Multivariable Calculus: Functions, Surfaces, and Vector Fields, Schemes and Mind Maps of Vector Analysis

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Multivariable Calculus

Oliver Knill

Math 21a, Fall 2017

These notes contain condensed ”two pages per lecture” notes with essential information.

Math 21a: Multivariable calculus Oliver Knill, Fall 2017

1: Geometry and Distance

A point in the plane R^2 has two coordinates P = (x, y) like P = (2, −3). A point in space is determined by three coordinates P = (x, y, z) like P = (1, 2 , 4). The plane is divided into 4 quadrants, and space is divided into 8 octants. The point P = (1, 2 , 4) is in the first octant. These regions by intersect at the origin O = (0, 0) or O = (0, 0 , 0) and are separated by coordinate axes {y = 0 } and {x = 0 } or coordinate planes {x = 0 }, {y = 0 }, {z = 0 }.

1 Describe the location of the points P = (− 1 , − 2 , −3), Q = (0, 0 , −5), R = (1, 2 , −3) in words.

Possible Answer: P = (1, 2 , 3) is in the negative octant of space, where all coordinates are negative. The point R = (0, 0 , −5) is on the negative z-axis. The point S = (1, 2 , −3) is below the xy-plane. When projected onto the xy-plane it is in the first quadrant.

2 Problem. Find the midpoint M of P = (1, 2 , 5) and Q = (− 3 , 4 , 9). Answer. The midpoint

is obtained by taking the average of the coordinates: M = (P + Q)/2 = (− 1 , 3 , 7).

The Euclidean distance between two points P = (√x, y, z) and Q = (a, b, c) in space is defined as d(P, Q) = (x − a)^2 + (y − b)^2 + (z − c)^2.

The Euclidean distance is motivated by the Pythagorean theorem. 1

3 Find the distance d(P, Q) between the points P = (1, 2 , 5) and Q = (− 3 , 4 , 7) and verify that

d(P, M) + d(Q, M) = d(P, Q). Answer: The distance is d(P, Q) =

42 + 2^2 + 2^2 =

The distance d(P, M) is

22 + 1^2 + 1^2 =

6. The distance d(Q, M) is

22 + 1^2 + 1^2 =

Indeed d(P, M) + d(M, Q) = d(P, Q).

A circle of radius r centered at P = (a, b) is the collection of points in the plane which have constant distance r from P.

A sphere of radius ρ centered at P = (a, b, c) is the collection of points in space which have constant distance ρ from P. The equation of a sphere is (x − a)^2 + (y − b)^2 + (z − c)^2 = ρ^2.

4 Is the point (3, 4 , 5) outside or inside the sphere (x − 2)^2 + (y − 6)^2 + (z − 2)^2 = 25? Answer:

The distance of the point to the center of the sphere is

1 + 4 + 9 which is smaller than 5, the radius of the sphere. The point is inside. (^1) It appears in an appendix to ”Geometry” of ”Discours de la m´ethode” from 1637, Ren´e Descartes (1596- 1650). More about Descartes in Aczel’s book ”Descartes Secret Notebook”.

Math 21a: Multivariable calculus Oliver Knill, Fall 2017

2: Vectors and Dot product

Two points P = (a, b, c) and Q = (x, y, z) in space define a vector P Q~ = ~v = 〈x−a, y−b−z−c〉 pointing from P to Q. The real numbers v 1 , v 2 , v 3 in ~v = 〈v 1 , v 2 , v 3 〉 are the components of ~v.

Similar definitions hold in two dimensions, where vectors have two components. Vectors can be drawn everywhere in space. Two vectors with the same components are considered equal. 1

The addition of two vectors is ~u + ~v = 〈u 1 , u 2 , u 3 〉 + 〈v 1 , v 2 , v 3 〉 = 〈u 1 + v 1 , u 2 + v 2 , u 3 + v 3 〉. The scalar multiple λ~u = λ〈u 1 , u 2 , u 3 〉 = 〈λu 1 , λu 2 , λu 3 〉. The difference ~u − ~v can be seen as the addition of ~u and (−1) · ~v.

The addition and scalar multiplication of vectors satisfy the laws you know from arithmetic. commutativity ~u + ~v = ~v + ~u, associativity ~u + (~v + w~) = (~u + ~v) + w~ and r ∗ (s ∗ ~v) = (r ∗ s) ∗ ~v as well as distributivity (r+s)~v = ~v(r+s) and r(~v+ w~) = r~v+r ~w, where ∗ is scalar multiplication.

The length or magnitude |~v| of a vector ~v = P Q~ is defined as the distance d(P, Q) from P to Q. A vector of length 1 is called a unit vector. A synonym is direction. Nonzero vectors have length and magnitude.

1 |〈 3 , 4 〉| = 5 and |〈 3 , 4 , 12 〉| = 13. Examples of unit vectors are |~i| = |~j| = ~k| = 1 and

〈 3 / 5 , 4 / 5 〉 and 〈 3 / 13 , 4 / 13 , 12 / 13 〉. The only vector of length 0 is the zero vector |~ 0 | = 0.

The dot product of two vectors ~v = 〈a, b, c〉 and w~ = 〈p, q, r〉 is defined as ~v · w~ = ap + bq + cr.

The dot product determines distance and distance determines the dot product.

Proof: Using the dot product one can express the length of ~v as |~v| =

~v · ~v. On the other hand, (~v + w~) · (~v + w~) = ~v · ~v + w~ · w~ + 2(~v · w~) allows to solve for ~v · w~:

~v · w~ = (|~v + w~|^2 − |~v|^2 − | w~|^2 )/ 2.

The Cauchy-Schwarz inequality tells |~v · w~| ≤ |~v|| w~|.

Proof. We only need to show it in the case | w~| = 1. Define a = ~v · w~ and estimate 0 ≤ (~v − a ~w) · (~v − a ~w) to get 0 ≤ (~v − (~v · w~) w~) · (~v − (~v · w~) w~) = |~v|^2 + (~v · w~)^2 − 2(~v · w~)^2 = |~v|^2 − (~v · w~)^2 which means (~v · w~)^2 ≤ |~v|^2. The Cauchy-Schwarz inequality allows us to define what an ”angle” is.

The angle between two nonzero vectors is defined as the unique α ∈ [0, π] which satisfies ~v · w~ = |~v| · | w~| cos(α).

(^1) We cover 2400 years of math from Pythagoras (500 BC), Al Kashi (1400), Cauchy (1800) to Hamilton (1850).

Al Kashi’s theorem: A triangle ABC with side lengths a, b, c and angle α opposite to c satisfies a^2 + b^2 = c^2 + 2ab cos(α).

Proof. Define ~v = AB, ~~ w = AC~. Because c^2 = |~v − w~|^2 = (~v − w~) · (~v − w~) = |~v|^2 + | w~|^2 − 2 ~v · w~, We know ~v · w~ = |~v| · | w~| cos(α) so that c^2 = |~v|^2 + | w~|^2 − 2 |~v| · | w~| cos(α) = a^2 + b^2 − 2 ab cos(α).

The triangle inequality tells |~u + ~v| ≤ |~u| + |~v|

Proof: |~u+~v|^2 = (~u+~v)·(~u+~v) = ~u^2 +~v^2 +2~u·~v ≤ ~u^2 +~v^2 +2|~u·~v| ≤ ~u^2 +~v^2 +2|~u|·|~v| = (|~u|+|~v|)^2.

Two vectors are called orthogonal or perpendicular if ~v · w~ = 0. The zero vector ~0 is orthogonal to any vector. For example, ~v = 〈 2 , 3 〉 is orthogonal to w~ = 〈− 3 , 2 〉.

Pythagoras theorem: if ~v and w~ are orthogonal, then |~v − w~|^2 = |~v|^2 + | w~|^2.

Proof: (~v − w~) · (~v − w~) = ~v · ~v + w~ · w~ + 2~v · w~ = ~v · ~v + w~ · w~. 2

The vector P(~v) = (^) |~v w~·^ w~| 2 w~ is called the projection of ~v onto w~. The scalar projection ~v | w·~^ w~| is plus or minus the length of the projection of ~v onto w~. The vector ~b = ~v − P (~v) is a vector orthogonal to w~.

2 Find the projection of ~v = 〈 0 , − 1 , 1 〉 onto w~ = 〈 1 , − 1 , 0 〉. Answer: P(~v) = 〈 1 / 2 , − 1 / 2 , 0 〉.

3 A wind force F~ = 〈 2 , 3 , 1 〉 is applied to a car which drives in the direction of the vector

w ~ = 〈 1 , 1 , 0 〉. Find the projection of F~ onto w~, the force which accelerates or slows down the car. Answer: w~( F~ · w/~ | w~|^2 ) = 〈 5 / 2 , 5 / 2 , 0 〉.

4 How can we visualize the dot product? Answer: the absolute value of the dot product

is the length of the projection. Positive dot product means ~v and w~ form an acute angle, negative if that angle is obtuse.

5 Given ~v = 〈 2 , 1 , 2 〉 and w~ = 〈 3 , 4 , 0 〉. Find a vector which is in the plane defined by ~v and w~

and which bisects the angle between these two vectors. Answer. Normalize the two vectors to make them unit vectors then add them to get 〈 13 , 17 , 10 〉/15.

6 Given two vectors ~v, ~w which are perpendicular. Under which condition is ~v+ w~ perpendicular

to ~v − w~? Answer: Find the dot product of ~v + w~ with ~v − w~ and set it zero.

7 Is the angle between 〈 1 , 2 , 3 〉 and 〈− 15 , 2 , 4 〉 acute or obtuse? Answer: the dot product is

1. Ah! Cute!

(^2) We have just proven Pythagoras and Al Kashi. Distance and angle were defined, not deduced.

The trigonometric sin-formula: if a, b, c are the side lengths of a triangle and α, β, γ are the angles opposite to a, b, c then a/ sin(α) = b/ sin(β) = c/ sin(γ.

Proof. Twice the area of the triangle is ab sin(γ) = bc sin(α) = ac sin(β) Divide the first equation by sin(γ) sin(α) to get one identity. Divide the second equation by sin(α) sin(β) to get the second identity.

2 If ~v = 〈a, 0 , 0 〉 and w~ = 〈b cos(α), b sin(α), 0 〉, then ~v × w~ = 〈 0 , 0 , ab sin(α)〉 which has length

|ab sin(α)|.

The scalar [~u, ~v, ~w] = ~u · (~v × w~) is called the triple scalar product of ~u, ~v, ~w. The number |[~u, ~v, ~w]| de- fines the volume of the parallelepiped spanned by ~u, ~v, ~w and the orientation of three vectors is the sign of [~u, ~v, ~w].

The value h = |~u · ~n|/|~n| is the height of the parallelepiped if ~n = (~v × w~) is a normal vector to the ground parallelogram of area A = |~n| = |~v × w~|. The volume of the parallelepiped is hA = (~u · ~n/|~n|)|~v × w~| which simplifies to ~u · ~n = |(~u · (~v × w~)| which is indeed the absolute value of the triple scalar product. The vectors ~v, ~w and ~v × w~ form a right handed coordinate system. If the first vector ~v is your thumb, the second vector w~ is the pointing finger then ~v × w~ is the third middle finger of the right hand.

3 Problem: Find the volume of a cuboid of width a length b and height c. Answer. The

cuboid is a parallelepiped spanned by 〈a, 0 , 0 〉 〈 0 , b, 0 〉 and 〈 0 , 0 , c〉. The triple scalar product is abc.

4 Problem Find the volume of the parallelepiped which has the vertices O = (1, 1 , 0), P =

(2, 3 , 1), Q = (4, 3 , 1), R = (1, 4 , 1). Answer: We first see that it is spanned by the vectors ~u = 〈 1 , 2 , 1 〉, ~v = 〈 3 , 2 , 1 〉, and w~ = 〈 0 , 3 , 1 〉. We get ~v × w~ = 〈− 1 , − 3 , 9 〉 and ~u · (~v × w~) = 2. The volume is 2.

5 Problem: find the equation ax + by + cz = d for the plane which contains the point

P = (1, 2 , 3) as well as the line which passes through Q = (3, 4 , 4) and R = (1, 1 , 2). To do so find a vector ~n = 〈a, b, c〉 normal to the and noting (~x − OP~ ) · ~n = 0. Answer: A normal vector ~n = 〈 1 , − 2 , 2 〉 = 〈a, b, c〉 of the plane ax + by + cz = d is obtained as the cross product of P Q~ and RQ~ With d = ~n · OP~ = 〈 1 , − 2 , 2 〉 · 〈 1 , 2 , 3 〉 = 3, we get the equation x − 2 y + 2z = 3.

The cross product appears in physics, like for the angular momentum, the Lorentz force or the Coriolis force. We will however mainly use the cross product for constructions like to get the equation of a plane through 3 points A, B, C.

Math 21a: Multivariable calculus Oliver Knill, Fall 2017

3: Lines and Planes

A point P = (x 0 , y 0 , z 0 ) and a vector ~v = 〈a, b, c〉 define the line

〈x, y, z〉〈x 0 , y 0 , z 0 〉 + t〈a, b, c〉.

It is called the parameterization of the line.

The line is obtained by adding a multiple of the vector ~v to the vector OP~ = 〈x 0 , y 0 , z 0 〉. Every vector contained inside the line is parallel to ~v. We think about the parameter t as ”time” and about ~v as the velocity. For t = 0, we are at P and for t = 1 we are at OP~ + ~v.

Given two points like (2, 3 , 4) with (3, 3 , 5), the line

〈x(t), y(t), z(t)〉 = 〈 2 , 3 , 4 〉 + t〈 2 , 1 , 1 〉

connects (2, 3 , 4) with (3, 3 , 5). If t is restricted to [0, 1] we get the line segment defined by the two points.

1 Problem. Get the line through P = (1, 1 , 2) and Q = (2, 4 , 6).

Solution. with ~v = P Q~ = 〈 1 , 3 , 4 〉 we get get the line

〈x, y, z〉 = 〈 1 , 1 , 2 〉 + t〈 1 , 3 , 4 〉

which is ~r(t) = 〈1 + t, 1 + 3t, 2 + 4t〉. Since 〈x, y, z〉 = 〈 1 , 1 , 2 〉 + t〈 1 , 3 , 4 〉 consists of three equations x = 1 + 2t, y = 1 + 3t, z = 2 + 4t we can solve each for t to get t = (x − 1)/2 = (y − 1)/3 = (z − 2)/4.

A point P and two vectors ~v, ~w define a plane ~r(t) = OP~ + t~v + s ~w, where t, s are real numbers. This is called the parametric description of a plane.

2 An example is Σ = {〈x, y, z〉 = 〈 1 , 1 , 2 〉 + t〈 2 , 4 , 6 〉 + s〈 1 , 0 , − 1 〉 }.

If a plane contains the two vectors ~v and w~, then the vector

~n = ~v × w~

is orthogonal to both ~v and w~. Because also the vector P Q~ = OQ~ − OP~ is perpendicular to ~n, we have (Q − P ) · ~n = 0. With Q = (x 0 , y 0 , z 0 ), P = (x, y, z), and ~n = 〈a, b, c〉, this means ax+by +cz = ax 0 by 0 +cz 0 = d. The plane is therefore described by a single equation ax+by +cz = d, where d is a constant obtained by plugging in a point. We have just shown

The equation of the plane ~x = ~x 0 + t~v + s ~w

ax + by + cz = d ,

where 〈a, b, c〉 = ~v × w~ and d is obtained by plugging in ~x 0.

Math 21a: Multivariable calculus Oliver Knill, Fall 2017

Lecture 4: Functions

A function of two variables f (x, y) is a rule which assigns to two numbers x, y a third number f (x, y). For example, the function f (x, y) = x^2 y+2x assigns to (3, 2) the number 3^2 2 + 6 = 24. The domain D of a func- tion is set of points where f is defined, the range is {f (x, y) | (x, y) ∈ D }. The graph of f (x, y) is the sur- face {(x, y, f (x, y)) | (x, y) ∈ D } in space Graphs allow to visualize functions.

1 The graph of f (x, y) =

1 − (x^2 + y^2 ) on the domain D = {x^2 + y^2 < 1 } is a half sphere. The range is the interval [0, 1].

The set f (x, y) = c = const is called a contour curve or level curve of f. For example, for f (x, y) = 4x^2 + 3y^2 , the level curves f = c are ellipses if c > 0. The collection of all contour curves {f (x, y) = c } is called the contour map of f.

2 For f (x, y) = x^2 − y^2 , the set x^2 − y^2 = 0 is the union of the lines x = y and x = −y. The

curve x^2 − y^2 = 1 is made of two hyperbola with with their ”noses” at the point (− 1 , 0) and (1, 0). The curve x^2 − y^2 = −1 consists of two hyperbola with their noses at (0, 1) and (0, −1).

3 For f (x, y) = (x^2 − y^2 )e−x^2 −y^2 , we can

not find explicit expressions for the con- tour curves (x^2 − y^2 )e−x^2 −y^2 = c. but we can draw the curves with the com- puter:

A function of three variables g(x, y, z) assigns to three variables x, y, z a real num- ber g(x, y, z). We can visualize it by contour surfaces g(x, y, z) = c, where c is constant. It is helpful to look at the traces, the intersections of the surfaces with the coordinate planes x = 0, y = 0 or z = 0.

4 For g(x, y, z) = z−f (x, y), the level surface g = 0 which is the graph z = f (x, y) of a function

of two variables. For example, for g(x, y, z) = z − x^2 − y^2 = 0, we have the graph z = x^2 + y^2 of the function f (x, y) = x^2 + y^2 which is a paraboloid. Most surfaces g(x, y, z) = c are not graphs.

5 If f (x, y, z) is a polynomial and f (x, x, x) is quadratic in x, then {f = c} is a quadric.

Sphere (^) Paraboloid Plane

x^2 + y^2 + z^2 = 1 x^2 + y^2 − c = z ax^ +^ by^ +^ cz^ =^ d

One sheeted Hyperboloid Cylinder Two sheeted Hyperboloid

x^2 + y^2 − z^2 = 1 x^2 + y^2 = r^2 x^2 + y^2 − z^2 = − 1

Ellipsoid Hyperbolic paraboloid Elliptic hyperboloid

x^2 /a^2 + y^2 /b^2 + z^2 /c^2 = 1 x^2 − y^2 + z = 1 x^2 /a^2 + y^2 /b^2 − z^2 /c^2 = 1

• 3 - 2 - 1 1 2 3
  • 3
  • 2
  • 1

1

2

3

If ~r(t) = 〈x(t), y(t), z(t)〉 is a curve, then ~r ′(t) = 〈x′(t), y′(t), z′(t)〉 = 〈 x,˙ y,˙ z˙〉 is called the velocity at time t. Its length |~r′(t)| is called speed and ~v/|~v| is called direction of motion. The vector ~r ′′(t) is called the acceleration. The third derivative ~r ′′′^ is called the jerk. Then come snap, crackle, pop and Harvard.

If ~r ′(t) 6 = ~0, any vector parallel to ~r ′(t) is called tangent to the curve at ~r(t).

The addition rule in one dimension (f +g)′^ = f ′^ +g′, the scalar multiplication rule (cf )′^ = cf ′ and the Leibniz rule (f g)′^ = f ′g + f g′^ and the chain rule (f (g))′^ = f ′(g)g′^ generalize to vector- valued functions because in each component, we have the single variable rule. The process of differentiation of a curve can be reversed using the fundamental theorem of calculus. If ~r′(t) and ~r(0) is known, we can figure out ~r(t) by integration ~r(t) = ~r(0)+

∫ (^) t 0 ~r

′(s) ds.

Assume we know the acceleration ~a(t) = ~r′′(t) at all times as well as initial velocity and position ~r′(0) and ~r(0). Then ~r(t) = ~r(0)+t~r′(0)+ R~(t), where R~(t) =

∫ (^) t 0 ~v(s)^ ds and ~v(t) =

∫ (^) t 0 ~a(s)^ ds.

The free fall is the case when acceleration is constant. In particular, if ~r′′(t) = 〈 0 , 0 , − 10 〉, ~r′(0) = 〈 0 , 1000 , 2 〉, ~r(0) = 〈 0 , 0 , h〉, then ~r(t) = 〈 0 , 1000 t, h + 2t − 10 t^2 / 2 〉.

If r′′(t) = F~ is constant, then ~r(t) = ~r(0) + t~r′(0) − F t~^2 /2.

50 100 150 200 250

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Math 21a: Multivariable calculus Oliver Knill, Fall 2017

6: Arc length and curvature

If ∫ t ∈ [a, b] 7 → ~r(t) is a curve with velocity ~r ′(t) and speed |~r ′(t)|, then L = b a |~r^

′(t)| dt is called the arc length of the curve. Written out in three dimensions, this is L =

∫ (^) b a

x′(t)^2 + y′(t)^2 + z′(t)^2 dt.

1 The arc length of the circle of radius R given by ~r(t) = 〈R cos(t), R sin(t)〉 parameterized

by 0 ≤ t ≤ 2 π is 2π because the speed |~r′(t)| is constant R. The answer is 2πR.

2 The helix ~r(t) = 〈cos(t), sin(t), t〉 has velocity ~r ′(t) = 〈− sin(t), cos(t), 1 〉 and constant

speed |~r ′(t)| = 〈− sin(t), cos(t), 1 〉 =

3 What is the arc length of the curve

~r(t) = 〈t, log(t), t^2 / 2 〉

for 1 ≤ t ≤ 2? Answer: Because ~r′(t) = 〈 1 , 1 /t, t〉, we have ~r′(t) =

1 + (^) t^12 + t^2 = | (^1) t + t| and L =

1

1 t +^ t dt^ = log(t) +^

t^2 2 |

(^21) = log(2) + 2 − 1 /2.

4 Find the arc length of the curve ~r(t) = 〈 3 t^2 , 6 t, t^3 〉 from t = 1 to t = 3.

5 What is the arc length of the curve ~r(t) = 〈cos^3 (t), sin^3 (t)〉, 0 ≤ t ≤ 2 π? Answer: We have

|~r′(t)| = 3

sin^2 (t) cos^4 (t) + cos^2 (t) sin^4 (t) = (3/2)| sin(2t)|. The absolute value forces us to split the integral into 4 intervals. Since

∫ (^) π/ 2 0 sin(2t)^ dt^ = 1, we have^

∫ (^2) π 0 (3/2)|^ sin(2t)|^ dt^ = (3/2)4 = 6.

6 Find the arc length of ~r(t) = 〈t^2 / 2 , t^3 / 3 〉 for − 1 ≤ t ≤ 1. This cubic curve satisfies

y^2 = x^38 /9 and is an example of an elliptic curve. The speed is |~r(t)| =

∫ t^2 +^ t^4. Because x

1 + x^2 dx = (1 + x^2 )^3 /^2 /3, the arc length integral can be evaluated using substitution by as

− 1 |t|

1 + t^2 dx = 2

0 t

1 + t^2 dt = 2(1 + t^2 )^3 /^2 / 3 |^10 = 2(

7 The arc length of an epicycle ~r(t) = 〈t + sin(t), cos(t)〉 parameterized by 0 ≤ t ≤ 2 π. We

have |~r′(t)| =

2 + 2 cos(t). so that L =

∫ (^2) π 0

2 + 2 cos(t) dt. A substitution t = 2u gives L =

∫ (^) π 0

2 + 2 cos(2u) 2du =

∫ (^) π 0

2 + 2 cos^2 (u) − 2 sin^2 (u) 2du =

∫ (^) π 0

4 cos^2 (u) 2du = 4

∫ (^) π 0 |^ cos(u)|^ du^ = 8.

8 Compute the arc length of the catenary ~r(t) = 〈t, et^ + e−t〉 on an interval [a, b] can be

computed as eb^ − ea^ − e−b^ + e−a. By the way, (et^ + e−t)/2 is called the hyperbolic cosine and denoted by cosh(t).

Because a parameter change t = t(s) corresponds to a substitution in the integration which does not change the integral, we immediately have

The arc length is independent of the parameterization of the curve.

Math 21a: Multivariable calculus Oliver Knill, Fall 2017

7: Polar and spherical coordinates

A point (x, y) in the plane has the polar coordinates r =

x^2 + y^2 , θ = arctan(y/x) leading to the relation (x, y) = (r cos(θ), r sin(θ)). Θ

r

x

y

The formula θ = arctan(y/x) defines the angle θ only up to an addition of π as (x, y) and (−x, −y) have the same θ value. It is custom to let arctan(y/x) be in (−π/ 2 , π/2] for x > 0 and define it to be π/2 for the positive y axes and arctan(y/x) + π for x < 0 and equal to −π/2 on the negative y-axes. For (x, y) = (0, 0), the polar angle θ is not defined.

A curve given in polar coordinates as r(θ) = f (θ) is called a polar curve. It can in Cartesian coordinates be described as ~r(t) = 〈f (t) cos(t), f (t) sin(t)〉.

1 Describe the curve√ r = θ in Cartesian coordinates. Solution A formal substitution gives

x^2 + y^2 = arctan(y/x) but we can do better. Remember that for the curve ~r(t) = 〈r cos(t), r sin(t) where r = t = θ. The curve is a spiral.

2 What is the curve r = |2 sin(θ)|? Solution: Lets ignore the absolute value for a moment

and look at r^2 = 2r sin(θ). This can be written as x^2 + y^2 = 2y which is x^2 + y^2 − 2 y + 1 = 1. The curve is a circle of radius 1 centered at (0, 1). Since we have the absolute value, the radius at θ and θ + π is the same and add the circle of radius 1 centered at (0, −1).

A representation (x, y, z) = (r cos(θ), r sin(θ), z) is called cylindrical coordinates.

Here are some surfaces described in cylindrical coordinates:

3 r = 1 is a cylinder,

4 r = |z| is a double cone

5 θ = 0 is a half plane

6 r = θ is a rolled sheet of paper

7 r = 2 + sin(z) is an example of a surface of revolution.

Spherical coordinates use the distance ρ to the origin as well as two angles θ and φ. The first angle θ is the polar angle in polar coordinates of the xy coordinates and φ is the angle between the vector OP~ and the z-axis. The relation is

(x, y, z) = (ρ cos(θ) sin(φ), ρ sin(θ) sin(φ), ρ cos(φ)).

There are two important figures to see the connection. The distance to the z axes r = ρ sin(φ) and the height z = ρ cos(φ) can be read off by the left picture the rz-plane, the coordinates x = r cos(θ), y = r sin(θ) can be seen in the right picture the xy-plane.

r

z

ϕ ρ r x^ =^ ρ^ cos(θ) sin(φ), y = ρ sin(θ) sin(φ), z = ρ cos(φ)

x

y

r θ

Here are some level surfaces described in spherical coordinates.

8 ρ = 1 is a sphere,

9 The surface φ = π/4 is a single cone. The surface φ = π/2 is the xy-plane.

10 The surface sin(θ) = cos(θ) is a plane if we take the liberty to allow on the z-axes any θ

value.

11 ρ = φ is an apple shaped surface.

12 ρ = 2 + cos(7θ) sin(11φ) is an example of a bumpy sphere. The radius ρ depends on the

position.

13 Problem: Write x^2 + y^2 − 5 x = z^2 in cylindrical coordinates! Answer: r^2 − 5 r cos(θ) = z^2.

14 Match the surfaces with ρ = | sin(3φ)|, ρ = | sin(3θ)| in spherical coordinates (ρ, θ, φ). It

helps to see this in the rz plane or the xy plane.

II Spheres. Remember spherical coordinates? ρ = 1 was a sphere. If we vary the other parameters θ, φ we get ~r(θ, φ) = 〈cos(θ) sin(φ), sin(θ) sin(φ), cos(φ)〉. We can modify this to get parametrizations of ellipsoids. An example is ~r(θ, φ) = 〈3 cos(θ) sin(φ), 2 sin(θ) sin(φ), 5 cos(φ)〉.

III Graphs can be parametrized as

~r(x, y) = 〈x, y, f (x, y)〉.

The parametrization of the paraboloid z = x^2 + y^2 is 〈x, y, x^2 + y^2 〉. The graphs can also be turned. The parametrization of the surface y = cos(xz) for example is ~r(x, z) = 〈x, cos(xz), z〉.

IV Surfaces of revolution parametrizations are based on cylin- drical coordinates. ~r(u, v) = 〈g(v) cos(u), g(v) sin(u), v〉 It belongs to the surface r = g(z) which can be written as x^2 + y^2 = g(z)^2.

1 Describe the surface ~r(u, v) = 〈v^5 cos(u), v^5 sin(u), v〉.

Solution. It is a surface of revolution. We have r = v^5 = z^5. We see that r = z^5. This can be drawn in the rz plane.

2 Find a parametrization for the plane through the three points P = (3, 7 , 1),Q = (6, 2 , 1) and

R = (0, 3 , 4). Solution. Take ~r(s, t) = OP~ + s QP~ + t RP~. ~r(s, t) = 〈 3 − 3 s − 3 t, 7 − 5 s − 4 t, 1 + 3t〉.

3 Parametrize the lower half of the ellipsoid x^2 /4 + y^2 /9 + z^2 /25 = 1, z < 0. Solution. We

could solve for z and get ~r(u, v) = 〈u, v, −

25 − 25 u^2 / 4 − 25 v^2 / 9 〉. Better is to deform the sphere ~r(θ, φ) = 〈2 sin(φ) cos(θ), 3 sin(φ) sin(θ), 5 cos(φ)〉 and restrict φ to [π/ 2 , π].

4 Parametrize the upper half of the hyperboloid (x−1)^2 +y^2 / 4 −z^2 = −1. Solution. First take

the round hyperboloid r^2 = z^2 −1 is parametrized by ~r(θ, z) = 〈

z^2 − 1 cos(θ),

z^2 − 1 sin(θ), z〉. Deform this now to get for x^2 +y^2 / 4 −z^2 = −1 the parametrization ~r(θ, z) = 〈

z^2 − 1 cos(θ), 2

z^2 − 1 Now move still the x coordinate to get ~r(θ, z) = 〈1 +

z^2 − 1 cos(θ), 2

z^2 − 1 sin(θ), z〉.

The surface ~r(u, v) = 〈 2 + sin(13u) + sin(17v))〈cos(u) sin(v), sin(u) sin(v), cos(v)〉 is a hedgehog and looks like one you have done in the homework. In spherical coordinates, it is ρ = (2 + sin(13θ) + sin(17φ)).

Math 21a: Multivariable calculus Oliver Knill, Fall 2017

9: Continuity

A function f (x, y) with domain R is is continuous at a point (a, b) ∈ R if f (x, y) → f (a, b) whenever (x, y) → (a, b). The function f is continuous on R, if f is continuous for every point (a, b) on R. If a function can be continued uniquely to a point not in the domain, it is still considered continuous. The function sin(x)/x for example is continuous everywhere because there is only one way to continue it.

1 a) f (x, y) = x^2 + y^4 + xy + sin(y + sin sin sin sin(x)^2 ) is continuous on the entire plane. It is

built up from functions which are continuous using addition, multiplication or composition of functions which are all continuous.

2 f (x, y) = 1/(x^2 +y^2 ) is continuous everywhere except at the origin, where it is not defined.

3 f (x, y) = y + sin(x)/|x| is continuous away from x = 0. At every point (0, y) it is discontin-

uous. f (1/n, y) → y + 1 and f (− 1 /n, y) → y − 1 for n → ∞.

4 f (x, y) = sin(1/(x + y)) is continuous except on the line x + y = 0.

5 f (x, y) = (x^4 − y^4 )/(x^2 + y^2 ) is continuous at (0, 0). You can divide out x^2 + y^2 to see that

the function is equivalent to x^2 − y^2 away from (0, 0). After defining f (0, 0) = 0 it becomes continuous.

6 The function f (x) = e−^1 /x^2 is continuous everywhere. Actually, even all derivatives are

continuous and are zero at 0. Still, the function is not constant zero.

7 There are three sources for discontinuous behavior: there can be jumps, there can be poles,

or the function can oscillate. An example of a jump appears with f (x) = sin(x)/|x|, a pole example is g(x) = 1/x leads to a vertical asymptote and the function going to infinity. An example of a function discontinuous due to oscillations is h(x) = sin(1/x). Its graph is the devil’s comb.

The prototypes in one dimensions are

Jump f (x) = sin(x)/|x| - 5 5

Diverge g(x) = 1/x - 1.0^ - 0.5^ 0.5^ 1.

• 10
  • 5

0

5

10