Week 7Stewart §12.4, 10.1, 13.1, 13.2, 13.3, 14.173 min full read
6 concepts18 worked examples26 exercises4 exam-level6 figures
What are you here for?
07Cross products, curves in space, arc length and curvature, and first functions of several variables
A drone logs its position ten times a second as three numbers: how far east, how far north, how high. From that table alone, with no compass and no speedometer bolted to the frame, the ground crew has to answer three things about the moment $t=2$ seconds: which way was it pointed, how many metres of air did it actually cut through on the way there, and was the turn sharp enough to spill a full cup of coffee. Every one of those answers is already hiding inside the position log.
By the end you can take a moving point given as a $\mathbf r(t)$ and produce, by hand, the exact direction it travels at any instant, its , the true length of its curved path, and a single number $\kappa$ that says how sharply it bends.
In 60 seconds
A curve in space is one vector function $\mathbf r(t)$; this section reads four things off it — a perpendicular direction and area from the cross product, the velocity $\mathbf r'(t)$, the path length, and the curvature.
putting one number on how sharply a space curve bends
Three most common mistakes
Writing $\mathbf a\times\mathbf b$ as a number. The cross product is a vector; the dot product is the number.
Dropping the minus sign on the middle component of the determinant.
Forgetting the cube in $|\mathbf r'|^{3}$ when computing curvature.
The only graded weights tied to this material are the quizzes (10% of the course) and the homework (10%). Expect short computational items — a cross product, a domain, an arc length or curvature setup — rather than long proofs.
How much time do you have?
10 minutes
You walk in able to compute a cross product, read a velocity off a vector function, and write the arc length and curvature formulas.
card, c-cross-product, c-vector-derivatives, formula card
45 minutes
You can also trace and de-parametrise a curve, find the domain of a vector function, set up an arc length integral, and get the domain and of a function of two variables.
card, c-cross-product, c-parametric-curves, c-vector-functions, c-vector-derivatives, c-arc-length-curvature, c-several-variables, scaffolding ladder, practice B
full read
Everything above plus the geometry behind each formula, the worked examples, the exam-style item, and the mixed practice that makes you choose the method without being told which one it is.
all concepts, contrast pairs, faded ladder, exam example, practice A, practice B, practice C, practice D
By the end of this section
Compute the cross product of two space vectors and use its length and direction to find areas and perpendicular vectors.
Trace a with its orientation and eliminate the parameter to recover the Cartesian equation, keeping the correct piece.
Interpret a vector function as a space curve, find its domain, and evaluate a limit component by component.
Differentiate a vector function into velocity, speed and unit tangent, and apply the dot and cross product rules.
Set up and evaluate an arc length integral and compute curvature from the first and second derivatives.
Find the domain of a function of two variables and sketch its level curves as a contour map.
Speed, arc length and the length of a cross product are all magnitudes.
Dot product and the angle
$\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a|\,|\mathbf b|\cos\theta$, and $\mathbf a\perp\mathbf b$ exactly when $\mathbf a\cdot\mathbf b=0$.
The dot product tests the perpendicularity we will check on every cross product, and the $\cos\theta$ here is the partner of the $\sin\theta$ in the cross product.
Unit vector
$\dfrac{\mathbf a}{|\mathbf a|}$ is the vector of length $1$ pointing the same way as $\mathbf a$.
The unit tangent $\mathbf T$ is built exactly this way from $\mathbf r'(t)$.
Product and chain rules
$\dfrac{d}{dt}\big(fg\big)=f'g+fg'$ and $\dfrac{d}{dt}f(g(t))=f'(g(t))\,g'(t)$.
Vector functions are differentiated one component at a time, and the two product rules for dot and cross products mirror the scalar one.
A point going around a circle of radius $c$ has constant distance yet obvious nonzero velocity, matching the conclusion.
Keep this fact: whenever a length is constant, the vector and its derivative are perpendicular. It returns for the unit tangent.
3§07.0 — one product rule●●○○○
Vector functions get differentiated one component at a time, so a single-variable product rule has to be automatic.
Given
$g(t)=t^{2}\sin t$
Find
Differentiate $g(t)=t^{2}\sin t$.
Hint 1/4
This is a product of two functions of $t$; no vectors yet.
Hint 2/4
Product rule: $\dfrac{d}{dt}(uv)=u'v+uv'$ with $u=t^{2}$ and $v=\sin t$.
Hint 3/4
Here $u'=2t$ and $v'=\cos t$, so the two pieces are $2t\sin t$ and $t^{2}\cos t$.
Hint 4/4
Adding them gives $g'(t)=2t\sin t+t^{2}\cos t$.
Show solutionName the two factors and their derivatives
$$u=t^{2},\ u'=2t;\qquad v=\sin t,\ v'=\cos t$$
splitting first keeps the two pieces from getting mixed
Assemble $u'v+uv'$
$$g'(t)=2t\sin t+t^{2}\cos t$$
keep both terms; a common slip is to write only $2t\cos t$
Answer $$g'(t)=2t\sin t+t^{2}\cos t$$
Check
At $t=0$ the formula gives $0$, matching that $g(t)=t^{2}\sin t$ is flat there since it starts like $t^{3}$.
Notation
symbol
reads as
means
watch out
$\mathbf a\times\mathbf b$
a cross b
the vector perpendicular to both $\mathbf a$ and $\mathbf b$ whose length is the area of the parallelogram they span
$\mathbf a\times\mathbf b=-\,\mathbf b\times\mathbf a$: swapping the order flips the sign.
$\mathbf r(t)$
r of t
a position vector whose tip traces a curve as $t$ runs over an interval
the output is a vector with several components, driven by the single input $t$.
$\mathbf r'(t)$
r prime of t
the velocity: the tangent vector pointing the way $t$ increases
differentiate each component on its own; the answer is again a vector.
$\mathbf T(t)$
the unit tangent T of t
$\mathbf r'(t)$ rescaled to length $1$
all speed is divided out; only the direction of travel survives.
$\kappa$
kappa
the curvature: how sharply the curve bends, measured in units of $1/\text{length}$
$\kappa=1/\rho$, so a large circle has small curvature, not large.
$f(x,y)$
f of x and y
a rule sending each point of a plane region to one number, pictured as a surface $z=f(x,y)$
its domain is a region in the plane, not an interval on a line.
Conventions used here
Two notations for one vector
We write $\mathbf a=\langle a_1,a_2,a_3\rangle$ for components and $\mathbf a=a_1\mathbf i+a_2\mathbf j+a_3\mathbf k$ for the same vector. The two are interchangeable; the boldface letter always means a vector.
Everything is in radians
Every angle, and every input to $\sin$ or $\cos$ or a parametrisation, is in radians. The derivative rules for the trig functions, and the arc length and curvature formulas, are only valid in radians.
Cross product lives in space
$\mathbf a\times\mathbf b$ is defined for vectors in three dimensions. A plane vector $\langle a_1,a_2\rangle$ is treated as $\langle a_1,a_2,0\rangle$ whenever a cross product is needed, and then the result points along $\mathbf k$.
Velocity is a vector, speed is a length
$\mathbf r'(t)$ is the velocity, a vector. Its magnitude $|\mathbf r'(t)|$ is the speed, a single number that is never negative. Arc length and curvature are built from the speed.
Smooth curve assumption
For the tangent direction, arc length and curvature we assume $\mathbf r'(t)$ is continuous and never $\mathbf 0$ on the interval, so the curve has a well defined direction at every point.
7.1The cross product
Turn two space vectors into a third that is perpendicular to both, with length equal to the area they span.
Last week the dot product answered every question about angle between two vectors and returned a single number. But a number cannot point anywhere, and two questions in space need a direction: given two edges of a patch in space, what is its area, and what direction is perpendicular to both? We need an operation that eats two vectors and returns a third.
Solvable with what we have
Find $|\mathbf a|$ and the angle between $\mathbf a$ and $\mathbf b$ with the dot product.
Test whether two vectors are perpendicular.
Not solvable yet
Produce a single vector perpendicular to two given directions.
Find the area of a triangle whose corners are three points in space.
The dot product is our only vector-times-vector tool so far. For the triangle with corners $P,Q,R$ we might reach for $\overrightarrow{PQ}\cdot\overrightarrow{PR}$ — but that returns a number tied to $\cos\theta$, and an area needs $\sin\theta$, so it is the wrong tool.
Why it fails
The dot product collapses two vectors to a scalar; it throws away the plane they define. Area and a normal direction are exactly the information it discards.
DefinitionDefinition 12.4: The cross product
Conditions
$\mathbf a=\langle a_1,a_2,a_3\rangle$ and $\mathbf b=\langle b_1,b_2,b_3\rangle$ are vectors in space
Read it as the symbolic determinant with $\mathbf i,\mathbf j,\mathbf k$ on the top row, $\mathbf a$ on the middle row and $\mathbf b$ on the bottom. Expanding along the top row, the $\mathbf i$ part is $a_2b_3-a_3b_2$, the $\mathbf j$ part is minus $(a_1b_3-a_3b_1)$, and the $\mathbf k$ part is $a_1b_2-a_2b_1$. The output is a vector, and it always comes out perpendicular to both $\mathbf a$ and $\mathbf b$; its length is $|\mathbf a|\,|\mathbf b|\sin\theta$, the area of the parallelogram they span.
The two input vectors $\textcolor{#1f6feb}{\mathbf a}$ and $\textcolor{#2f9e44}{\mathbf b}$ span a parallelogram; $\textcolor{#d1690a}{\mathbf a\times\mathbf b}$ stands perpendicular to it, and the length of that orange arrow is the shaded area.
Looks like this, but is not
Since the dot product obeys $\mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a$, it is tempting to assume $\mathbf a\times\mathbf b=\mathbf b\times\mathbf a$ as well.
They are equal in length but opposite in direction: $\mathbf b\times\mathbf a=-(\mathbf a\times\mathbf b)$. Swapping the two rows of the determinant flips its sign, which is why order matters for the cross product but not for the dot product.
Area of a triangle with corners in space
Find the area of the triangle whose vertices are $P(1,1,1)$, $Q(2,3,4)$ and $R(4,3,2)$, and give a vector perpendicular to its plane.
Given
$P(1,1,1)$, $Q(2,3,4)$, $R(4,3,2)$
Find
the triangle's area and a normal vector
SolutionBuild two edge vectors from the same corner
Check perpendicularity: $\langle -4,8,-4\rangle\cdot\langle 1,2,3\rangle=-4+16-12=0$ and $\cdot\langle 3,2,1\rangle=-12+16-4=0$. Both zero, so $\mathbf n$ really is perpendicular to both edges.
one cross product and one square root
The same three steps — two edges, one cross product, halve the length — give the area of any triangle in space, and the cross product itself is the normal you need for the plane through the three points.
Dot the answer with each input: $\langle 12,-6,-6\rangle\cdot\langle1,3,-1\rangle=12-18+6=0$ and $\cdot\langle2,0,4\rangle=24+0-24=0$. Perpendicular to both, as it must be.
The perpendicularity dot-check costs two multiplications per input and catches almost every sign slip; make it a habit.
⚠ Dropping the middle minus sign
the determinant expansion alternates $+,-,+$ and the middle sign is easy to lose
Let one parameter drive both coordinates, so the path can loop, cross itself, and carry a direction of travel.
A graph $y=f(x)$ can only ever assign one height to each $x$, so it can never be a full circle or a path that doubles back. The fix is to stop treating $x$ as the driver and hand that job to a separate parameter $t$: now $x$ and $y$ each follow their own rule, and the point is free to go anywhere.
DefinitionDefinition 10.1: Parametric curve
Conditions
$f$ and $g$ are functions defined on an interval $I$ of parameter values
As the parameter $t$ increases across the interval $I$, the point $(f(t),g(t))$ moves through the plane and drags out the curve behind it. The value of $t$ is often time; whatever it is, it also records the order in which the points are visited, which is the orientation of the curve.
The point $(3\cos t,\,2\sin t)$ traces an ellipse. The $\textcolor{#1f6feb}{\text{blue arrows}}$ give the orientation — counterclockwise — and the $\textcolor{#d1690a}{\text{marked}}$ parameter values show where the point is at $t=0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2}$.
Looks like this, but is not
The equations $x=t^{2}$, $y=t^{4}$ look like the recipe for a brand new curve.
Since $y=(t^{2})^{2}=x^{2}$, every point lies on the parabola $y=x^{2}$ — but $x=t^{2}$ is never negative, so the curve is only the right half, $x\ge0$. Eliminating the parameter recovered the shape but silently dropped the restriction the parameter was enforcing.
From an ellipse's parametrisation to its equation
The curve is $x=3\cos t,\ y=2\sin t$ for $0\le t\le2\pi$. Eliminate the parameter, name the curve, and give its orientation.
Given
$x=3\cos t$
$y=2\sin t$
$0\le t\le2\pi$
Find
a Cartesian equation, the curve's name, and its direction of travel
SolutionIsolate the sine and cosine so the identity can be used
Feed one parameter into three coordinates and the output vector's tip traces a one-dimensional curve through space.
A parametric curve in the plane needed two coordinate rules, $x(t)$ and $y(t)$. Add a third, $z(t)$, and the point is free to leave the page: it now traces a curve in space. Packaging the three rules into a single vector $\mathbf r(t)$ lets us treat the whole curve as one object.
DefinitionDefinition 13.1: Vector function and its limit
Conditions
$f,g,h$ are ordinary real-valued functions, the component functions
A vector function takes one number $t$ and returns a vector with three components. Its domain is the set of $t$ for which all three component functions are defined — the intersection of their domains. Limits, continuity and later derivatives are all done one component at a time: the limit of $\mathbf r$ is the vector of the three component limits, and exists only when every one of them exists.
The helix $\mathbf r(t)=\langle\cos t,\sin t,t\rangle$. Its $\textcolor{#2f9e44}{\text{shadow}}$ on the floor is the unit circle — the first two components — while the third component lifts it steadily, so it climbs a fixed height per turn.
Looks like this, but is not
Because $\mathbf r(t)$ lives in three dimensions, it is tempting to picture its graph as a surface.
There is only one input, $t$. One knob can only trace a one-dimensional object, so the graph of $\mathbf r(t)$ is a curve, not a surface. A surface would need two independent inputs, which is exactly what functions of several variables supply later in this section.
Domain of a vector function
Find the domain of $\mathbf r(t)=\big\langle \sqrt{4-t^{2}},\ \ln(t+1),\ \tfrac{1}{t}\big\rangle$.
the vector exists only where every component does, so we take the overlap, not the union
$$=(-1,0)\cup(0,2]$$
the strict $t>-1$ opens the left end, and $t\ne0$ punches out the middle
Answer $$\text{domain}=(-1,0)\cup(0,2]$$
Check
Spot-check the boundaries: $t=2$ gives $\sqrt0=0$ (legal) and is kept; $t=-1$ makes $\ln0$ undefined and is excluded; $t=0$ divides by zero and is excluded. All three match.
Domain of a vector function is always an intersection: a single illegal component rules the value out.
A curve where two surfaces meet
Find a vector function whose curve is the intersection of the cylinder $x^{2}+y^{2}=1$ with the plane $y+z=2$.
Given
cylinder $x^{2}+y^{2}=1$
plane $y+z=2$
Find
a vector function $\mathbf r(t)$ tracing the intersection
SolutionParametrise the cylinder first
$$x=\cos t,\quad y=\sin t,\quad 0\le t\le2\pi$$
any point on the unit cylinder has this form, since $\cos^{2}t+\sin^{2}t=1$ puts it on the circle for every height
Let the plane fix the remaining coordinate
$$y+z=2\ \Rightarrow\ z=2-y=2-\sin t$$
the plane equation solves for $z$ once $y$ is known
Differentiate a vector function component by component to get the velocity; rescale it to length one for the unit tangent.
Now the drone's log pays off. Its position is a vector function $\mathbf r(t)$; the question which way was it pointed is asking for the direction of $\mathbf r'(t)$, and how fast is asking for $|\mathbf r'(t)|$. Both come from differentiating, and differentiating a vector is as easy as doing each coordinate on its own.
DefinitionDefinition 13.2: Derivative and unit tangent
Conditions
the component functions are differentiable
$\mathbf r'(t)\ne\mathbf 0$ for the unit tangent to exist
To differentiate a vector function, differentiate each component and reassemble — nothing new from single-variable calculus is needed. The result $\mathbf r'(t)$ is the velocity: it is tangent to the curve and points the way $t$ increases, and its length $|\mathbf r'(t)|$ is the speed. Dividing by that length gives the unit tangent $\mathbf T(t)$, the pure direction of travel with the speed scaled away.
On the curve $\mathbf r(t)=\langle t,t^{2}\rangle$, the velocity $\textcolor{#d1690a}{\mathbf r'(1)=\langle 1,2\rangle}$ leaves the $\textcolor{#1f6feb}{\text{point }P=(1,1)}$ tangent to the curve and pointing the way the parameter grows.
Looks like this, but is not
The product rule for a dot product, $\tfrac{d}{dt}(\mathbf u\cdot\mathbf v)=\mathbf u'\cdot\mathbf v+\mathbf u\cdot\mathbf v'$, has an identical-looking partner for the cross product, $\tfrac{d}{dt}(\mathbf u\times\mathbf v)=\mathbf u'\times\mathbf v+\mathbf u\times\mathbf v'$.
It reads the same, but the order in each cross product may not be swapped: $\mathbf u'\times\mathbf v$ is not $\mathbf v\times\mathbf u'$. The dot-product rule survived reordering because the dot product commutes; the cross-product rule does not, so you must keep $\mathbf u$ before $\mathbf v$ in every term.
The drone's direction and speed at an instant
A drone's position is $\mathbf r(t)=\langle t,\ t^{2},\ t^{3}\rangle$. At $t=1$ find its velocity, its speed, the unit tangent, and the tangent line to its path.
Given
$\mathbf r(t)=\langle t,t^{2},t^{3}\rangle$
the instant $t=1$
Find
velocity, speed, unit tangent and the tangent line at $t=1$
SolutionDifferentiate each component for the velocity
$$\mathbf r'(t)=\langle 1,\ 2t,\ 3t^{2}\rangle$$
differentiate $t,\ t^{2},\ t^{3}$ separately
$$\mathbf r'(1)=\langle 1,2,3\rangle$$
evaluate at the instant asked for; this is the velocity vector
Unit-tangent check: $|\tfrac{1}{\sqrt{14}}\langle1,2,3\rangle|=\tfrac{\sqrt{14}}{\sqrt{14}}=1$, so $\mathbf T$ really has length one. The line passes through $\langle1,1,1\rangle$ at $s=0$, the correct point.
one derivative, one magnitude, one division
This is the whole hook resolved: position in, direction and speed out. Every 'which way, how fast' question is one derivative away.
Position recovered from velocity
A particle has velocity $\mathbf r'(t)=\langle 2t,\ 3t^{2},\ \cos t\rangle$ and starts at $\mathbf r(0)=\langle 1,0,2\rangle$. Find its position $\mathbf r(t)$.
Given
$\mathbf r'(t)=\langle 2t,3t^{2},\cos t\rangle$
$\mathbf r(0)=\langle1,0,2\rangle$
Find
the position vector $\mathbf r(t)$
SolutionAntidifferentiate each component, carrying a constant
Differentiate the answer back: $\tfrac{d}{dt}\langle t^{2}+1,t^{3},\sin t+2\rangle=\langle 2t,3t^{2},\cos t\rangle$, the given velocity, and at $t=0$ it reads $\langle1,0,2\rangle$.
Integrating velocity to position is the reverse of the first example, and the initial condition is what turns a family of antiderivatives into one curve.
⚠ Swapping the order in the cross-product rule
the dot-product rule commutes, so the reflex reorders the factors
Arc length integrates the speed; curvature measures how fast the direction turns, one over the best-fitting circle's radius.
The velocity told us the drone's direction and speed at an instant. Two whole-path questions remain: how far did it travel, and how sharp was the turn? The first adds up speed over time; the second measures how quickly the unit tangent swings around.
TheoremSection 13.3: Arc length and curvature
Conditions
the curve is smooth: $\mathbf r'(t)$ is continuous and never $\mathbf 0$ on $[a,b]$
Arc length adds up infinitely many tiny straight steps: over a slice $dt$ the point moves about $|\mathbf r'(t)|\,dt$, and the integral totals them into the true length $L$. Curvature $\kappa$ measures how fast the unit tangent changes direction per unit of length travelled; the cross-product formula is the practical way to compute it. Its value is $1/\rho$, where $\rho$ is the radius of the circle that best hugs the curve, so a gentle wide curve has small $\kappa$. For a plane curve $y=f(x)$ this reduces to $\kappa=\dfrac{|f''|}{(1+(f')^{2})^{3/2}}$.
At the vertex of $\textcolor{#1f6feb}{y=x^{2}}$ the $\textcolor{#d1690a}{\text{osculating circle}}$ has radius $\textcolor{#2f9e44}{\rho=\tfrac12}$, so the curvature there is $\kappa=1/\rho=2$. A bigger circle would mean a gentler bend and a smaller $\kappa$.
Looks like this, but is not
A curve that swings all the way across the page surely has large curvature.
Curvature is local bending, not overall size. A huge circle of radius $1000$ crosses the whole page yet has curvature $\kappa=1/1000$, almost a straight line at any point. What matters is the radius of the circle that fits the curve right there, not how far the curve wanders.
Arc length of one turn of a helix
Find the length of the helix $\mathbf r(t)=\langle\cos t,\ \sin t,\ t\rangle$ for $0\le t\le2\pi$.
a constant speed pulls straight out of the integral
$$=2\sqrt2\,\pi$$
evaluate the bracket at the two limits
Answer $$L=2\sqrt2\,\pi\approx8.89$$
Check
Sanity bound: one turn rises $2\pi\approx6.28$ in $z$ while circling a unit circle of circumference $2\pi$; the diagonal length $\sqrt{(2\pi)^2+(2\pi)^2}=2\sqrt2\,\pi$ matches exactly, since constant speed makes the helix unroll to a straight line.
one speed computation, one one-line integral
When the speed $|\mathbf r'(t)|$ turns out constant, the arc length is just speed times elapsed parameter — no real integration needed.
Curvature of the same helix
Find the curvature of $\mathbf r(t)=\langle\cos t,\ \sin t,\ t\rangle$.
the denominator is the speed cubed — do not forget the cube
$$\kappa=\dfrac{\sqrt2}{2\sqrt2}=\dfrac12$$
constant, so the helix bends the same everywhere
Answer $$\kappa=\dfrac12\ \text{(constant along the helix)}$$
Check
Dimensional check: the radius of curvature is $\rho=1/\kappa=2$, larger than the helix's own circular radius $1$ because the steady climb straightens the turn a little — reasonable.
The cross-product curvature formula is the reliable route for space curves; the only common slip is dropping the cube in $|\mathbf r'|^{3}$.
A function of two variables assigns each planar point one height — a surface to graph, or a contour map to read.
Every function so far had one input. A vector function had one input and several outputs. Now flip it: take two inputs $x$ and $y$ and return a single number $f(x,y)$. The input is now a point in a plane region, and the graph rises into a surface over that region.
DefinitionDefinition 14.1: Function of two variables
A function of two variables eats a point $(x,y)$ of a plane region $D$ and returns one number. Its graph is the surface of all points $(x,y,f(x,y))$ hovering above $D$. A level curve is the set of inputs that share one output value $k$: slice the surface with the horizontal plane $z=k$ and drop the slice straight down onto the $xy$-plane. A whole family of level curves for evenly spaced $k$ is a contour map, and where the curves crowd together the surface is steep.
A contour map of $\textcolor{#1f6feb}{f(x,y)=x^{2}+y^{2}}$: each circle collects the inputs with one height $\textcolor{#d1690a}{k}$. Equal steps in $k$ give circles that crowd together as $k$ grows, the signature of a surface getting steeper.
Looks like this, but is not
It is natural to draw the level curve $f(x,y)=k$ up on the surface, along the rim of the horizontal slice.
The level curve lives in the flat $xy$-plane, not on the surface. It is the shadow of that rim dropped straight down — its whole point is to describe a three-dimensional surface using only a flat picture, the way a topographic map shows a mountain on paper.
Domain and range of a square-root surface
Find the domain and range of $f(x,y)=\sqrt{9-x^{2}-y^{2}}$, and name the surface.
Given
$f(x,y)=\sqrt{9-x^{2}-y^{2}}$
Find
the domain (a region), the range, and the surface
SolutionForce the inside of the root to be nonnegative
squaring shows it is the top half of a sphere of radius $3$
Answer $$\text{domain}=\{x^{2}+y^{2}\le3^{2}\},\quad \text{range}=[0,3],\quad \text{upper hemisphere of radius }3$$
Check
Check a corner and the centre: at the origin $f=\sqrt9=3$ (the top of the dome); on the rim $x^{2}+y^{2}=9$, $f=0$ (ground level). Both sit inside the range $[0,3]$.
one inequality for the domain, one for the range
Domain of a function of two variables is a region in the plane; describe it by the curve that bounds it, not by intervals.
Level curves of a paraboloid
Describe and sketch the level curves of $f(x,y)=x^{2}+y^{2}$ for $k=1,4,9$.
Given
$f(x,y)=x^{2}+y^{2}$
levels $k=1,4,9$
Find
the level curves and what their spacing says
SolutionSet the function equal to each level
$$x^{2}+y^{2}=k$$
a level curve is $f=k$; here that is a circle for every positive $k$
Read off the circles
$$k=1,4,9\ \Rightarrow\ \text{radii } 1,2,3$$
the circle $x^{2}+y^{2}=k$ has radius $\sqrt{k}$
Interpret the spacing
$$\text{radii }1,2,3\ \text{are evenly spaced, but the heights }1,4,9\text{ are not}$$
equal radial steps carry ever-larger height jumps, so the bowl steepens outward
Answer $$\text{circles of radius }\sqrt{k}:\ r=1,2,3\ \text{for }k=1,4,9;\ \text{the surface is a bowl steepening outward}$$
Check
The value at the centre is $f(0,0)=0$, below every listed level, and each circle's points all return their own $k$ — for instance $(2,0)$ gives $4$, on the $k=4$ circle.
Reading a contour map: circles that get closer together for equal height steps mean the surface is getting steeper there.
⚠ Drawing level curves on the surface
the slice is taken up on the surface, so the curve seems to belong there
wrong$$f(x,y)=k\ \text{drawn on the surface}$$
right$$f(x,y)=k\ \text{drawn in the }xy\text{-plane}$$
⚠ Including a strict boundary
square-root and logarithm domains look alike but differ at the edge
wrong$$\ln(9-x^{2}-y^{2}):\ x^{2}+y^{2}\le9$$
right$$\ln(9-x^{2}-y^{2}):\ x^{2}+y^{2}<9$$
Checkpoint
§07.6 — domain of a logarithmic surface●●○○○
Thirty seconds: what does the logarithm demand?
Given
$f(x,y)=\ln(9-x^{2}-y^{2})$
Find
Find the domain of $f$ and say whether its boundary is included.
Hint 1/4
A logarithm needs its input strictly positive; set up that inequality in $x$ and $y$.
Hint 2/4
Require $9-x^{2}-y^{2}>0$, a strict inequality because $\ln 0$ is undefined.
Hint 3/4
Rearranged, that is $x^{2}+y^{2}<9$, the inside of the circle of radius $3$.
Hint 4/4
So the domain is the open disk $x^{2}+y^{2}<9$; the boundary circle is not included.
Show solutionPositivity, strictly
$$9-x^{2}-y^{2}>0\ \Rightarrow\ x^{2}+y^{2}<9$$
$\ln$ needs a strictly positive argument, so the inequality is strict
Contrast with the earlier square-root surface, whose domain $x^{2}+y^{2}\le9$ did include its rim: the root tolerates a zero inside, the logarithm does not.
⚠ Writing the domain as intervals
one-variable habit describes domains on a number line
you need $\mathbf a\times\mathbf b$ for an area, a normal vector, or a perpendicular direction
Set up the symbolic determinant
Write the $3\times3$ array with $\mathbf i,\mathbf j,\mathbf k$ on top, $\mathbf a$ in the middle row, $\mathbf b$ on the bottom.
Expand along the top row
$\mathbf i$-part $=a_2b_3-a_3b_2$; $\mathbf j$-part $=-(a_1b_3-a_3b_1)$ — keep that leading minus; $\mathbf k$-part $=a_1b_2-a_2b_1$.
Dot-check the answer
Confirm the result dotted with $\mathbf a$ and with $\mathbf b$ is $0$. Two cheap multiplications catch nearly every slip.
Where it goes wrong
Losing the middle minus sign.
Reporting a scalar instead of a vector.
Differentiation rules for vector functions
you differentiate combinations of vector functions rather than a single one
Componentwise and scalar multiples
$\dfrac{d}{dt}\langle f,g,h\rangle=\langle f',g',h'\rangle$, and constants pull out: $\dfrac{d}{dt}(c\,\mathbf u)=c\,\mathbf u'$.
Scalar-times-vector product
$\dfrac{d}{dt}\big(f(t)\,\mathbf u(t)\big)=f'\mathbf u+f\mathbf u'$, exactly the ordinary product rule.
Dot product
$\dfrac{d}{dt}(\mathbf u\cdot\mathbf v)=\mathbf u'\cdot\mathbf v+\mathbf u\cdot\mathbf v'$; order is free because the dot product commutes.
Cross product — keep the order
$\dfrac{d}{dt}(\mathbf u\times\mathbf v)=\mathbf u'\times\mathbf v+\mathbf u\times\mathbf v'$. Do not reorder any factor; the cross product is anticommutative.
Dot-check: $\langle1,-2,1\rangle\cdot\langle1,2,3\rangle=1-4+3=0$ and $\cdot\langle0,1,2\rangle=0-2+2=0$.
Same two vectors, two different products: the dot product returns the number $8$ (about the angle), the cross product returns the vector $\langle1,-2,1\rangle$ (perpendicular to both, about area and direction).
How to tell them apart
If the question asks 'how aligned / what angle', use the dot product and expect a number. If it asks 'what direction is perpendicular / what area', use the cross product and expect a vector.
Arc length is one number for the whole trip
Find the length of $\mathbf r(t)=\langle\cos t,\sin t,t\rangle$ over $0\le t\le4\pi$.
Given
$\mathbf r(t)=\langle\cos t,\sin t,t\rangle$
$0\le t\le4\pi$
Find
the arc length
SolutionConstant speed times elapsed parameter
$$|\mathbf r'(t)|=\sqrt2$$
speed is constant on a helix
$$L=\sqrt2\,(4\pi)=4\sqrt2\,\pi$$
length grows with how long you travel
Answer $$L=4\sqrt2\,\pi$$
Check
Twice the one-turn length $2\sqrt2\,\pi$, since the interval is twice as long.
Curvature is a local rate, blind to the interval
Find the curvature of the same helix $\mathbf r(t)=\langle\cos t,\sin t,t\rangle$.
curvature is the same at every point, no matter how far you go
Answer $$\kappa=\tfrac12$$
Check
Independent of the interval $[0,4\pi]$ — it is a property of the shape, not the trip length.
Arc length is a single number for the chosen interval and doubles when you double the interval; curvature is a function of position that does not change when the interval does.
How to tell them apart
'How far' totals the speed over an interval and depends on the endpoints. 'How sharp' is $\kappa$ at a point and ignores the endpoints entirely.
Scaffolding comes off
The common skeleton
Differentiate $\mathbf r(t)$ component by component to get the velocity $\mathbf r'(t)$.
Evaluate $\mathbf r'(t)$ at the required value $t_0$.
Compute the speed $|\mathbf r'(t_0)|$ by squaring each component, adding, and taking the root.
Divide the velocity by the speed: $\mathbf T(t_0)=\mathbf r'(t_0)/|\mathbf r'(t_0)|$.
1 · fully worked
Unit tangent, fully worked
Find the unit tangent $\mathbf T(1)$ for $\mathbf r(t)=\langle t^{2},\ 2t,\ \ln t\rangle$.
Length check: $\tfrac13\sqrt{4+4+1}=\tfrac13\cdot3=1$, so $\mathbf T$ is a unit vector.
2 · you write the reasoning
Now a plane curve, so the arithmetic is lighter and you can spend the effort writing why each step is legal. Find $\mathbf T(1)$ for $\mathbf r(t)=\langle t^{2},\ t^{3}\rangle$.
reasoning
Differentiate each component separately; a two-component vector works the same way as a three-component one.
reasoning
Substitute the required value $t=1$ to turn the velocity function into a single vector.
reasoning
Speed is the magnitude: square the components, add, take the root. Here $\sqrt{13}$ does not simplify, and that is fine.
reasoning
Dividing by the speed rescales the velocity to length one without changing its direction.
3 · find the buried error
Back to space, and this solution has exactly two mistakes buried in it. Find $\mathbf T(1)$ for $\mathbf r(t)=\langle t,\ t^{2},\ t^{3}\rangle$. Which steps are wrong?
the two buried errors (2)
⚠ step 3
The components were added before squaring: it reads $1+2+3$ instead of $1^{2}+2^{2}+3^{2}$.
the squares are easy to skip when the numbers are small and already look addable.
right
$|\mathbf r'(1)|=\sqrt{1+4+9}=\sqrt{14}$.
⚠ step 4
The velocity was multiplied by the speed instead of divided by it.
'make it a unit vector' gets remembered as 'combine with the length' without pinning down the direction.
A particle moves along $\mathbf r(t)=\langle 3\cos t,\ 3\sin t,\ 4t\rangle$. Find (a) its speed, (b) the distance travelled for $0\le t\le2\pi$, (c) the curvature, and (d) the unit tangent at $t=0$.
Cross-check curvature with the helix formula $\kappa=\tfrac{a}{a^{2}+b^{2}}=\tfrac{3}{9+16}=\tfrac{3}{25}$; and $|\mathbf T(0)|=\tfrac15\sqrt{0+9+16}=1$.
four short parts, each one line once the speed is known
A constant-speed curve makes every part easy: the speed falls out, arc length is speed times time, and only the curvature needs the second derivative.
Practice
A · concept 3 questions
1§07.1 — what the cross product is●●○○○
Before any arithmetic, be sure what kind of object a cross product even is.
Given
$\mathbf a$ and $\mathbf b$ are nonzero, non-parallel vectors in space.
Find
Which statement about $\mathbf a\times\mathbf b$ is correct?
Hint 1/4
Ask two things: is the output a number or a vector, and where does it point?
Hint 2/4
The dot product is the one that returns a number; the cross product returns a vector perpendicular to both inputs.
Hint 3/4
Recall too that reversing the order flips the sign, so $\mathbf a\times\mathbf b$ and $\mathbf b\times\mathbf a$ differ.
Hint 4/4
The only correct description is: a vector perpendicular to both $\mathbf a$ and $\mathbf b$.
Show solutionRule out the scalar and in-plane options
the trinomial is a perfect square, and $1+t\ge0$ on the interval
Integrate
$$L=\int_0^1(1+t)\,dt=1+\tfrac12=\tfrac32$$
an ordinary single-variable integral once the root is gone
Answer $$L=\tfrac32$$
Check
The straight-line distance between the endpoints $\mathbf r(0)=\langle0,0,0\rangle$ and $\mathbf r(1)=\langle1,\tfrac{2\sqrt2}{3},\tfrac12\rangle$ is about $1.28<1.5$, and a curved path being longer than the chord is exactly right.
3§07.1 — plane through three points●●●●○
This mixes the cross product with last week's plane equations.
Given
$P(1,0,0)$
$Q(0,2,0)$
$R(0,0,3)$
Find
Find an equation of the plane through $P$, $Q$ and $R$.
Hint 1/4
A plane needs a point and a normal; the normal is perpendicular to two edges of the triangle.
Hint 2/4
Build $\overrightarrow{PQ}$ and $\overrightarrow{PR}$, cross them for a normal $\mathbf n$, then use $\mathbf n\cdot(\mathbf x-P)=0$.
Hint 3/4
$\overrightarrow{PQ}=\langle-1,2,0\rangle$, $\overrightarrow{PR}=\langle-1,0,3\rangle$, so $\mathbf n=\langle6,3,2\rangle$.
Close the page and write, from memory: the component formula for $\mathbf a\times\mathbf b$ and the two facts that make it useful; how to eliminate a parameter and what you must not lose when you do; the domain rule for a vector function; how to get velocity, speed and the unit tangent from $\mathbf r(t)$; the arc length and curvature formulas; and what a level curve is and where it is drawn. Then check each against the formula card.
Compute a cross product, use it for an area, and produce a vector perpendicular to two given ones?
c-cross-product
Trace a parametric curve with its orientation and eliminate the parameter without dropping the correct piece?
c-parametric-curves
Find the domain of a vector function and take a limit component by component?
c-vector-functions
Get velocity, speed and the unit tangent, and apply the dot and cross product rules?
c-vector-derivatives
Set up an arc length integral and compute curvature from $\mathbf r'$ and $\mathbf r''$?
c-arc-length-curvature
Find the domain of $f(x,y)$ and sketch its level curves as a contour map?
c-several-variables
Glossary (11 terms)
cross productvektörel çarpım
The vector $\mathbf a\times\mathbf b$ perpendicular to both $\mathbf a$ and $\mathbf b$, with length $|\mathbf a||\mathbf b|\sin\theta$ equal to the area of the parallelogram they span.
parametric curveparametrik eğri
A curve given as $x=f(t)$, $y=g(t)$, whose points are traced in the order of increasing $t$.
vector functionvektör değerli fonksiyon
A function $\mathbf r(t)=\langle f(t),g(t),h(t)\rangle$ that takes one number and returns a vector, whose domain is the intersection of the component domains.
space curveuzay eğrisi
The one-dimensional path traced by the tip of a vector function $\mathbf r(t)$ in three dimensions.
unit tangent vectorbirim teğet vektör
$\mathbf T(t)=\mathbf r'(t)/|\mathbf r'(t)|$, the direction of travel scaled to length one.
speedsürat
The magnitude $|\mathbf r'(t)|$ of the velocity, a single nonnegative number.
arc lengthyay uzunluğu
The true length $L=\int_a^b|\mathbf r'(t)|\,dt$ of a curve, the accumulated speed over the interval.
curvatureeğrilik
$\kappa=|\mathbf r'\times\mathbf r''|/|\mathbf r'|^{3}$, the rate at which the unit tangent turns per unit length; equal to one over the radius of the best-fitting circle.
osculating circledeğme çemberi
The circle of radius $1/\kappa$ that best matches a curve at a point, sharing its tangent and bending.
level curveseviye eğrisi
The set $f(x,y)=k$ of inputs sharing one output value $k$, drawn in the $xy$-plane.
contour mapeş yükselti haritası
A collection of level curves for evenly spaced values of $k$, used to read a surface from a flat picture.
What comes next
§08 · Partial derivatives
Next week the second input stops being a spectator: we differentiate $f(x,y)$ in the $x$-direction and the $y$-direction separately, and those partial derivatives assemble into the gradient.
Sources
James Stewart, Calculus, ninth edition — sections 12.4, 10.1, 13.1, 13.2, 13.3 and 14.1 The six sections named on this week's syllabus line; definitions, numbering and conventions follow this book.
Standard multivariable calculus practice Problem types follow the usual repertoire for this material; the numbers and wording are our own.