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08Quadric surfaces, limits in two variables, and partial derivatives
A thin metal plate is warmed at one corner, and you are standing somewhere on it with a thermometer. Step one centimetre east and the reading climbs; go back, step one centimetre north instead, and it falls. So the question how fast is it getting hotter right here has no single answer — the rate depends on which way you walk, and the two directions along the axes already hand you two different numbers.
By the end of this section you can take a formula in two or three variables and read off its rate of change along each axis, see each of those rates as the slope of a slice through a surface, and handle the four second-order rates including the one identity that saves half the work.
In 60 seconds
To differentiate a function of several variables, freeze every variable but one and use ordinary one-variable calculus; each variable you free gives a different rate.
Dropping the inner factor when a variable is buried: the x-partial of $\sin(xy)$ is $y\cos(xy)$, not $\cos(xy)$.
Treating the frozen variable as if it depended on the live one and bolting on a stray $dy/dx$ term; in a partial, the other variable is a plain constant.
Concluding a two-variable limit exists because it agrees along two paths. Agreement along every straight line still does not settle it.
The graded work tied to this material is quizzes (10% of the course) and homework (10%); the syllabus states no exam weight beyond that, so treat every type on this page as fair game rather than betting on one. The week covers three textbook sections at once, which is why the page is long.
How much time do you have?
10 minutes
You leave able to compute a partial derivative and to kill a limit with two paths. Enough for a find the partials question; not enough for the conceptual traps.
The 60 second card, Formula card, Partial derivatives: freeze one variable, Mistake ledger
45 minutes
You add the two ideas the computation rule cannot give you on its own: why a two-variable limit is a genuinely harder object than a one-variable one, and how the four second partials fit together.
The 60 second card, Reading a surface by its traces, When a two-variable limit fails to exist, Partial derivatives: freeze one variable, Second partials and Clairaut's theorem, Fading ladder, Practice B (computation), Check yourself
Full reading
Everything in the order it was built: the surfaces the rates live on, the limit theory that decides when a rate even makes sense, the rate itself, its second-order relatives, and the interleaved practice that makes you pick the tool yourself.
Hook, Recall first, Try it yourself first, All six concept blocks, Method boxes, Contrast pairs, Fading ladder, Full exam-style question, Practice A to D, Mistake ledger, Check yourself
By the end of this section
Identify a or a from its equation by slicing it into traces in the three coordinate planes.
Show that a two-variable limit does not exist by producing two paths to the point that give different values.
Decide whether a two-variable limit exists by bounding the function or passing to polar coordinates, and say where a function is continuous.
Compute the first partial derivatives of a function of two or three variables and read each one as the slope of a slice.
Compute the four second-order partial derivatives and use Clairaut's theorem to check them or to choose the cheaper order.
Verify that a given function of several variables satisfies a stated partial differential equation.
Syllabus coverage
12.6
Cylinders and quadric surfaces
Treated as the geometry the later rates live on: the trace method that identifies a surface is the same slicing that a partial derivative later differentiates.
covered
14.2
Limits and continuity in higher dimensions
Split over two blocks: showing a limit fails by two paths first, then bounding and polar coordinates for the limits that do exist, and what continuity means on a domain.
covered
14.3
Partial derivatives
Split over three blocks: first partials and their slope reading, then the four second partials with Clairaut's theorem, then three-or-more variables and verifying a partial differential equation.
covered
Recall first
One-variable derivative rules
Power $\frac{d}{dx}x^{n}=nx^{n-1}$, product $(fg)'=f'g+fg'$, quotient $\left(\frac fg\right)'=\frac{f'g-fg'}{g^{2}}$, and chain $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$.
A partial derivative is an ordinary derivative in one variable, so every rule from the one-variable course is used unchanged; only the meaning of the letters changes.
Derivatives of the standard functions
$(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, $(e^{x})'=e^{x}$, $(\ln x)'=1/x$, all with $x$ in radians.
These appear as the outer or inner layer of almost every worked function here, now with the argument built out of two variables.
Functions of several variables and their graphs
A function $z=f(x,y)$ assigns a number to each point of a region in the plane; its graph is a surface in space, and a level curve is the set where $f(x,y)=k$.
Introduced the previous week; the surfaces of this section are the graphs, and traces are the graph sliced by planes.
Two-sided limits in one variable
$\lim_{x\to a}g(x)=L$ requires the same value $L$ whether $x$ approaches $a$ from the left or from the right.
The two-variable limit generalises this, and the whole difficulty is that there are now infinitely many directions of approach, not two.
Try it yourself first (2 questions)
1§08.0 — a one-variable warm-up●○○○○
Before we let a second variable in, one line of ordinary differentiation. Treat every letter other than $x$ as a fixed number.
Given
$g(x)=5x^{3}-7x+2$
Find
(a) Find $g'(x)$.
(b) Evaluate $g'(1)$.
Hint 1/4
This is the plain power rule, term by term; nothing new yet.
Hint 2/4
$\frac{d}{dx}x^{n}=nx^{n-1}$, and a constant differentiates to $0$.
Hint 3/4
With $g(x)=5x^{3}-7x+2$: the cube gives $15x^{2}$, the linear term gives $-7$, the constant gives $0$.
Hint 4/4
So $g'(x)=15x^{2}-7$ and $g'(1)=8$.
Show solutionDifferentiate
$$g'(x)=15x^{2}-7$$
power rule on each term; the constant $2$ contributes nothing
$$g'(1)=15-7=8$$
substitute $x=1$
Answer $$\boxed{g'(x)=15x^{2}-7,\quad g'(1)=8}$$
Check
Order-of-magnitude check: near $x=1$ the cubic term dominates the slope, and $15$ against $-7$ leaves a positive $8$, consistent with $g$ rising there.
2§08.0 — a deliberate trap about approach●●○○○
One belief students carry in from the one-variable course, stated as a claim to judge. It is the exact belief this section has to dismantle, so getting it wrong here is the point.
Given
A function $f(x,y)$ is defined everywhere except at the origin.
Along the x-axis $f\to0$ as you approach the origin, and along the y-axis $f\to0$ as well.
Find
(a) True or false: these two facts already force $\lim_{(x,y)\to(0,0)}f(x,y)=0$.
Hint 1/4
Ask how many ways there are to approach the origin in the plane, versus on a line.
Hint 2/4
A two-variable limit must give the same value along every path, not just two chosen ones.
Hint 3/4
Two axes are two paths out of infinitely many; a third path, even a curved one, could disagree with them.
Hint 4/4
So the statement is false: two agreeing paths are evidence, never a proof.
a point in the plane can be approached along any curve, not only the two axes
$$\lim=0\ \Rightarrow\ \text{same value on every path}$$
the definition demands agreement along all of them, so two is never sufficient
Answer $$\boxed{\text{False}}$$
Check
Concrete refutation ahead: $xy/(x^{2}+y^{2})$ is $0$ on both axes yet $1/2$ along $y=x$, so the pattern of this claim genuinely fails.
Notation
symbol
reads as
means
watch out
$\partial f/\partial x$
the partial derivative of f with respect to x
the rate of change of $f$ as $x$ varies and every other variable is held fixed; the round $\partial$ marks that other variables are present and frozen.
The $\partial$ symbol is not a letter you can cancel: $\partial f/\partial x$ is one indivisible symbol, not a quotient of two quantities $\partial f$ and $\partial x$.
$f_x,\ f_y$
f sub x, f sub y
the same partial derivatives in subscript form; $f_x$ means differentiate with respect to $x$, holding the rest fixed.
The subscript names the variable you differentiate by, not a component or an index. $f_x$ and $f_y$ are usually different functions.
$f_{xy},\ \partial^{2}f/\partial y\,\partial x$
f sub x y, second partial of f
differentiate first with respect to $x$, then with respect to $y$; the two notations denote the same second partial derivative.
The letter order is reversed between the two notations. Subscripts run left to right, the $\partial$ denominators run right to left, and both describe x-then-y here.
$z=f(x,y)$
z equals f of x and y
a surface in space, one height $z$ above each point $(x,y)$ of the plane region where $f$ is defined.
A trace fixes one of $x$, $y$, $z$ to a constant and looks at the resulting curve; it is a slice of this surface, not the surface itself.
Conventions used here
Continuity is asked only on the domain
A function is continuous or not only at points where it is defined. At a point outside the domain we do not say the function is discontinuous there; we ask instead whether the limit exists and whether a value can be assigned to extend the function continuously. So $f(x,y)=xy/(x^2+y^2)$ is continuous at every point of its domain, and the only live question is at the origin, which is not in the domain until we assign a value there.
Angles are in radians
Every derivative of a trigonometric function on this page assumes its argument is in radians, exactly as in the one-variable course. The partial derivatives of $\sin(xy)$ or $\cos(x^2y)$ inherit that assumption through the inner function.
Reading the mixed-partial notation
The subscript $f_{xy}$ means differentiate with respect to $x$ first, then $y$: leftmost subscript first. The Leibniz form for the same object is $\partial^2 f/\partial y\,\partial x$, where the denominators are read right to left. The two notations put the letters in opposite order for the same derivative, and that is the single most common source of confusion in this section.
8.1Reading a surface by its traces
Slice a surface by the planes x = k, y = k, z = k; the cross-sections tell you which standard shape it is.
Last week a function $z=f(x,y)$ became a surface in space. Before we measure how fast it changes, we need to be able to see it, and no hand can plot a surface point by point.
Solvable with what we have
sketch $y=x^{2}$ in the plane by plotting a handful of points and joining them
read off from $y=f(x)$ where a plane curve rises, falls or turns
Not solvable yet
sketch $z=x^{2}+y^{2}$ by plotting points: a scatter of dots in space joins into nothing the eye can read
tell $z=x^{2}+y^{2}$ apart from $z^{2}=x^{2}+y^{2}$ from the formula alone
Make a table of $z$ over a grid of $(x,y)$ and plot the points in space. For $z=x^{2}+y^{2}$ you get a cloud that thins out as you leave the origin, and joining the dots by hand into a smooth surface is exactly the skill you do not have yet.
Why it fails
A surface is two-dimensional and a page is two-dimensional, but the surface lives in three dimensions and the projection throws away depth. The fix is not more dots; it is to cut the surface with flat planes and read the flat curves that result. Those curves are called traces, and three of them usually pin the surface down.
MethodIdentifying a surface by traces
Conditions
the equation is a second-degree (or simpler) relation among $x$, $y$, $z$
you slice with the coordinate planes and with planes parallel to them: set $z=k$, then $x=0$, then $y=0$
$$\boxed{\text{trace in the plane }z=k:\ \text{set }z=k\text{ and read the curve in }x,y;\ \text{likewise for }x=k,\ y=k}$$
To recognise a surface, hold one coordinate fixed at a constant and look at the curve the other two trace out; do this for a horizontal plane and for the two vertical coordinate planes, and the pattern of curves names the surface.
The $\textcolor{#1f6feb}{y=0}$ trace of $z=x^{2}+y^{2}$ is the parabola $z=x^{2}$, passing through $\textcolor{#d1690a}{(2,4)}$. A parabola on the vertical slice is what makes this a paraboloid; a cone $z=\sqrt{x^{2}+y^{2}}$ would slice into a straight V instead.
Looks like this, but is not
$z=x^{2}+y^{2}$ and the cone $z=\sqrt{x^{2}+y^{2}}$ both open upward and are round, so they look like the same shape at different steepness.
Their horizontal traces $z=k$ are circles in both cases, so that view cannot separate them. The vertical trace does: for the paraboloid $y=0$ gives $z=x^{2}$, a parabola, while for the cone it gives $z=|x|$, two straight rays. A curved vertical slice versus a straight one is the whole difference.
Identifying z = x² + y²
The plainest paraboloid, taken apart one plane at a time so the method is visible before the shapes get harder.
Given
$z=x^{2}+y^{2}$
Find
the surface, by its three traces
SolutionHorizontal traces z = k
$$z=k:\ x^{2}+y^{2}=k$$
holding $z$ fixed leaves a relation in $x,y$
$$k>0:\ \text{circle of radius }\sqrt{k};\quad k=0:\ \text{the point }(0,0);\quad k<0:\ \text{empty}$$
a sum of squares cannot be negative, so nothing lies below $z=0$
Vertical traces
$$y=0:\ z=x^{2}$$
one vertical slice, an upward parabola
$$x=0:\ z=y^{2}$$
the other vertical slice, the same parabola turned a quarter-turn
Answer $$\boxed{\text{circular paraboloid, opening upward, vertex at the origin}}$$
Check
Consistency of the two views: at height $z=4$ the circle has radius $2$, and the parabola $z=x^{2}$ reaches $z=4$ at $x=\pm2$ — the same width, so the round bowl and the parabola describe one surface.
Three traces, no algebra beyond reading a sum of squares.
Horizontal traces give the cross-sectional shape; vertical traces give whether the surface is a bowl, a cone or a saddle.
A hyperboloid of one sheet
When a minus sign appears among the squares, the horizontal traces survive at every height and the vertical traces open up into hyperbolas.
Given
$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-z^{2}=1$
Find
the surface, by its traces
SolutionHorizontal traces z = k
$$z=k:\ \frac{x^{2}}{4}+\frac{y^{2}}{9}=1+k^{2}$$
move the $z^{2}$ across; the right side is at least $1$
$$\text{ellipse for every }k,\ \text{smallest at }k=0$$
never empty, so the surface has no gap: a single connected piece
Vertical traces
$$y=0:\ \frac{x^{2}}{4}-z^{2}=1$$
a hyperbola opening left and right
$$x=0:\ \frac{y^{2}}{9}-z^{2}=1$$
a hyperbola opening front and back
Answer $$\boxed{\text{hyperboloid of one sheet, axis along }z}$$
Check
Sign test that names the family: exactly one of the three squared terms carries a minus and the constant is positive, which is the signature of the one-sheet hyperboloid; two minus signs would have given two sheets.
Three traces plus one sign count.
Count the minus signs among the squared terms: none is an ellipsoid or paraboloid, one is a one-sheet hyperboloid, two is a two-sheet hyperboloid.
A cylinder: the missing variable
An equation that names only two of the three variables is a cylinder, and the trap is to read it as a plane curve when it is a surface.
Given
$z=x^{2}$ in space (that is, with $y$ allowed any value)
whatever $(x,z)$ satisfies $z=x^{2}$, every $y$ is allowed with it
$$\text{each fixed }y\ \text{gives the same parabola }z=x^{2}$$
the trace does not change as $y$ moves
Sweep it
$$\text{parabola }z=x^{2}\ \text{swept parallel to the }y\text{-axis}$$
copies of one curve stacked along the free axis form a cylinder
Answer $$\boxed{\text{parabolic cylinder, a parabola extruded along the }y\text{-axis}}$$
Check
Reality check on dimension: a single equation in space cuts three-dimensional space down by one dimension, giving a surface, not the one-dimensional parabola you would get in the plane.
A missing variable is not a typo; it is the direction the cross-section is swept along.
Checkpoint
§08.1 — naming a surface from one minus sign●●○○○
Thirty seconds. You are handed an equation and asked only for the family, not a sketch.
Given
$x^{2}+y^{2}-z^{2}=1$
Find
(a) Which standard surface is this?
Hint 1/4
Count how many of the squared terms carry a minus sign, and check whether the horizontal traces ever run out.
Hint 2/4
Set $z=k$: the trace is $x^{2}+y^{2}=1+k^{2}$, and a positive right-hand side means a circle at every height.
Hint 3/4
One minus sign among the squares with a positive constant, and non-empty traces for all $z$, is the one-sheet hyperboloid.
Hint 4/4
It is a hyperboloid of one sheet.
Show solutionTraces
$$z=k:\ x^{2}+y^{2}=1+k^{2}$$
a circle for every $k$, so one connected piece
$$y=0:\ x^{2}-z^{2}=1$$
a hyperbola, confirming the family
Answer $$\boxed{\text{hyperboloid of one sheet}}$$
Check
One minus sign, positive constant, no empty traces: the signature is internally consistent.
⚠ Reading a cylinder as a plane curve
the equation looks exactly like a parabola from the one-variable course, so the missing variable is not noticed
wrong$$z=x^{2}\ \text{is a parabola}$$
right$$z=x^{2}\ \text{in space is a parabolic cylinder (a surface)}$$
⚠ Judging the family from horizontal traces alone
the circular traces of a paraboloid and a cone look identical, and the vertical slice is the one nobody draws
right$$\text{vertical trace parabola}\ \Rightarrow\ \text{paraboloid, not cone}$$
8.2When a two-variable limit fails to exist
In the plane a point can be approached along infinitely many paths; two paths with different values prove the limit does not exist.
In one variable a limit had two approaches to reconcile, left and right. A point in the plane can be reached along any curve at all, and that is the entire new difficulty.
RuleThe two-path test
Conditions
$C_1$ and $C_2$ are two paths that both end at the point $(a,b)$
$f$ tends to $L_1$ along $C_1$ and to $L_2$ along $C_2$
$$\boxed{\ L_1\neq L_2\ \Longrightarrow\ \lim_{(x,y)\to(a,b)}f(x,y)\ \text{does not exist}\ }$$
If the function heads for two different numbers along two different routes into the point, then no single number can be the limit, so the limit does not exist.
Two routes into the origin for $f=\dfrac{xy}{x^{2}+y^{2}}$: along $\textcolor{#1f6feb}{y=0}$ the value stays $0$, along $\textcolor{#d1690a}{y=x}$ it stays $\tfrac12$. Two destinations, so the limit does not exist.
Looks like this, but is not
For $f=\dfrac{x^{2}y}{x^{4}+y^{2}}$, take any straight line $y=mx$ into the origin; the value tends to $0$ every single time, which looks like overwhelming evidence that the limit is $0$.
Straight lines are not the only paths. Along the parabola $y=x^{2}$ the value is $\dfrac{x^{2}\cdot x^{2}}{x^{4}+x^{4}}=\dfrac12$ for all $x\neq0$. Every line says $0$, one parabola says $\tfrac12$, so the limit does not exist. Lines are never enough on their own.
The standard xy over x² + y²
The first limit worth killing, because both paths are lines and the arithmetic is short.
Given
$f(x,y)=\dfrac{xy}{x^{2}+y^{2}}$
approach $(0,0)$
Find
whether $\lim_{(x,y)\to(0,0)}f$ exists
SolutionPath along the x-axis
$$y=0:\ f=\frac{x\cdot0}{x^{2}+0}=0$$
choose the simplest path first; a zero in the numerator makes it vanish
pick a second line that keeps numerator and denominator the same order
$$\Rightarrow L_2=\tfrac12\neq L_1$$
two paths, two values
Answer $$\boxed{\text{the limit does not exist}}$$
Check
A third route confirms the disorder rather than a value: along $y=2x$, $f=\dfrac{2x^{2}}{5x^{2}}=\tfrac25$, yet another number, exactly as a non-existent limit should behave.
Two lines, two substitutions.
When the numerator and denominator are the same degree, try $y=mx$: if the value depends on $m$, the limit is already dead.
Lines agree but a parabola does not
The example that teaches why straight lines can never finish the job.
Answer $$\boxed{\text{the limit does not exist}}$$
Check
Why the parabola is the right probe: the denominator pairs $x^{4}$ with $y^{2}$, so $y\sim x^{2}$ makes the two terms the same size and stops the fraction from vanishing — the choice is forced by the exponents, not lucky.
One family of lines, one parabola.
If all lines give the same value, do not conclude anything yet; match the exponents in the denominator with a curved path before you trust it.
Checkpoint
§08.2 — kill a limit with two paths●●○○○
Thirty seconds, and only the two-path idea is being tested.
Given
$f(x,y)=\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}$
Find
(a) Show that $\lim_{(x,y)\to(0,0)}f$ does not exist.
Hint 1/4
You only need two routes into the origin that disagree; try the two axes.
Hint 2/4
Evaluate along $y=0$ and along $x=0$ and compare the two numbers.
Hint 3/4
Along $y=0$: $f=x^{2}/x^{2}=1$. Along $x=0$: $f=-y^{2}/y^{2}=-1$.
Hint 4/4
The two paths give $1$ and $-1$, so by the two-path test the limit does not exist.
Show solutionThe two axes
$$y=0:\ f=\frac{x^{2}}{x^{2}}=1$$
the $y$-terms vanish
$$x=0:\ f=\frac{-y^{2}}{y^{2}}=-1$$
the $x$-terms vanish
$$1\neq-1$$
two paths disagree
Answer $$\boxed{\text{limit does not exist}}$$
Check
A diagonal check keeps it honest: along $y=x$, $f=0$, a third distinct value.
⚠ Treating agreement along lines as a proof
checking a few lines feels exhaustive, and the definition's word every is easy to skip
wrong$$\text{all lines give }0\ \Rightarrow\ \lim=0$$
right$$\text{all lines give }0\ \Rightarrow\ \text{try a parabola; not yet decided}$$
⚠ Plugging in the point directly
it is the first move in one-variable limits, but here it gives $0/0$ and hides the path-dependence
wrong$$f(0,0)=\tfrac{0}{0}=0$$
right$$\tfrac{0}{0}\ \text{is indeterminate; test paths instead}$$
8.3Limits that do exist, and continuity
To prove a limit exists, bound the function by something that vanishes regardless of direction; polar coordinates make the bound easy.
The two-path test can only ever say no. To say yes — that a limit really is some number — you have to control the function along all paths at once, and a bound does that where a list of paths cannot.
TheoremSqueeze and polar coordinates for existence
Conditions
you want $\lim_{(x,y)\to(0,0)}f=0$
there is a bound $|f(x,y)|\le g(r)$ after writing $x=r\cos\theta,\ y=r\sin\theta$, with $g(r)\to0$ as $r\to0$ and $g$ not depending on $\theta$
If, once distance from the origin is called r, the size of the function is trapped under something that shrinks to zero with r no matter what the angle is, then the function is squeezed to zero and the limit exists.
Looks like this, but is not
$f(x,y)=\dfrac{xy}{x^{2}+y^{2}}$, defined to be $0$ at the origin, looks continuous everywhere: it is a tidy ratio of polynomials and we filled the one hole.
In polar form $f=\cos\theta\sin\theta$, a value that depends only on the angle and not on $r$. Approaching along $\theta=\tfrac\pi4$ gives $\tfrac12$, not the assigned $0$, so the limit at the origin does not exist and no assigned value can make $f$ continuous there. Filling a hole is not the same as the hole closing.
A limit that exists, by polar coordinates
The positive counterpart of the previous block: here the bound survives, so the limit is a number.
Independent path check: along $y=0$ the function is exactly $0$, and along $y=x$ it is $3r\cos^{2}\theta\sin\theta$ at $\theta=\tfrac\pi4$, which also runs to $0$ — consistent with the squeeze rather than a coincidence.
One polar substitution, one bound.
A leftover factor of $r$ after cancelling is the signal that the limit is $0$; a value with no $r$ in it is the signal that it is not.
Where is (x² - y²)/(x - y) continuous
A function with an apparent break that is really removable, and the discipline of answering on the domain.
Extension check: since $f=x+y$ off the line, defining $f=x+y$ on $x=y$ too gives a function continuous on the whole plane, so the break is removable — but the original $f$ is answered only on its domain.
Answer continuity on the domain first; whether a break is removable is a separate, second question.
Checkpoint
§08.3 — a limit that survives●●○○○
Thirty seconds. One polar substitution decides it.
Given
$f(x,y)=\dfrac{x^{3}}{x^{2}+y^{2}}$
Find
(a) Find $\lim_{(x,y)\to(0,0)}f$, or show it does not exist.
Hint 1/4
Try polar coordinates and watch whether a factor of $r$ is left over on top.
Hint 2/4
With $x=r\cos\theta,\ y=r\sin\theta$ the denominator is $r^{2}$; simplify.
Hint 3/4
$f=\dfrac{r^{3}\cos^{3}\theta}{r^{2}}=r\cos^{3}\theta$, and $|f|\le r$.
Hint 4/4
Since $r\to0$ bounds it regardless of $\theta$, the limit is $0$.
right$$\text{it is undefined on}\ x=y;\ \text{continuous on its domain}$$
8.4Partial derivatives: freeze one variable
The partial derivative in x is the ordinary derivative you get by treating every other variable as a constant.
We can now say when a function of two variables is continuous. The hook's question — how fast is it changing here — splits into one rate per direction, and the two axis directions are the place to start.
The x-partial is the rate of change of f as x alone moves, with y pinned at b; the y-partial is the rate as y alone moves, with x pinned at a. In practice: hold the other variable fixed as a constant and differentiate normally.
Holding $y=b$ cuts the surface into a single $\textcolor{#1f6feb}{\text{trace curve}}$ in the $x$ direction. The partial derivative $f_x(a,b)$ is the $\textcolor{#128a5a}{\text{slope of the tangent}}$ to that curve at $x=a$ — an ordinary one-variable slope on a one-variable slice.
Looks like this, but is not
Differentiating $x^{2}y$ in $x$, it is tempting to treat $y$ as a function of $x$ and write a product-rule term $2xy+x^{2}\dfrac{dy}{dx}$, echoing implicit differentiation.
In a partial derivative $y$ is an independent variable held fixed, not a function of $x$, so $\dfrac{dy}{dx}$ has no meaning here and there is no such term. The correct partial is simply $\dfrac{\partial}{\partial x}(x^{2}y)=2xy$, with $y$ riding along as a constant.
Both partials of x³y² - 2x + 5y
The first computation, done twice, so the freeze-the-other-variable move is unmistakable.
Given
$f(x,y)=x^{3}y^{2}-2x+5y$
Find
$f_x$, $f_y$, and their values at $(1,2)$
SolutionPartial in x: freeze y
$$f_x=3x^{2}y^{2}-2$$
treat $y$ as a constant: $y^{2}$ is a constant multiplier on $x^{3}$, the $5y$ term is constant so its derivative is $0$
$$f_x(1,2)=3\cdot1\cdot4-2=10$$
substitute
Partial in y: freeze x
$$f_y=2x^{3}y+5$$
now $x^{3}$ is the constant multiplier; $-2x$ is constant in $y$ so it drops
$$f_y(1,2)=2\cdot1\cdot2+5=9$$
substitute
Answer $$\boxed{f_x(1,2)=10,\qquad f_y(1,2)=9}$$
Check
Slope check by slicing: fix $y=2$, so $g(x)=4x^{3}-2x+10$ and $g'(1)=12-2=10$; fix $x=1$, so $h(y)=y^{2}+5y-2$ and $h'(2)=4+5=9$ — both partials reproduced from genuine one-variable derivatives.
Two ordinary derivatives, one per variable.
Every term that does not contain the live variable is a constant and differentiates to zero; that is what makes a partial cheaper than it looks.
Partials of x e^{xy}: product meets frozen chain
Now the frozen variable sits inside an exponential, so the chain rule fires with $y$ as a constant coefficient.
Point check at $(1,0)$: $f_x=e^{0}(1+0)=1$ and $f_y=1\cdot e^{0}=1$. Independently, the slice $x=1$ is $h(y)=e^{y}$ with $h'(0)=1$, matching $f_y(1,0)$.
One product rule and one chain rule, each with a frozen constant.
A frozen variable inside a composite becomes a constant factor when the chain rule reaches it; it never disappears.
Partials of sin(x²y)
A pure chain-rule case, and the one where the inner factor is dropped most often.
Ratio check: $f_x/f_y=2xy/x^{2}=2y/x$, which is exactly the ratio of the two inner derivatives $\partial_x(x^{2}y)$ and $\partial_y(x^{2}y)$ — as it must be, since the shared outer factor cancels.
One chain rule per variable.
The outer derivative is the same for both partials; only the inner derivative changes, so compute the outer once and swap the inside factor.
Checkpoint
§08.4 — first partials at a point●○○○○
Thirty seconds: two short derivatives and a substitution.
Given
$f(x,y)=x^{2}+3xy+y^{2}$
Find
(a) Find $f_x$ and $f_y$.
(b) Evaluate both at $(1,1)$.
Hint 1/4
Differentiate once with $y$ held constant, once with $x$ held constant.
Hint 2/4
The cross term $3xy$ contributes $3y$ to $f_x$ and $3x$ to $f_y$.
Hint 3/4
$f_x=2x+3y$ and $f_y=3x+2y$; now put $x=y=1$.
Hint 4/4
Both equal $2+3=5$ at $(1,1)$.
Show solutionBoth partials
$$f_x=2x+3y$$
$y^{2}$ is constant in $x$
$$f_y=3x+2y$$
$x^{2}$ is constant in $y$
$$f_x(1,1)=f_y(1,1)=5$$
substitute
Answer $$\boxed{f_x(1,1)=f_y(1,1)=5}$$
Check
Symmetry note: $f$ is unchanged if $x$ and $y$ swap, so equal partials at the symmetric point $(1,1)$ is expected.
⚠ Adding a dy/dx term to a partial derivative
implicit differentiation is fresh from last term, and the two operations look alike
Differentiate twice to get four second partials; the two mixed ones are equal whenever they are continuous, so pick the easier order.
Each partial $f_x$ and $f_y$ is itself a function of $x$ and $y$, so it can be differentiated again. Doing that in every combination gives four second-order partials, not two.
TheoremClairaut's theorem on mixed partials
Conditions
$f_{xy}$ and $f_{yx}$ are both continuous on a disk around $(a,b)$
$$\boxed{\ f_{xy}(a,b)=f_{yx}(a,b)\ }$$
If the two mixed second partials are continuous near a point, then differentiating x-then-y gives the same result as y-then-x there; the order of the two differentiations does not matter.
From $f$ there are two routes to the mixed second partial: differentiate by $x$ then by $y$, or by $y$ then by $x$. Clairaut's theorem says both routes end at the same function, $f_{xy}=f_{yx}$, when it is continuous.
Looks like this, but is not
The notations $f_{xy}$ and $\partial^{2}f/\partial x\,\partial y$ put $x$ and $y$ in opposite orders, so they look like they must mean opposite orders of differentiation.
They do not. $f_{xy}$ reads left to right — $x$ first, then $y$ — while $\partial^{2}f/\partial x\,\partial y$ reads the denominators right to left — $y$ first, then $x$. So $f_{xy}=\partial^{2}f/\partial y\,\partial x$, and the object that matches $\partial^{2}f/\partial x\,\partial y$ is $f_{yx}$. When the function is nice they are equal anyway, which is the only reason the clash is survivable.
All four second partials, and Clairaut checked
The full second-order picture on one function, with the mixed pair computed both ways to see them agree.
Given
$f(x,y)=x^{3}y^{2}+e^{xy}$
Find
the four second partials, and a check that the mixed ones agree
SolutionFirst partials
$$f_x=3x^{2}y^{2}+y\,e^{xy}$$
differentiate in $x$; the inner derivative of $e^{xy}$ is $y$
$$f_y=2x^{3}y+x\,e^{xy}$$
differentiate in $y$; the inner derivative of $e^{xy}$ is $x$
Pure second partials
$$f_{xx}=6xy^{2}+y^{2}e^{xy}$$
differentiate $f_x$ again in $x$; the $y\,e^{xy}$ term gives $y\cdot y e^{xy}$
$$f_{yy}=2x^{3}+x^{2}e^{xy}$$
differentiate $f_y$ again in $y$; the $x\,e^{xy}$ term gives $x\cdot x e^{xy}$
Mixed second partials, both orders
$$f_{xy}=(f_x)_y=6x^{2}y+e^{xy}+xy\,e^{xy}$$
differentiate $f_x$ in $y$; product rule on $y\,e^{xy}$ gives $e^{xy}+y\cdot x e^{xy}$
$$f_{yx}=(f_y)_x=6x^{2}y+e^{xy}+xy\,e^{xy}$$
differentiate $f_y$ in $x$; product rule on $x\,e^{xy}$ gives $e^{xy}+x\cdot y e^{xy}$
The check is built in: the two mixed partials were computed by independent routes and landed on the same expression, which is exactly what Clairaut's theorem promises for this .
Two first partials, then four second derivatives.
Computing both mixed orders is the cheapest self-check you have: if they disagree, there is an arithmetic slip in one of them.
Using Clairaut to pick the cheaper order
When one order of differentiation is much shorter, Clairaut lets you take it and trust the answer.
Given
$f(x,y)=x^{2}\ln(1+y^{4})$
Find
the mixed partial $f_{xy}$
SolutionChoose the order y then x
$$f_y=x^{2}\cdot\frac{4y^{3}}{1+y^{4}}$$
differentiate in $y$ first; the messy $\ln$ derivative happens once, with $x^{2}$ a constant
Sanity of the other order: $f_x=2x\ln(1+y^{4})$, and differentiating in $y$ gives $2x\cdot\dfrac{4y^{3}}{1+y^{4}}$ — the same, confirming the order swap was legitimate.
One quotient-free chain in the easy order.
Differentiate in the order that isolates the ugly factor from the variable you differentiate last; Clairaut guarantees the answer is order-independent.
Checkpoint
§08.5 — one mixed partial●○○○○
Thirty seconds. Differentiate twice, in whichever order you like.
Given
$f(x,y)=x^{2}y^{3}$
Find
(a) Find $f_{xy}$.
Hint 1/4
Take one partial, then differentiate the result by the other variable.
Hint 2/4
Either order works by Clairaut; start with $f_x=2xy^{3}$.
Hint 3/4
Differentiate $f_x=2xy^{3}$ in $y$: bring down the $3$ from $y^{3}$.
Hint 4/4
$f_{xy}=6xy^{2}$.
Show solutionTwo derivatives
$$f_x=2xy^{3}$$
differentiate in $x$
$$f_{xy}=6xy^{2}$$
then in $y$
Answer $$\boxed{f_{xy}=6xy^{2}}$$
Check
Other order: $f_y=3x^{2}y^{2}$, then $f_{yx}=6xy^{2}$ — equal, as Clairaut requires.
⚠ Reading the mixed-partial notation in the wrong order
the subscript and the Leibniz forms genuinely reverse the letters, so one of them is always counter-intuitive
every classroom example is continuous, so the continuity condition is easy to forget it is a condition at all
wrong$$f_{xy}=f_{yx}\ \text{always}$$
right$$f_{xy}=f_{yx}\ \text{when both are continuous}$$
8.6Three or more variables, and checking a PDE
Nothing needed exactly two variables; freeze all but one. A partial differential equation is verified by plugging the function's partials in.
The freeze-the-others rule never counted the variables. With three or more, freeze all but the live one and differentiate as before; the only new use is checking equations built from partials.
NotePartials in more variables, and PDEs
Conditions
for $f(x,y,z)$, each partial holds the other two variables constant
a partial differential equation (PDE) relates a function to its partial derivatives
A partial with respect to one variable freezes all the others, no matter how many there are; and to check that a function solves a PDE, compute the partials the equation names and see whether the two sides match.
Looks like this, but is not
Verifying that $u$ satisfies Laplace's equation might look like it means solving the equation, or deriving $u$ from it.
Verifying is only substitution: compute $u_{xx}$ and $u_{yy}$ from the given $u$ and check the sum is zero. It confirms $u$ is a solution; it does not find $u$, and it says nothing about the other solutions the equation has.
Partials of a three-variable function
The rule unchanged, now with two variables frozen at a time.
Symmetry check: $f$ is unchanged under the cycle $x\to y\to z\to x$, so at the symmetric point $(1,1,1)$ the three partials must coincide — and they do.
Three ordinary derivatives, two variables frozen each time.
More variables cost more partials, not a harder rule; each one is still a one-variable derivative.
Verifying Laplace's equation
A named PDE checked by pure substitution, the way every such problem on the course is.
Structural check: $u_{xx}$ and $u_{yy}$ share a denominator and have exactly opposite numerators, so their sum is forced to zero without re-expanding anything.
One chain rule, one quotient rule, one symmetry shortcut.
When a function is symmetric in $x$ and $y$, compute one second partial and read the other off by swapping the letters.
Verifying the wave equation
A time variable now, so the second partial in $t$ meets the second partial in $x$.
Answer $$\boxed{u_{tt}=c^{2}u_{xx},\ \text{so }u\ \text{solves the wave equation}}$$
Check
Dimensional read: each $t$-derivative pulls out a factor $-c$, so two of them give $c^{2}$, which is exactly the constant the wave equation puts in front of $u_{xx}$.
Two derivatives in each of two variables.
A travelling shape $f(x-ct)$ solves the wave equation for any $f$; here $f=\sin$, but the factor-of-$c$ bookkeeping is the general reason.
Checkpoint
§08.6 — a three-variable partial●○○○○
Thirty seconds. Freeze two variables, differentiate in the third.
Given
$f(x,y,z)=xyz+z^{2}$
Find
(a) Find $f_z$.
(b) Evaluate $f_z$ at $(2,3,1)$.
Hint 1/4
Hold $x$ and $y$ fixed; only $z$ is live.
Hint 2/4
$xyz$ is a constant $xy$ times $z$, and $z^{2}$ differentiates in the usual way.
Hint 3/4
$f_z=xy+2z$; now put $x=2,\ y=3,\ z=1$.
Hint 4/4
$f_z(2,3,1)=6+2=8$.
Show solutionPartial in z
$$f_z=xy+2z$$
$xy$ is the constant coefficient of $z$
$$f_z(2,3,1)=6+2=8$$
substitute
Answer $$\boxed{f_z(2,3,1)=8}$$
Check
Cross-check: $f_x=yz=3$ and $f_y=xz=2$ at the point, all small and consistent with a mostly-linear $f$.
⚠ Thinking a PDE check requires solving the equation
the word equation suggests solving, but here the solution is handed to you
right$$\text{substitute the given}\ u\ \text{and check}\ u_{xx}+u_{yy}=0$$
⚠ Losing a constant factor from the inner derivative
with a constant like $c$ inside, two derivatives multiply two copies of it and one is easily dropped
wrong$$u_{tt}=-c\sin(x-ct)$$
right$$u_{tt}=-c^{2}\sin(x-ct)$$
Computing a partial derivative
you need the rate of change of a function of several variables in one variable's direction.
Pick the live variable
Decide which variable you are differentiating by. Every other variable is now a frozen constant for the whole calculation.
Rewrite hidden layers
Turn roots and reciprocals into powers, exactly as in one-variable work, so the rules see what they need.
Differentiate normally
Apply the ordinary power, product, quotient and chain rules in the live variable. Frozen variables ride along as constant coefficients, and any term with no live variable in it differentiates to zero.
Repeat and check
Do the same for the next variable. Check one partial by slicing: fix the other variable at a number and differentiate the resulting one-variable function at a point.
Where it goes wrong
Adding a stray dy/dx term as if the frozen variable depended on the live one.
Dropping the inner factor when the live variable is buried inside a composite.
Deciding a two-variable limit
a limit at a point gives 0/0 and you must decide whether it exists.
Try two easy paths
Substitute $y=0$ and $x=0$ (or $y=mx$). If two paths give different values, stop: the limit does not exist.
If lines agree, try a curve
Match the exponents in the denominator with a path like $y=x^{2}$. Lines agreeing is never a proof.
To prove existence, bound
Go to polar coordinates $x=r\cos\theta,\ y=r\sin\theta$. If $|f|\le g(r)$ with $g(r)\to0$ and no $\theta$ left, the limit is 0.
Read the r that survives
A leftover positive power of $r$ means the limit is 0; a value with only $\theta$ in it means it depends on direction and usually does not exist.
Where it goes wrong
Concluding a limit from agreement along lines only.
Leaving a $\theta$ in the bound and still claiming the limit is 0.
Same degree top and bottom: xy/(x²+y²)
A ratio whose numerator and denominator have the same degree.
the leftover $r$ squeezes it to zero regardless of angle
Answer $$\boxed{\text{limit }=0}$$
Check
The bound $3r$ carries no $\theta$, so every path is controlled at once.
Both are 0/0 ratios with $x^{2}+y^{2}$ underneath, but one leaves only an angle after cancelling and the other leaves a power of $r$.
How to tell them apart
In polar form, a surviving positive power of $r$ forces the limit to 0; if the $r$ cancels completely and only $\theta$ remains, the value depends on direction and the limit does not exist.
Scaffolding comes off
The common skeleton
Decide which variable is live; freeze the others as constants.
Rewrite any hidden layer (roots, reciprocals) as a power.
Differentiate in the live variable with the ordinary one-variable rules, keeping every inner factor.
Repeat for the other variable.
Simplify, then check by slicing: fix the other variable at a number and differentiate that one-variable function.
1 · fully worked
Both partials of x⁴ + 6xy³ - y
The skeleton run in full, with every reason spelled out.
Given
$f(x,y)=x^{4}+6xy^{3}-y$
Find
$f_x$, $f_y$, and their values at $(1,1)$
SolutionPartial in x: freeze y
$$f_x=4x^{3}+6y^{3}$$
$x^{4}$ gives $4x^{3}$; $6xy^{3}$ has $6y^{3}$ as a constant multiplier on $x$; $-y$ is constant in $x$
$$f_x(1,1)=4+6=10$$
substitute
Partial in y: freeze x
$$f_y=18xy^{2}-1$$
$x^{4}$ is constant in $y$; $6xy^{3}$ gives $18xy^{2}$; $-y$ gives $-1$
Easier algebra, and it is already done for you. Your job is the missing column: write why each line is allowed, in your own words, before opening the model answers. Writing the reason is the part that transfers; copying the algebra is not.
$f(x,y)=x^{2}y+3y$
reasoning
Line 1 is the function. For $f_x$ we will read $y$ as a fixed number; for $f_y$ we will read $x$ as one.
$f_x=2xy$
reasoning
Line 2 ($f_x$): freeze $y$. Then $x^{2}y$ is (constant)$\cdot x^{2}$, whose derivative is $2xy$, and $3y$ has no $x$ so it contributes $0$.
$f_y=x^{2}+3$
reasoning
Line 3 ($f_y$): freeze $x$. Then $x^{2}y$ is (constant)$\cdot y$ giving $x^{2}$, and $3y$ gives $3$.
3 · find the buried error
Harder than the rung above, and nobody is checking your work now. The solution below computes both partials of $f(x,y)=x^{2}\sin(xy)$ and then evaluates $f_x$ at $(1,\pi)$. It contains exactly two errors. Each line follows in form from the one before, so you cannot spot them by looking for a jump.
The mixed partials were reached by two independent routes and matched; that agreement is the check, and it also catches any slip in either first partial.
Two first partials, two second partials, one substitution.
On a multi-part partials question, the mixed-partial agreement is a free correctness check on everything above it.
Practice
A · concept 4 questions
1§08.1 — a cone hiding among the paraboloids●●○○○
A quick classification, testing whether the vertical trace is what you reach for.
Given
$z^{2}=x^{2}+y^{2}$
Find
(a) Which surface is this?
Hint 1/4
Look at the vertical trace $y=0$ before anything else.
Hint 2/4
Set $y=0$: $z^{2}=x^{2}$, so $z=\pm x$, two straight lines through the origin.
Hint 3/4
Straight vertical traces meeting at a point, with circular horizontal traces $x^{2}+y^{2}=z^{2}$, is a cone.
Hint 4/4
It is a (double) cone with vertex at the origin.
Show solutionTraces
$$y=0:\ z^{2}=x^{2}\Rightarrow z=\pm x$$
two lines, not a parabola
$$z=k:\ x^{2}+y^{2}=k^{2}$$
circles shrinking to a point at $k=0$
Answer $$\boxed{\text{a cone}}$$
Check
Contrast: the paraboloid $z=x^{2}+y^{2}$ has the curved vertical trace $z=x^{2}$ instead.
2§08.2 — do axis limits settle it●●○○○
A statement to judge, aimed at the belief the two-path test exists to break.
Given
$f$ tends to the same value $L$ along the x-axis and along the y-axis into $(0,0)$.
Find
(a) True or false: this forces $\lim_{(x,y)\to(0,0)}f=L$.
Hint 1/4
Ask whether two axes exhaust the ways of approaching the origin.
Hint 2/4
A two-variable limit must agree along every path, and the axes are only two of them.
Hint 3/4
A curved path like $y=x^{2}$ can disagree with both axes, as the example $x^{2}y/(x^{4}+y^{2})$ shows.
Hint 4/4
So the statement is false: two agreeing paths are evidence, not proof.
Show solutionReason
$$\text{limit}=L\ \Rightarrow\ f\to L\ \text{on every path}$$
the definition quantifies over all approaches
$$\text{two axes}\subsetneq\text{all paths}$$
so two agreeing paths cannot force it
Answer $$\boxed{\text{False}}$$
Check
$x^{2}y/(x^{4}+y^{2})$ is $0$ on both axes yet $1/2$ along $y=x^{2}$.
3§08.4 — do partials imply continuity●●●○○
A claim that feels like the one-variable theorem but is not.
Given
Both $f_x(0,0)$ and $f_y(0,0)$ exist.
Find
(a) True or false: therefore $f$ is continuous at $(0,0)$.
Hint 1/4
Recall that partials only probe the two axis directions, while continuity is about every direction.
Hint 2/4
Existence of $f_x$ and $f_y$ says the two axis slices are differentiable, nothing about diagonal or curved approaches.
Hint 3/4
The function $xy/(x^{2}+y^{2})$ with value $0$ at the origin has $f_x(0,0)=f_y(0,0)=0$, yet it is not continuous there.
Hint 4/4
So the statement is false: partials existing does not give continuity.
Path check along $y=x$: $f=\dfrac{x^{4}}{2x^{2}}=\dfrac{x^{2}}{2}\to0$.
3§08.5 — a mixed partial●●○○○
Exam-style: one mixed second partial of a product.
Given
$f(x,y)=x^{2}e^{y}$
Find
(a) Which is $f_{xy}$?
Hint 1/4
Differentiate in $x$ first, then in $y$.
Hint 2/4
$f_x=2x e^{y}$ (with $e^{y}$ constant in $x$).
Hint 3/4
Now differentiate $f_x=2x e^{y}$ in $y$: the $2x$ is constant.
Hint 4/4
$f_{xy}=2x e^{y}$.
Show solutionTwo derivatives
$$f_x=2x e^{y}$$
$e^{y}$ constant in $x$
$$f_{xy}=2x e^{y}$$
$2x$ constant in $y$
Answer $$\boxed{f_{xy}=2x e^{y}}$$
Check
Other order: $f_y=x^{2}e^{y}$, then $f_{yx}=2x e^{y}$ — equal, as Clairaut says.
4§08.1 — two minus signs●●●○○
Exam-style classification from the sign pattern.
Given
$\dfrac{x^{2}}{4}-y^{2}-z^{2}=1$
Find
(a) Which surface is this?
Hint 1/4
Count the minus signs among the squared terms and test whether some heights give empty traces.
Hint 2/4
Set $x=k$: $\dfrac{k^{2}}{4}-1=y^{2}+z^{2}$, which needs $k^{2}\ge4$, so no trace for $|k|<2$.
Hint 3/4
Two minus signs and a forbidden band of $x$-values is the two-sheet hyperboloid, axis along $x$.
Hint 4/4
It is a hyperboloid of two sheets.
Show solutionTraces
$$x=k:\ y^{2}+z^{2}=\tfrac{k^{2}}{4}-1$$
a circle only when $k^{2}\ge4$
$$|k|<2:\ \text{empty}$$
a gap splits the surface into two pieces
Answer $$\boxed{\text{hyperboloid of two sheets}}$$
Check
Two minus signs among the squares is the signature, matching the empty band.
D · interleaved 3 questions
1§08.4 — reading a rate off a formula●●○○○
A partial derivative with a physical reading, mixing in the surface-and-level-curve language from the previous week. Decide which way the plate is heating.
Given
A plate has temperature $T(x,y)=100-x^{2}-2y^{2}$ degrees at the point $(x,y)$.
Find
(a) Find $T_x$ and $T_y$.
(b) At the point $(3,1)$, is the temperature rising or falling as you step in the $+x$ direction?
Hint 1/4
The rate in the $+x$ direction is exactly $T_x$; its sign answers part (b).
Hint 2/4
Differentiate with the other variable frozen: $T_x=-2x$, $T_y=-4y$.
Hint 3/4
At $(3,1)$: $T_x=-6$, which is negative.
Hint 4/4
A negative $T_x$ means the temperature falls as $x$ increases there.
Show solutionPartials
$$T_x=-2x,\quad T_y=-4y$$
freeze the other variable
$$T_x(3,1)=-6<0$$
negative rate in $x$
Answer $$\boxed{T_x=-2x,\ T_y=-4y;\ \text{falling, since }T_x(3,1)=-6}$$
Check
The level curves $T=\text{const}$ are ellipses $x^{2}+2y^{2}=\text{const}$ shrinking toward the hot centre $(0,0)$, so moving away from centre must cool — consistent with $T_x<0$ at $(3,1)$.
2§08.1 — a surface meets a plane●●○○○
Identify a surface and then read the trace a given plane cuts, tying the quadric to the plane language from the vectors week.
Given
$z=x^{2}+y^{2}$
the horizontal plane $z=9$
Find
(a) What curve is the trace in the plane $z=9$?
Hint 1/4
Substitute the plane's equation into the surface and read the resulting relation in $x,y$.
Hint 2/4
Setting $z=9$ gives $x^{2}+y^{2}=9$.
Hint 3/4
That is a circle in the plane $z=9$; find its radius.
Hint 4/4
A circle of radius $3$ centred on the $z$-axis.
Show solutionTrace
$$9=x^{2}+y^{2}$$
put $z=9$ into the surface
$$\text{circle, radius }3$$
a sum of squares equal to $9$
Answer $$\boxed{\text{circle of radius }3\ \text{in the plane}\ z=9}$$
Check
Consistent with the paraboloid widening as $\sqrt{z}$: at $z=9$ the radius is $\sqrt{9}=3$.
3§08.6 — a decaying wave and the heat equation●●●○○
Verify a PDE for a function that mixes an exponential decay in time with a sine in space, pulling together the exponential from earlier weeks with the new partials.
Given
$u(x,t)=e^{-t}\sin x$
Find
(a) Show that $u$ satisfies the heat equation $u_t=u_{xx}$.
Hint 1/4
Compute $u_t$ (freeze $x$) and $u_{xx}$ (freeze $t$) separately, then compare.
Close the page. On one sheet, from memory: how you tell a paraboloid from a cone using one trace; the one sentence that says why two agreeing paths never prove a limit; the rule for computing a partial derivative in a single line; what Clairaut's theorem says and the one condition it needs; and how you check that a function solves a PDE. Then open the formula card and mark what was missing. What you missed is your revision list; the rest of this section is not.
Name a surface from its equation by reading its horizontal and vertical traces, and say which trace separates a paraboloid from a cone?
c-quadric-surfaces
Show a two-variable limit does not exist by producing two paths with different values, and explain why lines alone are not enough?
c-limits-2d
Prove a limit is zero with a polar bound, and state where a function like $(x^{2}-y^{2})/(x-y)$ is continuous?
c-continuity-2d
Compute both first partials of a function built from products, quotients and composites, and read one as the slope of a slice?
c-partial-derivatives
Find all four second partials and use Clairaut's theorem to check the mixed pair or to pick the easier order?
c-higher-partials
Take partials of a three-variable function and verify a given function satisfies Laplace's, the wave, or the heat equation?
c-more-variables-pde
Glossary (9 terms)
tracekesit
The curve obtained by cutting a surface with a plane, usually a coordinate plane or one parallel to it; the tool for identifying a surface by hand.
quadric surfacekuadrik yüzey
A surface given by a second-degree equation in the three variables, such as an ellipsoid, paraboloid, cone or hyperboloid.
cylindersilindir
A surface whose equation omits one variable entirely, so a fixed plane curve is swept parallel to the axis of the missing variable.
partial derivativekısmi türev
The derivative of a function of several variables with respect to one of them while all the others are held fixed as constants.
two-path test
A method that proves a two-variable limit does not exist by finding two approaches to the point along which the function tends to different values.
continuous functionsürekli fonksiyon
A function whose limit at a point equals its value there; for several variables this must hold along every approach to the point.
karma kısmi türev
A second derivative taken with respect to two different variables in turn, such as first x then y.
Clairaut's theorem
The result that the two mixed second partials are equal wherever both of them are continuous, so the order of differentiation does not matter.
Laplace's equationLaplace denklemi
The partial differential equation stating that the sum of the pure second partials is zero, satisfied by many steady-state fields.
What comes next
§09 · The chain rule and directional derivatives in several variables
This section could find the rate of change along each axis, but never in a slanted direction, and never how a change in $x$ that itself depends on time reaches $f$. Next comes the multivariable chain rule, which threads rates through several linked variables at once, and the directional derivative, which gives the rate in any direction you point, not just along the axes.
Sources
James Stewart, Calculus, Metric Version, Ninth Edition — sections 12.6, 14.2 and 14.3 The topic split follows the week 8 line of the course syllabus, and section numbers are quoted only as they appear there.
Course syllabus: assessment weights Quizzes 10% of the course grade and homework 10%; the syllabus states no further exam weight, so no exam distribution is claimed here.
Conventions used on this page Continuity is discussed only on the domain; trigonometric arguments are radians; the mixed-partial notation is read left to right in subscripts and right to left in the Leibniz denominators.