gradient the direction of steepest ascent.
Equation of a plane from a normal
The plane through $P_0=(x_0,y_0,z_0)$ with normal vector $\mathbf n=\langle a,b,c\rangle$ is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$.
The tangent plane to a level surface uses the gradient as this normal vector, so the plane's equation is immediate once you have the gradient.
Try it yourself first (2 questions)
1§10.0 — turning a direction into a unit vector
A deliberate trap. Many first answers skip the one step this whole section depends on.
Find(a) Is $\mathbf v$ a unit vector? If not, give the unit vector $\mathbf u$ pointing the same way.
Giventhe direction vector $\mathbf v=\langle 3,4\rangle$
Hint 1/4
A unit vector has length exactly $1$. First just measure the length of $\mathbf v$.
Hint 2/4
Length is $|\mathbf v|=\sqrt{v_1^{2}+v_2^{2}}$, and the unit vector is $\mathbf v/|\mathbf v|$.
Hint 3/4
$|\langle 3,4\rangle|=\sqrt{9+16}=5\neq1$, so $\mathbf v$ is not a unit vector. Divide by $5$.
pair the first components and the second components
$$=\tfrac{12}{5}+\tfrac{4}{5}=\tfrac{16}{5}$$
common denominator $5$
Answer $$\mathbf a\cdot\mathbf u=\tfrac{16}{5}$$
Check
Order check: $\mathbf a\cdot\mathbf u=\mathbf u\cdot\mathbf a$, and indeed $\tfrac35\cdot4+\tfrac45\cdot1=\tfrac{16}{5}$ as well.
This is exactly the directional derivative of a function whose gradient is $\langle4,1\rangle$ in the direction $\mathbf u$.
Notation
symbol
reads as
means
watch out
$D_{\mathbf u}f$
the directional derivative of f in the direction u
the rate of change of $f$ per unit distance as you move from the point in the unit direction $\mathbf u$ — a number
The subscript must be a unit vector. $D_{\mathbf u}f$ with a non-unit $\mathbf u$ is not the directional derivative.
$\nabla f$
grad f, or del f
the gradient vector $\langle f_x,f_y\rangle$ (or $\langle f_x,f_y,f_z\rangle$ in three variables) — a vector, evaluated at a point
$\nabla f$ is a vector field, one arrow at each point; $\nabla f(a,b)$ is the single arrow at $(a,b)$.
$|\nabla f|$
the length of the gradient
the magnitude of $\nabla f$; it equals the maximum directional derivative, i.e. the steepest rate of increase
A number, never a direction. The steepest direction is $\nabla f/|\nabla f|$.
$D$
the discriminant D
$D=f_{xx}f_{yy}-(f_{xy})^{2}$, the quantity whose sign classifies a critical point
This $D$ has nothing to do with $D_{\mathbf u}f$; the reused letter is unfortunate but standard. The $f_{xy}$ term is squared.
Conventions used here
A direction must be a unit vector
A directional derivative is a rate per unit distance, so $\mathbf u$ must have length $1$. Handed a direction like $\langle 3,4\rangle$, divide by its length $5$ first. An angle $\theta$ already gives a unit vector $\langle\cos\theta,\sin\theta\rangle$.
The gradient is a vector; its rate is a number
$\nabla f(a,b)=\langle f_x,f_y\rangle$ is a vector. The fastest rate of increase is the number $|\nabla f(a,b)|$. Do not report a vector where a rate is wanted, or a length where a direction is wanted.
A critical point is a candidate, not a verdict
$\nabla f=\mathbf 0$ only makes $(a,b)$ eligible. It may be a maximum, a minimum, or a saddle; the Second Derivatives Test decides. A point where a partial derivative fails to exist is also critical.
When the discriminant is zero the test is silent
$D=0$ means the Second Derivatives Test gives no information — not that there is no extremum. You must then argue directly, by inspecting values near the point.
Absolute extrema need a closed and bounded region
The Extreme Value Theorem guarantees an absolute maximum and minimum only when the region is closed (it contains its boundary) and bounded. On an open or unbounded region a continuous function may attain neither.
10.1The directional derivative
Turns the two axis rates into the rate along any chosen direction; use it whenever a direction or angle is named.
We already know the rate of change of $f$ along the two axes: that is exactly what $f_x$ and $f_y$ are. The question now is the rate along any direction.
Solvable with what we have
the rate of change of $f$ heading due east — it is $f_x$
the rate heading due north — it is $f_y$
Not solvable yet
the rate heading northeast, along $\langle1,1\rangle$
the rate along a direction given by an angle, say $30^\circ$
For the northeast rate, just add the two axis rates: $f_x+f_y$.
Why it fails
A northeast unit step covers only $1/\sqrt2$ of a unit east and $1/\sqrt2$ north, not a full unit of each. Adding $f_x+f_y$ is the change over the vector $\langle1,1\rangle$, which is $\sqrt2$ units long — so it overstates the rate per unit distance by a factor of $\sqrt2$.
The rate of change in the unit direction $\langle a,b\rangle$ is the east rate weighted by how much of the step goes east, plus the north rate weighted by how much goes north.
Proof
Slice the surface along the line through $(x,y)$ in direction $\mathbf u$: let $g(h)=f(x+ha,\,y+hb)$.
By the chain rule, $g'(0)=f_x\cdot a+f_y\cdot b$, and $g'(0)$ is exactly the rate per unit $h$, i.e. per unit distance because $\mathbf u$ is a unit vector.
You stand at a point $P$ and may step off in any direction. The two partial derivatives $f_x$ and $f_y$ only report the rates for the east and north steps (the blue arrows); every other direction — the whole dashed circle of unit steps — needs a new number.
Looks like this, but is not
For the northeast direction, $D_{\mathbf u}f=f_x+f_y$.
That uses $\langle1,1\rangle$, whose length is $\sqrt2$. The unit vector is $\langle1/\sqrt2,1/\sqrt2\rangle$, so the correct rate is $(f_x+f_y)/\sqrt2$.
Rate of $x^{2}y$ at $(3,2)$ toward $\langle3,4\rangle$
Find the rate of change of $f(x,y)=x^{2}y$ at the point $(3,2)$ in the direction of $\mathbf v=\langle3,4\rangle$.
Find$D_{\mathbf u}f(3,2)$
Given
$f(x,y)=x^{2}y$
point $(3,2)$
direction $\mathbf v=\langle3,4\rangle$
Solution
Get the partials at the point
$$f_x=2xy,\quad f_y=x^{2}$$
differentiate holding the other variable fixed
$$f_x(3,2)=12,\quad f_y(3,2)=9$$
substitute $x=3,\,y=2$
Turn the direction into a unit vector
$$|\mathbf v|=\sqrt{3^{2}+4^{2}}=5$$
the direction is not yet unit length, so it must be rescaled
The gradient length is $|\langle12,9\rangle|=\sqrt{225}=15$, and $14.4\le15$: no directional derivative can beat the gradient's length, so the size is plausible.
One normalisation and one dot product.
Closing the northeast loop for $x^{2}y$ at $(3,2)$
Return to the northeast direction that stumped us. Find the rate of $f(x,y)=x^{2}y$ at $(3,2)$ along $\langle1,1\rangle$, and compare with the naive answer $f_x+f_y$.
Find$D_{\mathbf u}f(3,2)$ and the size of the naive error
Cross-check: $14.85\le|\nabla f|=15$, and it is slightly below the $14.4$ of the $\langle3,4\rangle$ direction, consistent with $\langle1,1\rangle$ pointing a little closer to the steepest direction $\langle12,9\rangle$.
The whole section exists to stop that factor of $\sqrt2$: always normalise first.
Checkpoint
§10.1 — a directional derivative in 30 seconds
Thirty seconds. One function, one point, one direction — the core computation.
Find(a) Find $D_{\mathbf u}f(1,3)$.
Given
$f(x,y)=xy+y^{2}$
point $(1,3)$
direction $\mathbf v=\langle4,-3\rangle$
Hint 1/4
You want a rate in a named direction, so you will need the two partials at the point and a unit direction.
Hint 2/4
$D_{\mathbf u}f=f_x a+f_y b$ with $\mathbf u=\mathbf v/|\mathbf v|$.
Hint 3/4
$f_x=y=3$, $f_y=x+2y=7$ at $(1,3)$; $|\mathbf v|=5$ so $\mathbf u=\langle4/5,-3/5\rangle$.
Packs both partials into one vector so a single dot product delivers every directional derivative at a point.
The expression $f_x a+f_y b$ is a dot product waiting to happen. Collect the partials into one vector and the pattern becomes a single, reusable object.
Toward the origin the surface $2x^{2}+y^{2}$ drops, so a negative rate is right; and $-3\sqrt2\approx-4.24\ge-|\nabla f|=-\sqrt{20}\approx-4.47$, within the allowed range.
One gradient answers every direction at this point — no need to recompute partials per direction.
A nearly flat direction for $x^{2}+xy+y^{2}$
For $f(x,y)=x^{2}+xy+y^{2}$, find $D_{\mathbf u}f(1,2)$ in the direction $\langle4,-3\rangle$.
The tiny result is expected: $\nabla f\cdot\langle4,-3\rangle=16-15=1$ is small, so $\mathbf u$ is nearly along a level curve, where the rate is near zero.
Checkpoint
§10.2 — one gradient, any direction
Thirty seconds. Build the vector once, then use it.
Find(a) Find $\nabla f(3,1)$ and then $D_{\mathbf u}f(3,1)$ in the direction $\langle1,2\rangle$.
Given
$f(x,y)=x^{2}-y^{2}$
point $(3,1)$
direction $\langle1,2\rangle$
Hint 1/4
First produce the gradient vector at the point; the directional derivative is then a single dot product.
Hint 2/4
$\nabla f=\langle f_x,f_y\rangle$ and $D_{\mathbf u}f=\nabla f\cdot\mathbf u$ with $\mathbf u$ of unit length.
Hint 3/4
$\nabla f=\langle2x,-2y\rangle=\langle6,-2\rangle$ at $(3,1)$; $|\langle1,2\rangle|=\sqrt5$ so $\mathbf u=\langle1,2\rangle/\sqrt5$.
10.3The steepest direction and the length of the gradient
Names the single direction of fastest increase (the gradient) and its size (the gradient's length), and why level curves are the flat directions.
Write the dot product with its angle: $D_{\mathbf u}f=\nabla f\cdot\mathbf u=|\nabla f|\cos\theta$. The whole geometry of the hillside falls straight out of that one line.
TheoremFastest increase, and the level-curve angle
Conditions
$f$ is differentiable and $\nabla f(a,b)\ne\mathbf 0$
Because $D_{\mathbf u}f=|\nabla f|\cos\theta$, the rate is largest ($+|\nabla f|$) when you walk along the gradient, most negative ($-|\nabla f|$) straight against it, and exactly zero when you walk at right angles to it — that is, along the level curve.
Proof
$D_{\mathbf u}f=\nabla f\cdot\mathbf u=|\nabla f|\,|\mathbf u|\cos\theta=|\nabla f|\cos\theta$ since $|\mathbf u|=1$.
$\cos\theta$ ranges over $[-1,1]$, hitting $1$ at $\theta=0$ (along $\nabla f$) and $0$ at $\theta=90^\circ$ (perpendicular to $\nabla f$).
The level curves of $f(x,y)=x^{2}+y^{2}$. The $\textcolor{#1f6feb}{\text{gradient}}$ crosses each curve at a right angle and points toward the closer, higher curves — the steepest way up — while stepping $\textcolor{#d1690a}{\text{along the curve}}$ changes $f$ by nothing.
Looks like this, but is not
The steepest way up is always the diagonal $\langle1,1\rangle$.
Steepest is $\nabla f/|\nabla f|$, which changes from point to point. It equals $\langle1,1\rangle/\sqrt2$ only where $f_x=f_y$.
direction angle θ
rate = 10·cos θ
0°
10.00 — steepest ascent
30°
8.66
45°
7.07
60°
5.00
90°
0 — along the level curve
120°
−5.00
180°
−10.00 — steepest descent
The rate follows $\cos\theta$: its full value straight up the gradient, exactly zero at $90^\circ$ where you move along the level curve, and the negative of the maximum when you head straight downhill.
Steepest ascent for $x^{2}+y^{2}$ at $(3,4)$
For $f(x,y)=x^{2}+y^{2}$, find the maximum rate of increase at $(3,4)$, the direction that achieves it, and a direction in which the rate is zero.
Findthe maximum rate, its direction, and a zero-rate direction
10.4Three variables: the gradient normal to a level surface
In space the gradient stands perpendicular to a level surface, which hands you the normal for the tangent plane at once.
Nothing above used the fact that there were only two variables. With three, the gradient gains a component — and the perpendicularity we just saw becomes perpendicularity to a whole surface.
At any point of a level surface the gradient sticks out perpendicular to the surface, so it is the normal vector of the tangent plane — and a plane is fixed by a point and a normal.
A level surface $F(x,y,z)=k$ and the flat tangent plane that touches it at $P$. The $\textcolor{#1f6feb}{\text{gradient}}$ $\nabla F$ stands at a right angle to the surface, so it is the normal vector of that plane — which is all you need to write the plane's equation.
Looks like this, but is not
For a surface written $z=f(x,y)$, the normal is $\langle f_x,f_y\rangle$.
Rewrite it as $F=f(x,y)-z=0$. Then $\nabla F=\langle f_x,f_y,-1\rangle$: the $-1$ third component is the one everybody forgets.
A directional derivative of $xyz$ in space
For $f(x,y,z)=xyz$, find the rate of change at $(2,1,3)$ in the direction of $\mathbf v=\langle1,2,2\rangle$.
$|\nabla f|=\sqrt{9+36+4}=7$, and $\tfrac{19}{3}\approx6.33\le7$: still bounded by the gradient length, exactly as in two variables.
Tangent plane to the sphere $x^{2}+y^{2}+z^{2}=14$ at $(1,2,3)$
Find the tangent plane to the sphere $x^{2}+y^{2}+z^{2}=14$ at the point $(1,2,3)$.
Findthe equation of the tangent plane
Given
surface $F=x^{2}+y^{2}+z^{2}=14$
point $(1,2,3)$
Solution
Gradient as the normal
$$\nabla F=\langle2x,2y,2z\rangle$$
the gradient is normal to the level surface
$$\nabla F(1,2,3)=\langle2,4,6\rangle$$
substitute — this is the plane's normal vector
Point-normal plane
$$2(x-1)+4(y-2)+6(z-3)=0$$
normal $\cdot\,(\mathbf r-\mathbf r_0)=0$
$$x+2y+3z=14$$
divide by $2$ and collect constants
Answer $$x+2y+3z=14$$
Check
The distance from the origin to $x+2y+3z=14$ is $14/\sqrt{1+4+9}=14/\sqrt{14}=\sqrt{14}$, exactly the sphere's radius — so the plane just touches the sphere.
For any implicit surface the recipe is identical: gradient at the point is the normal, then write the point-normal plane.
The normal line to the hyperboloid $x^{2}+y^{2}-z^{2}=4$ at $(2,2,2)$
Write parametric equations for the line through $(2,2,2)$ that is perpendicular to the surface $x^{2}+y^{2}-z^{2}=4$, and give the tangent plane there as well.
Findparametric equations of the normal line, and the tangent plane
Given
surface $F(x,y,z)=x^{2}+y^{2}-z^{2}=4$
point $P=(2,2,2)$, which satisfies $4+4-4=4$
Solution
One gradient, two answers
$$\nabla F=\langle 2x,\,2y,\,-2z\rangle$$
differentiate $F$ once in each variable; the $z$ term carries the minus sign
Test the normal against two curves that actually lie on the surface. Holding $y=2$ leaves $x^{2}-z^{2}=0$, so $\mathbf c_1(t)=(2+t,\,2,\,2+t)$ stays on it and $\mathbf c_1'=\langle1,0,1\rangle$; holding $x=2$ gives $\mathbf c_2(t)=(2,\,2+t,\,2+t)$ with $\mathbf c_2'=\langle0,1,1\rangle$. Both dot to zero against $\langle1,1,-1\rangle$, so the direction really is perpendicular to the surface in two independent directions — which is all perpendicularity to a surface means.
The plane and the line come out of the same vector: the gradient is the plane's normal and the line's direction. Compute $\nabla F(P)$ once and you have answered both questions.
Checkpoint
§10.4 — tangent plane to a paraboloid
Thirty seconds. The paraboloid $z=x^{2}+y^{2}$ can be read as a level surface of $F=x^{2}+y^{2}-z$ at level $0$.
Find(a) Find the tangent plane at $(1,1,2)$.
Given
surface $z=x^{2}+y^{2}$, i.e. $F=x^{2}+y^{2}-z=0$
point $(1,1,2)$
Hint 1/4
You need a point (given) and a normal vector; the gradient of $F$ supplies the normal.
Hint 2/4
$\nabla F$ is normal to $F=0$, and the plane is $\nabla F(P)\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0$.
Hint 3/4
$\nabla F=\langle2x,2y,-1\rangle=\langle2,2,-1\rangle$ at $(1,1,2)$; note the $-1$ from $-z$.
the tangent plane came out as a single equation, so the line is expected to look the same
wrong$$x+y-z=2\ \text{(a plane, not a line)}$$
right$$x=2+t,\ y=2+t,\ z=2-t$$
10.5Critical points: where a surface can peak or dip
Locates the only places a smooth surface can peak, dip, or pass — where both partials vanish at once.
To find the top of a hill you first find the places where the ground is momentarily flat in every direction. In gradient language, that is where $\nabla f=\mathbf 0$.
TheoremLocal extrema live at critical points
Conditions
$f$ has a local maximum or minimum at $(a,b)$
the first partials exist there
$$\boxed{\ \text{local max or min at }(a,b)\ \Rightarrow\ f_x(a,b)=0\ \text{ and }\ f_y(a,b)=0\ }$$
At a genuine peak or valley the surface is level in both the $x$- and the $y$-direction, so both partials vanish. Such a point — where the gradient is zero, or a partial fails to exist — is called a critical point.
Proof
Freeze $y=b$: the single-variable function $x\mapsto f(x,b)$ has a local extremum at $x=a$, so its derivative $f_x(a,b)=0$.
By the same argument with $x=a$ frozen, $f_y(a,b)=0$.
Why $\nabla f=0$ is not enough. At one critical point, the slice $\textcolor{#1f6feb}{\text{along }x}$ is a valley (a minimum) while the slice $\textcolor{#d1690a}{\text{along }y}$ is a hill (a maximum). Neither wins: the point is a saddle.
Looks like this, but is not
If $\nabla f=\mathbf 0$ then $f$ has a maximum or a minimum there.
$f=x^{2}-y^{2}$ has $\nabla f=\mathbf 0$ at the origin, yet it climbs along the $x$-axis and falls along the $y$-axis. It is a saddle — neither.
The one critical point of $x^{2}+y^{2}-4x+6y+13$
Find all critical points of $f(x,y)=x^{2}+y^{2}-4x+6y+13$.
Findall critical points
Given$f=x^{2}+y^{2}-4x+6y+13$
Solution
Set both partials to zero
$$f_x=2x-4=0\Rightarrow x=2$$
solve the $x$-equation
$$f_y=2y+6=0\Rightarrow y=-3$$
solve the $y$-equation
One critical point
$$(a,b)=(2,-3)$$
the only simultaneous solution
Answer $$(2,-3)$$
Check
Complete the square: $f=(x-2)^{2}+(y+3)^{2}$, whose smallest value $0$ sits exactly at $(2,-3)$ — so this critical point is a genuine minimum.
For a sum of squares the critical point is obviously a minimum; in general we will need a test.
Two critical points of $x^{3}-3x+y^{2}$
Find all critical points of $f(x,y)=x^{3}-3x+y^{2}$.
Findall critical points
Given$f=x^{3}-3x+y^{2}$
Solution
Solve the system
$$f_x=3x^{2}-3=0\Rightarrow x=\pm1$$
a quadratic in $x$: two roots
$$f_y=2y=0\Rightarrow y=0$$
the $y$-equation is independent here
Pair the solutions
$$(1,0)\ \text{and}\ (-1,0)$$
each $x$-root with the single $y$-root
Answer $$(1,0)\ \text{and}\ (-1,0)$$
Check
Substitute back: $\nabla f(1,0)=\langle0,0\rangle$ and $\nabla f(-1,0)=\langle0,0\rangle$, so both are genuinely critical (their type waits for the next concept).
Two simultaneous equations, two solutions.
Checkpoint
§10.5 — a coupled critical-point system
Thirty seconds. Here the two equations share both variables, so they must be solved together.
Find(a) Find every critical point of $f$.
Given$f(x,y)=x^{2}+xy+y^{2}-6y$
Hint 1/4
Critical points solve $f_x=0$ and $f_y=0$ at the same time; write both equations first.
Hint 2/4
$f_x=2x+y$ and $f_y=x+2y-6$; set each to zero and solve the pair.
Hint 3/4
From $f_x=0$, $y=-2x$. Put it in $f_y=0$: $x+2(-2x)-6=0$, i.e. $-3x-6=0$.
Hint 4/4
So $x=-2$, $y=4$: the single critical point $(-2,4)$.
Compute one number $D$. If it is positive the point is a genuine extremum and the sign of $f_{xx}$ says which — positive for a minimum, negative for a maximum; if $D$ is negative it is a saddle; if $D=0$ the test says nothing.
The Second Derivatives Test as a flowchart. Compute $D$ once; its sign sends you down one of three branches, and only the $D>0$ branch still asks a question — the sign of $f_{xx}$.
Looks like this, but is not
$D>0$ means a minimum.
$D>0$ only rules out a saddle. The sign of $f_{xx}$ then decides: a maximum also has $D>0$, with $f_{xx}<0$.
Classifying $(2,-1)$ for $x^{2}+xy+y^{2}-3x$
Classify the critical point $(2,-1)$ of $f(x,y)=x^{2}+xy+y^{2}-3x$.
Findthe type of the critical point
Given
$f=x^{2}+xy+y^{2}-3x$
critical point $(2,-1)$
Solution
Second partials
$$f_{xx}=2,\ f_{yy}=2,\ f_{xy}=1$$
differentiate the first partials again; $f_{xy}$ is the mixed one
Discriminant and verdict
$$D=f_{xx}f_{yy}-(f_{xy})^{2}=2\cdot2-1^{2}=3$$
$D>0$, so it is a genuine extremum
$$f_{xx}=2>0\Rightarrow\text{local minimum}$$
positive curvature means a valley
Answer $$\text{local minimum},\ f(2,-1)=-3$$
Check
Cross-check the value: $f(2,-1)=4-2+1-6=-3$, and $D>0$ with $f_{xx}>0$ is consistent with a bowl opening upward.
Three second partials, one discriminant.
A maximum: classifying $(1,2)$ for $-x^{2}-y^{2}+2x+4y$
Classify the critical point $(1,2)$ of $f(x,y)=-x^{2}-y^{2}+2x+4y$.
Findthe type of the critical point
Given
$f=-x^{2}-y^{2}+2x+4y$
critical point $(1,2)$
Solution
Second partials and discriminant
$$f_{xx}=-2,\ f_{yy}=-2,\ f_{xy}=0$$
the mixed partial is zero here
$$D=(-2)(-2)-0^{2}=4>0$$
positive, so not a saddle
Read the sign of the curvature
$$f_{xx}=-2<0\Rightarrow\text{local maximum}$$
negative curvature means a hilltop
Answer $$\text{local maximum},\ f(1,2)=5$$
Check
Complete the square: $f=5-(x-1)^{2}-(y-2)^{2}$, a dome peaking at $5$ over $(1,2)$ — matching $D>0,\ f_{xx}<0$.
Same test, opposite verdict: $D>0$ both times, and the sign of $f_{xx}$ is the whole difference.
Shortest distance from the origin to the plane $x+2y+3z=6$
Find the point of the plane $x+2y+3z=6$ closest to the origin, and the distance to it.
Findthe closest point and the distance
Given
the plane $x+2y+3z=6$
distance from the origin: $\sqrt{x^{2}+y^{2}+z^{2}}$
Solution
Choose what to minimise, and in how many variables
$$\text{minimise } d^{2}=x^{2}+y^{2}+z^{2}\ \text{rather than }d$$
$d\ge0$ and squaring is increasing, so both are smallest at the same point — and the square has no radical to differentiate
$$z=\tfrac{6-x-2y}{3}$$
the constraint is one equation, so it removes one variable and leaves an unconstrained problem in $x,y$
$$f(x,y)=x^{2}+y^{2}+\tfrac{(6-x-2y)^{2}}{9}$$
this is now exactly the two-variable minimisation we know how to run
$D>0$ with $f_{xx}>0$ is the Second Derivatives Test verdict: a local minimum
$$d^{2}=\tfrac{9+36+81}{49}=\tfrac{18}{7}$$
the question asked for a distance, not for the point, so convert back
Answer $$\text{closest point }\left(\tfrac37,\tfrac67,\tfrac97\right),\qquad d=\sqrt{\tfrac{18}{7}}=\tfrac{3\sqrt{14}}{7}$$
Check
The point-to-plane formula gives the distance independently: $\dfrac{|0+0+0-6|}{\sqrt{1^{2}+2^{2}+3^{2}}}=\dfrac{6}{\sqrt{14}}=\dfrac{3\sqrt{14}}{7}$, the same number. The point also checks out on the plane: $\tfrac37+\tfrac{12}{7}+\tfrac{27}{7}=\tfrac{42}{7}=6$.
The pattern is the whole of applied optimisation in this section: name the quantity, use the constraint to delete one variable, run the critical-point machinery, and convert back to what was asked.
Checkpoint
§10.6 — is it a saddle?
Thirty seconds. One critical point, one discriminant.
Find(a) Classify the critical point $(-1,0)$.
Given
$f(x,y)=x^{2}-y^{2}+2x$
critical point $(-1,0)$
Hint 1/4
You need the three second partials and then the discriminant $D$ — do not stop at the gradient.
Hint 2/4
$D=f_{xx}f_{yy}-(f_{xy})^{2}$; the sign of $D$ comes first, before any talk of min or max.
Hint 3/4
$f_{xx}=2$, $f_{yy}=-2$, $f_{xy}=0$, so $D=(2)(-2)-0=-4$.
Hint 4/4
$D<0$, so the point is a saddle — the sign of $f_{xx}$ is irrelevant here.
Show solution
Discriminant
$$f_{xx}=2,\ f_{yy}=-2,\ f_{xy}=0$$
the three second partials
$$D=(2)(-2)-0^{2}=-4$$
assemble; negative
Answer $$D=-4<0\Rightarrow\text{saddle point}$$
Check
Independent view: along $x$ the term $x^{2}$ opens up, along $y$ the term $-y^{2}$ opens down — the up-and-down signature of a saddle.
⚠ Forgetting to square the mixed partial
the other two terms are plain products, so the square on $f_{xy}$ slips
wrong$$D=f_{xx}f_{yy}-f_{xy}$$
right$$D=f_{xx}f_{yy}-(f_{xy})^{2}$$
⚠ Judging min versus max before checking D>0
the rule f_xx>0 means minimum gets memorised without its condition
Finds the true largest and smallest values on a closed region by comparing interior critical points against the whole boundary.
On a closed, bounded region a continuous function is guaranteed a largest and a smallest value. The catch: the winner may sit on the boundary, not at any interior peak.
MethodAbsolute extrema on a closed bounded region
Conditions
$f$ is continuous on a closed, bounded region $D$
$$\boxed{\ \text{abs.\ max/min}=\text{best/worst of }\{f\text{ at interior critical points}\}\cup\{f\text{ on the boundary}\}\ }$$
List the values of $f$ at every interior critical point, list the extreme values of $f$ along the boundary, put all these numbers together, and pick the largest and the smallest.
Finding an absolute extremum on a closed rectangle. There is one $\textcolor{#d1690a}{\text{interior critical point}}$; the winner might still sit on the $\textcolor{#1f6feb}{\text{boundary}}$, so both have to be checked and compared.
Looks like this, but is not
Just test the interior critical points and take the best.
On $[0,3]\times[0,2]$, the maximum of $x^{2}-2xy+2y$ is $9$ at the corner $(3,0)$, far above the interior critical value $1$ at $(1,1)$.
Absolute extrema of $x^{2}-2xy+2y$ on $[0,3]\times[0,2]$
Find the absolute maximum and minimum of $f(x,y)=x^{2}-2xy+2y$ on the closed rectangle $0\le x\le3,\ 0\le y\le2$.
Findthe absolute maximum and minimum and where they occur
Given
$f=x^{2}-2xy+2y$
rectangle $[0,3]\times[0,2]$
Solution
Interior critical points
$$f_x=2x-2y=0,\ f_y=-2x+2=0$$
set both partials to zero
$$x=1,\ y=1\Rightarrow f(1,1)=1$$
the one interior critical point and its value
Boundary: bottom and top edges
$$y=0:\ f=x^{2}\in[0,9]$$
on $0\le x\le3$: min $0$ at $(0,0)$, max $9$ at $(3,0)$
$$y=2:\ f=(x-2)^{2}\in[0,4]$$
min $0$ at $(2,2)$; endpoints give $4$ at $(0,2)$, $1$ at $(3,2)$
Boundary: left and right edges
$$x=0:\ f=2y\in[0,4]$$
min $0$ at $(0,0)$, max $4$ at $(0,2)$
$$x=3:\ f=9-4y\in[1,9]$$
max $9$ at $(3,0)$, min $1$ at $(3,2)$
Compare every candidate
$$\{1,\,0,\,9,\,0,\,4,\,1,\,4,\,1\}$$
interior value with all boundary extremes and corners
$$\max=9,\ \min=0$$
pick the largest and smallest
Answer $$\text{abs max }9\ \text{at }(3,0);\quad \text{abs min }0\ \text{at }(0,0)\ \text{and }(2,2)$$
Check
Spot-check two corners directly: $f(3,0)=9-0+0=9$ and $f(2,2)=4-8+4=0$, matching the max and a min.
The maximum was a corner and the interior critical point won nothing — which is exactly why the boundary is not optional.
Checkpoint
§10.7 — a linear function on a rectangle
Thirty seconds. A linear function has no interior critical point, so where can the extremes hide?
Find(a) Find the absolute maximum and minimum of $f$ on the rectangle.
Given
$f(x,y)=2x+y$
rectangle $0\le x\le2,\ 0\le y\le3$
Hint 1/4
First check the interior: is $\nabla f$ ever zero? If not, the extremes must live on the boundary.
Hint 2/4
For a linear function on a rectangle the largest and smallest values sit at the corners; evaluate all four.
Hint 3/4
$\nabla f=\langle2,1\rangle\ne\mathbf 0$, so no interior critical point. Corners: $(0,0),(2,0),(0,3),(2,3)$.
Hint 4/4
$f=0,4,3,7$ at those corners, so max $7$, min $0$.
Show solution
Interior
$$\nabla f=\langle2,1\rangle\ne\mathbf 0$$
never zero, so nothing in the interior
Corners
$$f(0,0)=0,\ f(2,0)=4,\ f(0,3)=3,\ f(2,3)=7$$
a linear function is extremal at corners
Answer $$\text{max }7\text{ at }(2,3),\ \text{min }0\text{ at }(0,0)$$
Check
Because $f$ increases in both $x$ and $y$, the corner with both largest is the max and both smallest is the min — consistent with the numbers.
⚠ Testing only the interior critical points
the earlier concepts were all about interior critical points, so the boundary is forgotten
the surface arrives as an equation $F(x,y,z)=k$ (sphere, ellipsoid, cone, hyperboloid), or as $z=f(x,y)$ that you can move onto one side
Put the surface in level form
Write it as $F(x,y,z)=k$. If it came as $z=f(x,y)$, use $F=f(x,y)-z$ with $k=0$ — that is where the third component $-1$ comes from.
One gradient, evaluated at the point
Compute $\nabla F$ and substitute $P$. Scale it down to small integers if you like; direction is all that matters.
Plane: normal dotted with the offset
$F_x(P)(x-x_0)+F_y(P)(y-y_0)+F_z(P)(z-z_0)=0$.
Line: the same vector as a direction
$x=x_0+F_x(P)t$, $y=y_0+F_y(P)t$, $z=z_0+F_z(P)t$. The plane and the line share one vector, used in the two different roles.
Where it goes wrong
leaving the surface as $z=f(x,y)$ and reporting a two-component normal
using the coordinates of $P$ as the normal or the direction
writing the line as an equation in $x,y,z$ instead of three parametric equations
Find and classify the critical points
any question that asks for local maxima, minima, or saddle points of a function of two variables
Find critical points
Solve $f_x=0$ and $f_y=0$ together. Watch for coupled equations.
Second partials
Compute $f_{xx}$, $f_{yy}$, and the mixed $f_{xy}$.
Discriminant
Form $D=f_{xx}f_{yy}-(f_{xy})^{2}$ — square the mixed term.
Read the verdict
$D>0$: min if $f_{xx}>0$, max if $f_{xx}<0$. $D<0$: saddle. $D=0$: no verdict.
Where it goes wrong
forgetting to square $f_{xy}$
reading min/max off $f_{xx}$ before confirming $D>0$
solving $f_x=0$ and $f_y=0$ separately for a coupled system
Absolute extrema on a closed bounded region
a largest or smallest value is asked over a closed region such as a rectangle or disk
Interior
Find interior critical points and evaluate $f$ at each.
Boundary
On each boundary piece reduce $f$ to one variable and find its extremes.
Corners
Include the endpoints where boundary pieces meet.
Compare
The largest value is the absolute max; the smallest is the absolute min.
Where it goes wrong
stopping at the interior
dropping the corners
not reducing each edge to a single variable
From a word problem to a critical point
a largest or smallest value is asked about a physical or geometric quantity that carries a side condition — a fixed sum, a fixed area, a point tied to a surface
Name the objective and the constraint
Write one formula for the quantity to be optimised and one equation for the condition it must satisfy. Both in the same letters.
Spend the constraint
Solve the constraint for one variable and substitute, leaving a function of exactly two free variables. Squaring a distance here is legal and saves a radical.
Run the machinery
Solve $f_x=0$ and $f_y=0$ together, then apply the Second Derivatives Test at each critical point.
Convert back
Answer the question that was asked — the distance, not the point; the volume, not the dimensions — and check the constraint still holds.
Where it goes wrong
reporting the critical point when the question asked for the value
differentiating with the constraint still unspent, so the variables are not independent
forgetting to verify the constraint at the answer
$(1,1)$ for $x^{4}+y^{4}-4xy+1$: a minimum
Classify the critical point $(1,1)$ of $f(x,y)=x^{4}+y^{4}-4xy+1$.
Findthe type
Given
$f=x^{4}+y^{4}-4xy+1$
critical point $(1,1)$
Solution
Discriminant at (1,1)
$$f_{xx}=12x^{2},\ f_{yy}=12y^{2},\ f_{xy}=-4$$
second partials in general
$$D=12\cdot12-(-4)^{2}=128>0$$
at $(1,1)$; positive, so an extremum
$$f_{xx}=12>0\Rightarrow\text{min}$$
positive curvature
Answer $$\text{local minimum},\ f(1,1)=-1$$
Check
$f(1,1)=1+1-4+1=-1$; a nearby value $f(1.1,1.1)\approx-0.98>-1$ confirms a dip.
$(0,0)$ for the same $f$: a saddle
Classify the critical point $(0,0)$ of the same $f(x,y)=x^{4}+y^{4}-4xy+1$.
Findthe type
Given
$f=x^{4}+y^{4}-4xy+1$
critical point $(0,0)$
Solution
Discriminant at (0,0)
$$f_{xx}=0,\ f_{yy}=0,\ f_{xy}=-4$$
at the origin the pure fourth-power curvatures vanish
$$D=0\cdot0-(-4)^{2}=-16<0$$
negative, so a saddle
Answer $$\text{saddle point at }(0,0)$$
Check
Along $y=x$, $f=2x^{4}-4x^{2}+1$ dips below $1$; along $y=-x$, $f=2x^{4}+4x^{2}+1$ rises above $1$ — up one way, down the other.
Both points have $\nabla f=\mathbf 0$ and share the same $f_{xy}=-4$; only the pure second partials differ, and that flips the sign of $D$.
How to tell them apart
Compute $D$: positive at $(1,1)$ gives an extremum, negative at $(0,0)$ gives a saddle. The zero gradient alone cannot tell them apart.
Scaffolding comes off
The common skeleton
Find. Solve $f_x=0$ and $f_y=0$ together to locate every critical point.
Curvature. Compute $f_{xx}$, $f_{yy}$, and the mixed $f_{xy}$.
Discriminant. Form $D=f_{xx}f_{yy}-(f_{xy})^{2}$, squaring the mixed term.
Verdict. $D>0$ with $f_{xx}>0$ is a minimum, with $f_{xx}<0$ a maximum; $D<0$ is a saddle; $D=0$ is no verdict.
1 · fully worked
Classify the critical point of $x^{2}+y^{2}+xy-6x$
Find and classify the critical point of $f(x,y)=x^{2}+y^{2}+xy-6x$.
Findthe critical point and its type
Given$f=x^{2}+y^{2}+xy-6x$
Solution
Find the critical point
$$f_x=2x+y-6=0,\ f_y=2y+x=0$$
set both partials to zero
$$x=-2y\Rightarrow 2(-2y)+y-6=0$$
substitute $x=-2y$ from $f_y=0$
$$y=-2,\ x=4$$
solve, then back-substitute
Classify
$$f_{xx}=2,\ f_{yy}=2,\ f_{xy}=1$$
second partials
$$D=2\cdot2-1^{2}=3>0,\ f_{xx}>0$$
positive discriminant, positive curvature
$$\Rightarrow\text{local minimum}$$
a valley
Answer $$\text{local minimum at }(4,-2),\ f=-12$$
Check
$f(4,-2)=16+4-8-24=-12$; $D>0$ with $f_{xx}>0$ agrees with a minimum.
2 · you write the reasoning
Classify the critical point of $f(x,y)=x^{2}+y^{2}-6x+2y+5$. The steps are filled in; you supply the reason for each before revealing ours. (Easier than rung 1: the equations are already uncoupled.)
$f_x=2x-6=0,\quad f_y=2y+2=0$
reasoning
Critical points need both partials zero; here each equation involves only one variable.
$x=3,\quad y=-1$
reasoning
The two equations are independent, so each solves on its own — no substitution needed.
$f_{xx}=2,\ f_{yy}=2,\ f_{xy}=0$
reasoning
Differentiate the first partials again; the mixed partial is zero because there is no $xy$ term.
$D=2\cdot2-0^{2}=4$
reasoning
Assemble the discriminant; $D=4>0$ rules out a saddle.
$f_{xx}>0\Rightarrow$ local minimum
reasoning
With $D>0$ and $f_{xx}>0$ the surface curves up both ways: a local minimum, value $f(3,-1)=-5$.
3 · find the buried error
A student classifies the critical point $(2,-1)$ of $f(x,y)=x^{2}+4xy+y^{2}-6y$. Exactly two steps are wrong. Find them. (Harder than rung 2: a nonzero mixed partial, and the true answer is not a minimum.)
Step 1. $f_x=2x+4y=0$ and $f_y=4x+2y-6=0$ give the critical point $(2,-1)$.
Step 2.$f_{xx}=2,\ f_{yy}=2,\ f_{xy}=4$, so $D=2\cdot2-4=0$: the test is inconclusive.
Step 3. Since the test is inconclusive and $f_{xx}=2>0$, the point is a local minimum.
the two buried errors (2)
⚠ step 2
the mixed partial is not squared: $D$ should be $f_{xx}f_{yy}-(f_{xy})^{2}=2\cdot2-4^{2}=-12$, not $0$.
the first two terms of $D$ are plain products, so the square on the third term is the easy one to drop
right
$D=2\cdot2-(4)^{2}=4-16=-12<0$.
⚠ step 3
a verdict is drawn from an inconclusive test, and it leans on $f_{xx}$ — which is only allowed once $D>0$ is known. With the correct $D=-12<0$ the point is a saddle.
the rule f_xx>0 means minimum is memorised without its precondition $D>0$
right
$D<0$ forces a saddle point, regardless of $f_{xx}$.
4 · the bare problem
§10.6 — classify, no scaffolding
No steps this time. Find the critical point and classify it.
Find(a) Find and classify the critical point of $f$.
Given$f(x,y)=x^{2}+2y^{2}+2xy-6x-10y$
Hint 1/4
Two tasks: locate the critical point by solving $f_x=0,f_y=0$, then run the discriminant.
Hint 2/4
Critical point from $\nabla f=\mathbf 0$; classify with $D=f_{xx}f_{yy}-(f_{xy})^{2}$.
Hint 3/4
$f_x=2x+2y-6=0$ and $f_y=4y+2x-10=0$ give $(1,2)$; $f_{xx}=2,f_{yy}=4,f_{xy}=2$.
Hint 4/4
$D=2\cdot4-2^{2}=4>0$ and $f_{xx}>0$, so $(1,2)$ is a local minimum.
Show solution
Critical point
$$f_x=2x+2y-6=0\Rightarrow x+y=3$$
first partial
$$f_y=2x+4y-10=0\Rightarrow x+2y=5$$
second partial
$$y=2,\ x=1$$
subtract the equations
Classify
$$f_{xx}=2,\ f_{yy}=4,\ f_{xy}=2$$
second partials
$$D=2\cdot4-2^{2}=4>0,\ f_{xx}>0$$
minimum
Answer $$\text{local minimum at }(1,2),\ f=-13$$
Check
$f(1,2)=1+8+4-6-20=-13$, and $D>0,f_{xx}>0$ matches a minimum.
Full exam-style question
Find and classify every critical point of $x^{4}+y^{4}-4xy+1$exam format
Find all critical points of $f(x,y)=x^{4}+y^{4}-4xy+1$ and classify each one.
The perpendicular choice gives $0$ and the opposite gives $-|\nabla f|$, both smaller.
4§10.4 — where the normal line points
One vector plays two roles at a point of a level surface, and the roles are opposite. This question checks that you have not merged them.
Find(a) True or False: the line through $P$ in the direction $\nabla F(P)$ lies inside the tangent plane at $P$.
Given
a level surface $F(x,y,z)=k$ with $\nabla F(P)\ne\mathbf 0$
$P$ a point on that surface
Hint 1/4
Ask what angle the gradient makes with the tangent plane, not what angle it makes with the surface.
Hint 2/4
$\nabla F(P)$ is the normal vector of that plane, and a plane's normal is perpendicular to every direction lying in it.
Hint 3/4
On $x^{2}+y^{2}+z^{2}=14$ at $(1,2,3)$ the plane is $x+2y+3z=14$ and the gradient is $\langle2,4,6\rangle$; the point $(1,2,3)+\langle2,4,6\rangle=(3,6,9)$ gives $3+12+27=42\ne14$, so it left the plane at once.
Hint 4/4
The line leaves the plane immediately: it is the normal line, perpendicular to the plane, not inside it.
Show solution
Use the defining property of the plane
$$\nabla F(P)\cdot(\mathbf r-\mathbf r_0)=0\ \text{ for }\mathbf r\text{ in the plane}$$
the gradient is not the zero vector, so only the point $P$ itself survives
Answer $$\text{False — the line meets the plane only at }t=0$$
Check
The same conclusion from the geometry: a plane's normal is by definition perpendicular to every direction inside the plane, so it can lie in the plane only if it is the zero vector, which the hypothesis excludes.
B · computation 6 questions
1§10.1 — a directional derivative
A clean directional-derivative computation, the bread-and-butter of the section.
Find(a) Find $D_{\mathbf u}f(2,1)$.
Given
$f(x,y)=x^{3}-xy^{2}$
point $(2,1)$
direction $\mathbf v=\langle3,-4\rangle$
Hint 1/4
You will need the two partials at the point and a unit direction.
Hint 2/4
$D_{\mathbf u}f=f_x a+f_y b$ with $\mathbf u=\mathbf v/|\mathbf v|$.
Hint 3/4
$f_x=3x^{2}-y^{2}=11$, $f_y=-2xy=-4$ at $(2,1)$; $|\mathbf v|=5$ so $\mathbf u=\langle3/5,-4/5\rangle$.
Along $x$, the cubic $x^{3}-3x$ has its local min at $x=1$ and local max at $x=-1$; adding the upward $y^{2}+2y$ turns the second into a saddle — consistent.
5§10.4 — tangent plane to an ellipsoid
A tangent plane straight from the gradient normal.
Find(a) Find the tangent plane to the surface at $(1,1,1)$.
Given
surface $x^{2}+2y^{2}+3z^{2}=6$
point $(1,1,1)$
Hint 1/4
The gradient of $F=x^{2}+2y^{2}+3z^{2}$ is the normal you need.
Take the line's point at $t=1$, namely $(6,6,6)$, and check it against the plane: $30+24+18=72\ne22$, so the line is not inside the plane; and $(6,6,6)-(1,2,3)=\langle5,4,3\rangle$ is the plane's normal, so the line is perpendicular to it. The point $P$ itself gives $5+8+9=22$, so the plane does pass through it.
Whenever a question asks for both, do the gradient once: the answer to (b) is the vector you already used in (a).
C · exam level 4 questions
1§10.6 — classify the origin
A short classification with four verdicts on offer. Work the test; do not guess from the shape.
Find(a) The origin is:
Given
$f(x,y)=x^{2}+xy+y^{2}$
critical point $(0,0)$
Hint 1/4
Confirm the origin is critical, then compute the discriminant before naming a type.
Hint 2/4
$D=f_{xx}f_{yy}-(f_{xy})^{2}$; only after its sign do you look at $f_{xx}$.
Hint 3/4
$f_{xx}=2,\ f_{yy}=2,\ f_{xy}=1$, so $D=4-1=3$.
Hint 4/4
$D>0$ and $f_{xx}>0$, so the origin is a local minimum.
Show solution
Test
$$f_{xx}=2,\ f_{yy}=2,\ f_{xy}=1$$
second partials (constants here)
$$D=2\cdot2-1^{2}=3>0,\ f_{xx}>0$$
positive discriminant, positive curvature
Answer $$\text{local minimum}$$
Check
$f=\left(x+\tfrac y2\right)^{2}+\tfrac34 y^{2}\ge0$ with equality only at the origin — a genuine minimum.
2§10.7 — absolute extrema on a square
A full closed-region problem: interior plus boundary, then compare.
Find(a) Find the absolute maximum and minimum of $f$ on the square and where they occur.
Given
$f(x,y)=x^{2}+xy+y^{2}$
square $-1\le x\le1,\ -1\le y\le1$
Hint 1/4
Do the interior first (critical points), then each of the four edges as a one-variable problem, then the corners.
Hint 2/4
Interior: $\nabla f=\mathbf 0$. Boundary: substitute the fixed coordinate and optimise in the other.
Hint 3/4
Interior gives $(0,0)$, $f=0$. On $x=1$: $f=1+y+y^{2}$ on $[-1,1]$, largest $3$ at $y=1$. Symmetry handles the rest.
Hint 4/4
Absolute min $0$ at $(0,0)$; absolute max $3$ at $(1,1)$ and $(-1,-1)$.
Moving the opposite way, along $+\nabla T$, would heat the bug at the same rate $\sqrt{41}$ — the sign is the only difference.
4§10.8 — a box of largest volume
A workshop has $12\ \mathrm{m^{2}}$ of cardboard and wants the roomiest open-topped box it can fold, so the material is fixed and the volume is free. This is the shape almost every applied optimisation question takes.
Find(a) Find the dimensions that maximise the volume, and the maximum volume.
Given
open-topped rectangular box with base $x$ by $y$ and height $z$, all positive
surface used: $xy+2xz+2yz=12$
volume: $V=xyz$
Hint 1/4
Three unknowns and one constraint leave two free variables — exactly the setting the Second Derivatives Test was built for.
Hint 2/4
Solve the surface equation for $z$, substitute into $V=xyz$, then set both partials of the resulting $V(x,y)$ to zero.
Hint 3/4
From $xy+2xz+2yz=12$ you get $z=\dfrac{12-xy}{2(x+y)}$, so $V(x,y)=\dfrac{12xy-x^{2}y^{2}}{2(x+y)}$; its numerator after differentiating in $x$ factors as $y^{2}(12-x^{2}-2xy)$.
Hint 4/4
The two conditions $x^{2}+2xy=12$ and $y^{2}+2xy=12$ force $x=y=2$, then $z=1$ and $V=4\ \mathrm{m^{3}}$.
Show solution
Reduce to two variables
$$z=\frac{12-xy}{2(x+y)}$$
the constraint is linear in $z$, so $z$ is the cheapest variable to eliminate
$$V(x,y)=\frac{12xy-x^{2}y^{2}}{2(x+y)}$$
substitute into $V=xyz$; the domain is $x,y>0$ with $xy<12$
The constraint is satisfied exactly: $4+4+4=12$. And the point really beats its neighbours — widening the base to $x=y=2.4$ forces $z=0.65$ and gives $V=3.74<4$, while narrowing to $x=y=1.6$ forces $z=1.5$ and gives $V=3.84<4$.
No boundary case needs checking here: as the base grows the height is squeezed to zero and the volume tends to zero, so the single interior critical point has to be the maximum.
D · interleaved 3 questions
1§10.2 — a rate along a path
This mixes the gradient with the chain rule for a moving point from the previous section. The type is for you to spot.
Find(a) Find $\dfrac{dz}{dt}$ at $t=1$.
Given
$z=x^{2}y$
$x=t^{2}$, $y=t^{3}$
evaluate at $t=1$
Hint 1/4
You are tracking $z$ along a curve, so the rate is the gradient dotted with the velocity of the point.
$F$ differentiable at $P$ and $\nabla F(P)\ne\mathbf 0$
Constrained optimisation in two variables
$$\text{substitute the constraint, then }f_x=f_y=0\text{ and test }D$$
the constraint solves for one variable and the region has no boundary to check
Check yourself
Close the page. On one sheet write, from memory: the formula for $D_{\mathbf u}f$ (and why $\mathbf u$ must be a unit vector), the two facts about the gradient's direction and length, how to get the tangent plane to $F(x,y,z)=k$, the two equations that locate a critical point, and the full Second Derivatives Test including the $D=0$ case. Then open the formula card and mark what was missing — that is your revision list, and the rest of the section is not.
take $D_{\mathbf u}f$ in a given direction, normalising first?
c-directional-derivative
build $\nabla f$ and use $D_{\mathbf u}f=\nabla f\cdot\mathbf u$?
c-gradient
name the steepest direction and give the rate $|\nabla f|$?
c-steepest-ascent
write the tangent plane to a level surface from $\nabla F$?
c-gradient-normal
solve $f_x=0,f_y=0$ together for every critical point?
c-critical-points
classify a critical point with $D$ and the sign of $f_{xx}$?
c-second-derivative-test
find absolute extrema by comparing interior and boundary values?
c-absolute-extrema
turn a constrained word problem into a two-variable critical-point problem and finish with the value that was asked for?
c-second-derivative-test
Glossary (12 terms)
directional derivativeyönlü türev
The rate of change of $f$ per unit distance in a chosen unit direction $\mathbf u$, equal to $\nabla f\cdot\mathbf u$.
gradient vectorgradyan
The vector $\langle f_x,f_y\rangle$ of first partial derivatives; it points in the direction of fastest increase and its length is that fastest rate.
unit vectorbirim vektör
A vector of length $1$; any nonzero $\mathbf v$ becomes one as $\mathbf v/|\mathbf v|$.
level curveseviye eğrisi
The set $f(x,y)=k$ of inputs sharing one output value $k$, drawn in the $xy$-plane. The gradient at a point of the curve is perpendicular to it.
level surfacedüzey yüzeyi
The set where $F(x,y,z)$ equals a constant; the gradient $\nabla F$ is normal to it, giving the tangent plane.
critical pointkritik nokta
A point where $\nabla f=\mathbf 0$ or a partial fails to exist; the only candidates for local extrema.
saddle pointeyer noktası
A critical point that is a maximum along one direction and a minimum along another, so neither a local max nor min.
discriminantdiskriminant
The number $D=f_{xx}f_{yy}-(f_{xy})^{2}$ whose sign classifies a critical point in the Second Derivatives Test.
local maximumyerel maksimum
A point where $f$ is at least as large as at every nearby point.
absolute maximummutlak maksimum
The single largest value of $f$ over an entire region, as opposed to a merely local peak.
normal linenormal doğru
The line through a point of a surface in the direction $\nabla F(P)$; it is perpendicular to the tangent plane and meets it only at that point.
kısıt denklemi
A side condition the variables must satisfy, such as a fixed surface area; solving it for one variable turns a constrained problem into an ordinary two-variable one.
What comes next
§11 · Double integrals over rectangles
Every question so far asked about a single point. Next we add up a function over a whole region — the double integral — starting on a rectangle where the setup is cleanest.
Sources
textbookJames Stewart, Calculus, Ninth Edition — sections 14.6 and 14.7 Section numbers are quoted exactly as they appear on the week 10 line of the course syllabus.
course materialCourse syllabus, week 10 line: Partial Derivatives 14.6, 14.7 Fixes the two topics covered: directional derivatives with the gradient, and maximum and minimum values.
standard resultExtreme Value Theorem for continuous functions on closed, bounded sets The guarantee behind the closed-region procedure for absolute extrema.