Calculus III Final Review Notes

Chapter 13: Vectors and the Geometry of space

§13.1: 3-dimensional coordinate systems. Points in 3-D coordinate sys-tem

A point P is represented by a triple (a, b, c), or P (a, b, c).

a, b, c are the x-, y-, and z-coordinates of the point P .

In the 3-dimensional coordinate system, we have octants.

The first octant has all coordinates (+). The other octants aren’t numbered.

Axes and Planes

In 3-D coordinate system the axes are:

• The x-axis is set of all points (a, 0, 0) (only the x-coordinate can be non-zero).• The y-axis is the set of all points (0, b, 0).• The z-axis is the set of all points (0, 0, c).

Several important planes are:

• The xy-plane is the set of all points (a, b, 0) (all points with z-coordinate0).• The xz-plane is the set of all points (a, 0, c).• The yz-plane is the set of all points (0, b, c).

Distance from planes

The magnitude of the x-coordinate of P (a, b, c) is the distance of the point P fromthe yz-plane.

|b| is the distance of P from the xz-plane, and |c| is the distance of P from thexy-plane.

Projections onto planes

For P (a, b, c), the projection of P onto the xy-plane is the point Q on the xy-planeand closest to P . This point is Q(a, b, 0).

The projection of P onto the xz-plane is R(a, 0, c) and is the closest point on thatplane to P . The projection of P onto the yz-plane is S(0, b, c) and is the closestpoint on that plane to P .

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Example 1. Let P be the point P (2,−3, 5). Draw P , and find the projection ofP onto each of the xy-, xz-, and yz-planes. What is the distance of P from each ofthese planes?

Cartesian product The Cartesian product R3 = R×R×R = {(x, y, z) | x, y, z ∈R} is the set of all triples of real numbers. This corresponds to the set of points in3-dimensional space.

The first octant corresponds to the set of triples of positive reals.

Curves vs Surfaces In 2-D, the set of all points (a, b) satisfying an equation in xand y is a curve.

In 3-D, the set of all points (a, b, c) satisfying an equation in 3 variables x, y, andz is a surface.

We will often need to indicate whether this should be done in 2 or 3 dimensions.

Example 2. The equation x2 + y2 = 25 gives rise to both a curve and a surface.What are they?

Distance

For points P1(x1, y1, z1) and P2(x2, y2, z2) the distance between P1 and P2 is

|P1P2| =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

Example 3. Find the distance between P (0, 7,−5) and Q(−2,−2, 1).

Spheres

A sphere is the set of all points P (x, y, z) which are a fixed distance r (the radius)from a fixed point C(h, k, l) (the center).

A sphere is an example of a surface, the equation of which is

(x− h)2 + (y − k)2 + (z − l)2 = r2.

Example 4. What surface has the equation x2 + y2 + z2 − 4x+ 6z = 12?

Example 5. What is the equation for a sphere which has a diameter with endpoints(1, 4, 1) and (3, 0,−1)?

Example 6. What region is represented by the inequalities 4 ≤ (x − 1)2 + (y −2)2 + z2 ≤ 9 and z ≤ 0?

§13.2: Vectors. Vectors A vector is a quantity that has magnitude (size) anddirection.

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They are represented by arrows (directed line segments). The length of the arrowshould be proportional to the magnitude of the vector, and the arrow should pointin the direction of the vector.

Vectors are written several different ways: v or −→v .

Displacement vectors Vectors talk about movement. If an object starts at pointA (initial point) and ends at B (terminal point), the displacement vector v is the

vector v =−−→AB.

A vector doesn’t contain information about starting or end position, only direc-tion and magnitude. Two displacement vectors can have the same magnitude anddirection.

Equal vectors

If v =−−→AB and u =

−−→CD have the same magnitude and direction, then they are

equivalent/equal, written u = v.

The zero vector, denoted by 0, is the vector of magnitude 0. It does not need adirection.

Adding vectors

Combine vectors the same way you combine displacements of an object/particle.

To add−−→AB and

−−→BC we think about the total displacement of a particle moving

from A to B, and then to C; so the vector−→AC.

−→AC is the sum of

−−→AB and

−−→BC, or

−→AC =

−−→AB +

−−→BC.

Adding vectors more generally To add general vectors u and v, reposition vso that its tail (start) meets the tip (end) of the vector u.

u + v is the vector from the initial point of u to the terminal point of v, once theyare lined up correctly.

Vector addition is commutative: u + v = v + u

Scalar multiplication

We can also multiply a vector by a real number c, or a scalar.

(1) If c > 0 and v 6= 0 then cv is c times as long and has the same direction asv.

(2) If c < 0 and v 6= 0 then cv is |c|v with the direction flipped.(3) If c = 0 or v = 0, then cv = 0.

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Vectors are parallel ⇐⇒ they are scalar multiples of each other

Vector subtraction

We can subtract too.

u− v = u + (−v) = u + (−1v)

Representing vectors Represent vectors as triples 〈a, b, c〉.

The meaning is that the displacement is by a in the x direction, b in the y direction,and c in the z direction.

The vector v = 〈a, b, c〉 is the displacement vector when moving from the origin O

to the point P (a, b, c); so v =−−→OP .

Vector components, position vectors

a, b and c are the components of the vector 〈a, b, c〉.

A different placement of a vector is a different representation. The position vectorof a point P is the representation of a vector from the origin O to point P .

If v has representation−−→AB for pts A(x1, y1, z1) and B(x2, y2, z2), then v =

−−→AB is

v = 〈x2 − x1, y2 − y1, z2 − z1〉.

Length of a vector The length of v = 〈a, b, c〉 is |v| =√a2 + b2 + c2.

In 2 dimensions, the length of v = 〈a, b〉 is |v| =√a2 + b2.

Vector operations using components

For u = 〈x1, y1, z1〉 and v = 〈x2, y2, z2〉 and c ∈ R

(1) u + v = 〈x1 + x2, y1 + y2, z1 + z2〉(2) u− v = 〈x1 − x2, y1 − y2, z1 − z2〉(3) cu = 〈cx1, cy1, cz1〉

Vn is the set of all n-dimensional vectors v = 〈x1, x2, . . . , xn〉.

e.g. V3 = {〈x, y, z〉 | x, y, z,∈ R}.

Properties of vectors

For vectors a,b, c in Vn and c, d ∈ R:

(1) a + b = b + a(2) a + (b + c) = (a + b) + c

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(3) a + 0 = a(4) a + (−a) = 0(5) c(a + b) = ca + cb(6) (c+ d)a = ca + da(7) (cd)a = c(da)(8) 1a = a

Basis vectors The standard basis vectors for V3 are i = 〈1, 0, 0〉, j = 〈0, 1, 0〉 andk = 〈0, 0, 1〉.

Each of these has length 1, and direction of positive x-, y-, and z-axis respectively.

For V2, i = 〈1, 0〉 and j = 〈0, 1〉 are the standard basis vectors.

We can write any vector in V3 as sum of scalar multiples of i, j,k.

Example 7. Write u = 〈2,−5, 1〉 and v = 〈0, 2,− 12 〉 in terms of i, j, and k What

is u− 2v?.

Unit vectors Unit vectors are vectors of length 1, e.g. i, j,k.

If a 6= 0, then there is a unit vector 1|a|a whose direction is the same as a.

Example 8. Find a unit vector with the same direction as v = 〈4, 4,−2〉.

Application to physics Like vectors, a force has magnitude and direction. Werepresent forces as vectors.

The resultant force is the sum of forces, calculated using the sum of vectors.

Example 9. Consider the vector v = i + 3j− 2k. Find a vector whose direction isopposite to v, and whose length is 3.

Relation to trigonometry If you know the magnitude of a vector, and the di-rection (in terms of an angle), what can you say about the components of thevector?

Example 10. A man is on a raft that is moving due south at a speed of 15 mph.He is facing northeast and walking at a speed of 2

√2 mph. What is his speed

relative to the water?

§13.3: The Dot Product. Dot product We use the dot product to “multiply”two vectors.

The dot product of a = 〈a1, a2, a3〉 and b = 〈b1, b2, b3〉 is

a · b = a1b1 + a2b2 + a3b3

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This is also know as the inner product or scalar product.

Note that this is a scalar, not a vector.

We can define a dot product similarly for any Vn.

Example 11. Compute 〈 12 ,−3, 4〉 · 〈4, 1,−1〉.

Example 12. Find the dot product of −2i + 3j + k and 3i− j + k.

Properties of dot product

For vectors a,b, c in V3:

(1) a · a = |a|2(2) a · b = b · a(3) a · (b + c) = a · b + a · c(4) (ca) · b = c(a · b) = a · (cb)(5) 0 · a = 0

If θ is the angle formed by a and b, then a · b = |a||b| cos θ.

For parallel vectors a and b, note that θ = 0 or θ = π.

For perpendicular/orthogonal vectors, we have a · b = 0.

The dot product will be positive if the angle is acute, negative if the angle isobtuse.

Example 13. What is the angle formed by the vectors i + j and i− 2j?

Direction angles and cosines

The direction angles of a vector v = 〈a, b, c〉 6= 0 are the angles α, β, γ ∈ [0, π] thatlie between v and the positive x-, y-, and z-axis respectively.

The direction cosines of v = 〈a, b, c〉 are cosα, cosβ, and cos γ.

(1) cosα = a|v|

(2) cosβ = b|v|

(3) cos γ = c|v|

We can now rewrite v = 〈|v| cosα, |v| cosβ, |v| cos γ〉.

Example 14. Consider the vector v = 〈3, 2,−2〉. What are the direction cosinesof v? What are the direction angles of v?

Vector projections

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The vector projection of b onto a is

projab =

(a · b|a|

)a

|a|=

a · b|a|2

a

The vector projection of b onto a is a vector c parallel to a and such that b− c isperpendicular to a.

Scalar projections

The scalar projection of b onto a is

compab =a · b|a|

The scalar projection of b onto a is the signed magnitude of the vector projection,that is |b| cos θ where θ is the angle between a and b.

Example 15. Find the scalar and vector projections of 〈−3,−4〉 onto 〈2, 1〉.

In physics, the work done by a constant force F in moving an object along displace-ment vector D is calculated by W = F ·D.

Example 16. A tall man is pushing a shopping cart at a 45 degree angle fromthe floor. He is pushing with a force of 10 N, and moves the cart a distance of 20meters. How much work has been done?

§13.4: The Cross Product. Cross product The cross product of a = 〈a1, a2, a3〉and b = 〈b1, b2, b3〉 is

a× b = 〈a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1〉

Note that this is a vector, unlike the dot product.

a× b is (usually) nonzero.

Understanding the cross product

a× b is perpendicular/orthogonal to a and b.

There is no cross-product in 2 dimensions (why?).

The cross product is the determinant

a× b =

∣∣∣∣∣∣i j ka1 a2 a3b1 b2 b3

∣∣∣∣∣∣|a× b| = |a||b| sin θ

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Graphical representation

The orientation of a× b is determined by right-hand rule.

a and b are parallel if and only if a× b = 0.

The length of a× b is the area of the parallelogram with sides given by a and b.

Example 17. Find all possible cross products among i, j, and k.

This gives us examples of commutativity and associativity not holding.

If a,b, c are vectors and c a scalar:

(1) a× b = −b× a(2) (ca)× b = c(a× b) = a× (cb)(3) a× (b + c) = a× b + a× c(4) (a + b)× c = a× c + b× c(5) a · (b× c) = (a× b) · c(6) a× (b× c) = (a · c)b− (a · b)c

Example 18. Without computing anything, what will be the sign of the x-, y-,and z-components of the vector 〈1,−2,−1〉 × 〈3, 2,−2〉?

Triple products

The scalar triple product of a,b, and c is

a · (b× c) =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣The magnitude of the scalar triple product is the volume of the parallelepiped withsides a,b, and c. That is V = |a · (b× c)|.

If this volume/triple product is 0, then the vectors are coplanar.

The vector triple product is a× (b× c).

Example 19. Find the scalar triple product of i− j−k, −i+2j+k, and 3i+ i+k.

Example 20. What is the volume of the parallelepiped defined by 〈2, 1, 2〉, 〈3,−1,−2〉and 〈2, 2, 2〉?

Example 21 (from homework). Find two unit vectors orthogonal to both 〈1,−1, 1〉and 〈0, 4, 4〉.

Example 22. Find a unit vector orthogonal to both 〈0, 2, 3〉 and 〈0,−4,−6〉.

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§13.5: Equations of Lines and Planes. Lines

A 3-D line L is determined by a point on the line P0(x0, y0, z0) and the directionof L, given as a vector v parallel to L.

Equations of vectors

Let P (x, y, z) be arbitrary on L, and r0 and r be the position vectors of P0 and P

(i.e. with representations−−→OP0 and

−−→OP respectively). Then there is a t such that

r = r0 + tv.

This is called the vector equation for L. Each value of the parameter t gives theposition vector r of a point on L.

Parametric equations for a line

We can write the vector equation in component form: 〈x, y, z〉 = 〈x0 + ta, y0 +tb, z0 + tc〉.

Broken into components, we have 3 parametric equations for L:

(1) x = x0 + at(2) y = y0 + bt(3) z = z0 + ct

Example 23. What are the vector and parametric equations for the line thatpasses through the point (0, 2,−2) and is parallel to the vector 〈−1,−2, 3〉?

Example 24. Find another point on the line.

Example 25. What are the vector and parametric equations for the line thatpasses through the points A(−1, 10,−4) and B(3, 2,−2)?

Components

When L is described with a direction vector v = 〈a, b, c〉, we call the componentsof v the direction numbers of L.

Any line will have many sets of direction numbers, since anything parallel to v canbe used instead.

If no direction number of L is 0, we can write a symmetric equation for L byeliminating the parameter t:

x− x0a

=y − y0b

=z − z0c

Example 26. Write a symmetric equation for the line that passes through thepoint (0, 2,−2) and is parallel to the vector 〈−1,−2, 3〉?

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Example 27. Where does this line intersect the xz-plane?

Line segments

To get a line segment instead of a line, we restrict the values of the parameter t.

The line segment from r0 to r1 is given by

r(t) = (1− t)r0 + tr1 0 ≤ t ≤ 1

Lines or line segments are skew if they do not intersect.

Example 28. Do the lines L1 and L2 defined by the parametric equations

x = 1 + t y = −2 + 3t z = 4− tand

x = 2s y = 3 + s z = −3 + 4s

intersect in a point, are they parallel, or are they skew?

Planes

A plane is determined by a point and the direction of a vector which is orthogonalto the plane. Note that a single vector in the plane would not tell us which planeit is.

The vector orthogonal to the plane is called the normal vector, n.

Equations of planes

If r0 =−−→OP0 is a fixed position vector of a point on the plane, and r =

−−→OP is an

arbitrary position vector of a point on the plane, then r − r0 =−−→P0P lies on the

plane and so is orthogonal to n. Then n · (r− r0) = 0.

The vector equation of a plane is either

n · (r− r0) = 0 or n · r = n · r0

Scalar equations for planes

Let n = 〈a, b, c〉, r = 〈x, y, z〉, r0 = 〈x0, y0, z0〉. In terms of components, we get thescalar equation of a plane:

a(x− x0) + b(y − y0) + c(z − z0) = 0

This is the the scalar equation of the plane through P0(x0, y0, z0) with normalvector n = 〈a, b, c〉.

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We can rewrite the equation of a plane by grouping the terms, and get the linearequation:

ax+ by + cz + d = 0

Two planes are parallel if and only if their normal vectors are parallel.

Two planes that are not parallel intersect in a line. The angle between two planesis the acute angle formed by their normal vectors.

Example 29. What plane contains the point (−2, 2,−2) and has normal vector〈3,−1, 2〉? What are the intercepts of this plane?

Example 30. What plane contains the point (1, 0, 0) and is parallel to the planex− y + 5z = 10?

Example 31. What plane contains the points (1, 1, 1), (2, 2, 2), and (3, 4, 5)?

Example 32. Find an equation for the line formed by the intersection of the planes2x+ 3y − z = 1 and 4x− y − 2z = 2.

Distance from a point to a plane

The distance from a point P (x1, y1, z1) to the plane ax+ by+ c+ d = 0 is given by

D =|ax1 + by1 + cz1 + d|√

a2 + b2 + c2

§13.6: Cylinders and Quadratic Surfaces. §13.6: Cylinders and QuadraticSurfaces

Tracings, rulings, and cylinders The tracings of a surface are the curves thatresult when you intersect a surface with many given planes.

A cylinder is a type of surface defined to be the set of all lines (rulings) parallel toa given line and passing through a given plane curve.

Example 33. x2 + y2 = 25 defines a cylinder. It is the set of all lines parallelto the line x = y = 0 and passing through a certain circle in the xy-plane. Thetracings of the surface in the planes z = k are circles.

Example 34. Graph the surface z = sin y.

Parabolic cylinders and quadratic surfaces A parabolic cylinder is a cylindermade up of infinitely many translated copies of a given parabola.

A quadratic surface is the graph of any 2nd degree equation in x, y, and z.

Types of quadratic surfaces

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• Ellipsoidx2

a2+y2

b2+z2

c2= 1

Traces are ellipses.• Elliptic Paraboloid

z

c=x2

a2+y2

b2

Traces are parabolas (vertically) and ellipses (horizontally).


• Hyperbolic Paraboloidz

c=x2

a2− y2

b2

Traces are parabolas (vertically) and hyperbolas (horizontally).• Cone

x2

a2+y2

b2=z2

c2

Traces are ellipses (horizontal), hyperbolas (vertical), and pairs of inter-secting lines (vertical).


• Hyperboloid of one sheet

x2

a2+y2

b2− z2

c2= 1

Traces are ellipses (horizontal) and hyperbolas (verical).• Hyperboloid of two sheets

−x2

a2− y2

b2+z2

c2= 1

Traces are ellipses (horizontal), nothing (horizontal), or hyperbolas (verti-cal).

Example 35. Sketch traces of and classify the surface whose equation is 100x −4y2 + 25z2 = 0.

Example 36. Classify and sketch the surface whose equation is 4x2 + 4y2 + z2 −8x− 16y + 2z + 17 = 0.

Review of chapter 13. Some review questions

Example 37. What is the distance between a plane and a line that does notintersect that line?

Example 38. What is the distance between two parallel lines?

Example 39. What is the equation of a plane that contains two given lines?

Example 40. At what angle does a given line intersect a plane?

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Chapter 14: Vector Functions

§14.1: Vector Functions and Space Curves. Functions A function is anymethod or rule for assigning to every element in the domain a single element in therange.

If f is a function which takes real numbers as input, the domain of f is the set ofall values of t for which f(t) is defined (or exists).

The range of f is the set of all possible outputs of f , that is {f(t) | t ∈ R}.

Vector functions A vector-valued function is a function whose domain is a set ofreal numbers and whose range is a set of vectors.

If r is a vector function, we can write

r(t) = 〈f(t), g(t), h(t)〉

where f, g, h are called the component functions of r.

Example 41. What is the domain of the function

r(t) =

⟨√t2 − 11

t,t+ 4

t+ 12, cot t

⟩?

Limits The limit of a vector function is defined by the limits of the componentfunctions: when the individual limits exist, we define

limt→a

r(t) =⟨

limt→a

f(t), limt→a

g(t), limt→a

h(t)⟩.

A vector function r is continuous at a if limt→a r(t) = r(a).

Example 42. Compute limt→1 r(t) where

r(t) =

⟨t2 + t− 2

t2 − 1, 5√t− 1,

1

t

⟩.

Is this function continuous at 1?

Space curves If f, g, h are continuous real-valued functions on an interval I, thenthe set C of all points (x, y, z) = (f(t), g(t), h(t)) for t ∈ I is called a space curve.

The functions f, g, h are called the parametric equations of C and t is called theparameter.

A space curve is like the path of a particle whose position at time t is given by(f(t), g(t), h(t)).

Example 43. Describe and draw the space curve given by the vector function

r(t) = 〈cos t, sin t, t〉.

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Example 44. Draw the space curve whose vector function is

r(t) = 〈cos t,− cos t, sin t〉

Parametrizing curves If we have an arbitrary space curve C, we often will wantto find a vector equation which describes C. To do so is to parametrize C.

Example 45. Parametrize the curve which is the intersection of the cone z =√x2 + y2 and the plane z = y + 1.

Example 46. Parametrize the line passing through the points (−4, 2, 3) and (0, 1, 2).

§14.2: Derivatives and Integrals of Vector Functions. Derivatives of vec-tor functions If r is a vector function, then the derivative of r is the vector functiondefined by

r′(t) =dr

dt= limh→0

r(t+ h)− r(t)

hwhen this limit exists.

We call r′(t) the tangent vector to the curve C at the point P (the point whoseposition vector is P ). It tells you the direction of C at the point P .

Tangent lines, components of derivatives The tangent line to C at point P isthe line through P and parallel to r′(t).

The unit tangent vector is the unit vector whose direction is the same as r′(t), that

is T(t) = r′(t)|r′(t)| .

Derivatives and components

If we write our vector function as r(t) = 〈f(t), g(t), h(t)〉, then we can also just takederivatives in terms of components:

r′(t) = 〈f ′(t), g′(t), h′(t)〉 = f ′(t)i + g′(t)j + h′(t)k

We can also iterate our derivatives:

r′′ = (r′)′

Example 47. Find the derivative of r(t) = (t3 − t)i + (tet)j + (cos t)k.

Example 48. Find the unit tangent vector and tangent line at the point wheret = 0.

Example 49. The curves r1(t) = 〈3t, t2 + t,−t3〉 and r2(t) = 〈t5 − t,−t, t2 + t〉intersect at the origin (0, 0, 0). How could we find their angle of intersection?

Properties of derivatives If u and v are differentiable vector functions, c is ascalar, and f is a real-valued function, then:

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(1) ddt (u(t) + v(t)) = u′(t) + v′(t)

(2) ddt (cu(t)) = cu′(t)

(3) ddt (f(t)u(t)) = f ′(t)u(t) + f(t)u′(t)

(4) ddt (u(t) · v(t)) = u′(t) · v(t) + u(t) · v′(t)

(5) ddt (u(t)× v(t)) = u′(t)× v(t) + u(t)× v′(t)

(6) ddt (u(f(t))) = f ′(t)u′(f(t))

Integrals

The definite integral of a continuous vector function

r(t) = 〈f(t), g(t), h(t)〉can be defined similarly, in terms of the integrals of the components:∫ b

a

r(t)dt =

(∫ b

a

f(t)dt

)i +

(∫ b

a

g(t)dt

)j +

(∫ b

a

h(t)dt

)k

Example 50. Compute∫ 2

0

((cosπt)i + (et)j + (t2 + 1)k)dt

§14.3:Arc Length and Curvature. Curve length If a space curve has vectorequation r(t) = 〈f(t), g(t), h(t)〉 for a ≤ t ≤ b and f ′, g′, h′ are all continuous onthis interval, the length of the curve∗ is given by

L =

∫ b

a

√[f ′(t)]2 + [g′(t)]2 + [h′(t)]2 dt

=

∫ b

a

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt

*If the curve never retraces its steps or backtracks.

Simpler statement of curve length Note that both of these forms could berewritten as

L =

∫ b

a

|r′(t)|dt

It is useful to note that the length of a curve is independent of the parametrizationused to describe the curve.

Example 51. Find length of curve

r(t) = 〈cos t, sin t, t〉for 0 ≤ t ≤ 4π.

Example 52. Find length of curve

r(t) =

⟨2t, t2,

1

3t3⟩

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for 0 ≤ t ≤ 1.

Arc length function If a space curve is given by a vector function r(t) = 〈f(t), g(t), h(t)〉for a ≤ t ≤ b and f ′, g′, h′ are all continuous on this interval, the arc length functions is defined as

s(t) =

∫ t

a

|r′(u)|du

=

∫ t

a

√[f ′(u)]2 + [g′(u)]2 + [h′(u)]2 du

The function s tells us the length of the curve r between a and t.

Using the fundamental theorem of calculus, we have

ds

dt= |r′(t)|

Sometimes we can parametrize a curve with respect to arc length. This will oftenjust require a substitution for t, using s(t).

Example 53. Reparametrize

r(t) = 〈cos t, sin t, t〉

with respect to arc length from (1, 0, 0) in the direction of increasing t.

Example 54. Reparametrize the curve

r(t) = 2ti + (1− 3t)j + (5 + 4t)k

with respect to arc length measured from t = 0 in the direction of increasing t.

Smooth curves A parametrization r(t) is called smooth on an interval I if r′(t) 6= 0on I.

A curve is smooth if it has a smooth parametrization.

If T is the unit tangent vector, the curvature of a curve is

κ =

∣∣∣∣dTds∣∣∣∣ .

Rewriting curvature Note that κ is expressed in terms of arc length. A moreuseful form is

κ =

∣∣∣∣dT/dtds/dt

∣∣∣∣ .Expressed as a function, we can write κ(t) =

|T′(t)||r′(t)| .

17

Another useful form of the curvature is

κ(t) =|r′(t)× r′′(t)||r′(t)|3

Example 55. Find the curvature of the circle

r(t) = 〈5 cos t, 5 sin t, 0〉.

Example 56. Find the curvature of the curve given by

r(t) = 〈t, t2, t3〉.

Normal and binormal vectors

The principal unit normal vector is

N(t) =T′(t)

|T′(t)|.

This vector is orthogonal to the unit tangent vector T(t).

The binormal vector is B(t) = T(t)×N(t) and is perpendicular to both T and N .

Example 57. Find the unit normal and binormal vectors for the curve

r(t) = 〈cos t, sin t, t〉both for arbitrary t and at the point given by t = π.

Example 58. Find the unit normal and binormal vectors for the curve

r(t) = 〈t, 5, ln cos t〉for arbitrary t ∈ (−π/2, π/2) and at the point given by t = 0.

Normal and osculating plane Since the vectors N and B are not parallel, theydetermine a plane at any point P on the curve. This is called the normal plane ofC at P

The plane determined by N and T is called the osculating plane of C at P .

Osculating circle

The osculating circle of C at P is the circle that lies in the osculating plane of Cat P , has the same tangent as C at P , lies on the concave side of C and has radiusρ = 1/κ.

The osculating circle is the circle whose behavior most resembles that of C, in thatthey share the same tangent and normal vectors, and have the same curvature atthe point P .

Example 59. Consider the parabola in the xy-plane given by y = 12x

2. Find the

osculating circle of the parabola at the point (1, 12 ).

18

§14.4: Motion in Space: Velocity and Acceleration. Particles in space

If r(t) describes the position of a particle in space at time t, then the velocity vectorat time t is given by v(t) = r′(t).

The speed of the particle at time t is the magnitude of the velocity vector, that is

|v(t)| = |r′(t)| = ds

dt.

The acceleration of the particle is defined as a(t) = v′(t) = r′′(t).

Example 60. A particle’s position at any time t ≥ 0 is given by r(t) = ti+t2j+t3k.Find the velocity, acceleration, and speed of the particle for arbitrary t and for t = 1.

Example 61. A particle’s position at any time t ≥ 0 is given by r(t) = 5 cos ti +5 sin tj + tk. Find the velocity, acceleration, and speed of the particle for arbitraryt and for t = π.

Recovering velocity and position Using integrals, we can recover velocity fromacceleration. That is,

v(t) = v(t0) +

∫ t

t0

a(u)du

In addition, we can recover position from velocity, using

r(t) = r(t0) +

∫ t

t0

v(u)du

Example 62. A particle starts at position r(0) = 〈1, 0, 0〉 with velocity v(0) =i− j + k. Its acceleration is given by the function a(t) = 4ti + 6tj + k. Find vectorequations for its velocity and position at any time t.

Components of acceleration

If r(t) describes the path of a particle, it is always true that the acceleration a(t)at a fixed point lies in the osculating plane at that point (the plane determined byT and N).

So we can decompose acceleration into two components: one in the direction of thetangent, and one in the direction of the normal vector.

Decomposing acceleration If v is the speed of the particle, then a = v′T+κv2N.

We call aT = v′ and aN = κv2 the tangential component and normal component ofacceleration.

19

Two useful simplifications are:

aT =r′(t) · r′′(t)|r′(t)|

aN =|r′(t)× r′′(t)

|r′(t)|.

Example 63. Find the tangential and normal components of acceleration for aparticle whose position is given by r(t) = 〈1, t, t2〉.Example 64. Find the tangential and normal components of acceleration for aparticle whose position is given by r(t) = 〈t2, cos t, sin t〉.

Chapter 15: Partial Derivatives

§15.1: Functions of Several Variables. Functions of two variables

A function f of two variables is a rule assigning to pairs (x, y) of real numbers aunique real number f(x, y).

The domain is a subset of R× R = R2, and the range is a subset of R.

We will often write z = f(x, y). In this notation, we say x and y are independentvariables, while z is the dependent variables.

Example 65. Sketch the domains and ranges of the functions

f(x, y) =√x2 − y

g(x, y) = exy

Graphing functions

If f is a function of two variables whose domain is D, then the graph of f is the setof all points (x, y, z) such that z = f(x, y) and (x, y) ∈ D.

In other words, the graph is the set of all points (x, y, f(x, y)).

A function of the form f(x, y) = ax+ by + c is called a linear function. The graphof a linear function is a plane.

Example 66. Sketch the function f(x, y) = x2 + y2.

Level curves

The level curves of a function f of two variables are the curves with equationsf(x, y) = k for some constant k (in the range of f).

A level curve can be thought of as the set of (x, y) in the domain of f for which ftakes a given fixed value.

20

Real world level curves include contour lines (topographical maps), isothermals(temperature), isobars (pressure), isobaths (depth), . . .

Example 67. Sketch the level curves of the function f(x, y) = x2 + 9y2.

Functions of 3 variables

A function f of three variables is a rule assigning to ordered triples (x, y, z) of realnumbers in the domain D a unique real number f(x, y, z).

Example 68. Find the domain and range of f(x, y, z) =√

25− x2 − y2 − z2.

Visualizing functions of 3 variables

Note that it is very hard to visualize a function f of 3 variables, since the graph off lives in 4 dimensional space.

However, we can at least visualize the level surfaces of f , that is the surfaces withequations f(x, y, z) = k for fixed k in the range.

Example 69. What are the level surfaces of f(x, y, z) = x+ y + z?

Functions of n real variables

A function f of n variables is a rule assigning to ordered triples (x1, . . . , xn) of realnumbers in the domain D ⊂ Rn a unique real number z = f(x1, . . . , xn).

Since we can identify Rn with V n, we can also think about a function of n realvariables x1, . . . , xn as a function of a single vector variable x = 〈x1, . . . , xn〉.

§15.2: Limits and Continuity. Limits of functions of 2 variables

If f is a function of two variables with domain D including points arbitrarily closeto (a, b), then the limit of f(x, y) as (x, y) approaches (a, b) is defined to be

lim(x,y)→(a,b)

f(x, y) = L

if for every ε > 0, there is a δ > 0 such that if (x, y) ∈ D and 0 <√

(x− a)2 + (y − b)2 <δ, then |f(x, y)− L| < ε.

Example 70. If f(x, y) = x2 + y2 − x, what is lim(x,y)→(0,0) f(x, y)?

Another way to think about limits

Explained another way, the limit of f(x, y) as (x, y) → (a, b) is defined to be thecommon limit of f along any curve passing through (a, b).

So if f(x, y)→ L1 as (x, y)→ (a, b) along a path C1, but f(x, y)→ L2 as (x, y)→(a, b) along C2 (and L1 6= L2), then the limit does not exist.

21

Example 71. If f(x, y) = x2−y2x2+y2 , what is lim(x,y)→(0,0) f(x, y)?

Continuity

A function f of two variables is continuous at (a, b) if

lim(x,y)→(a,b)

f(x, y) = f(a, b).

A function f is continuous on D if f is continuous at every point (a, b) ∈ D.

Examples of continuous functions

A function f of two variables is called a polynomial if it is a sum of terms of theform axmyn for nonnegative integers m,n.

As with polynomials of 1 variable, polynomials are continuous.

A function f of two variables is called a rational function if it has the form f(x, y) =g(x, y)/h(x, y) where g and h are polynomials.

As with rational functions of 1 variable, rational functions are continuous when theyare defined.

Example 72. Let f(x, y) = x3y2 − 8xy3 + 2x2 − y2 + xy + 1. At what points is itcontinuous? Find lim(x,y)→(1,1) f(x, y).

Example 73. Let f(x, y) = x−yx2−y2 . At what points is it continuous? Find

lim(x,y)→(2,1) f(x, y) and lim(x,y)→(0,0) f(x, y).

Example 74. Let f(x, y) = tan(x+y)ln(x2+y2+1) . At what points is f continuous?

Limits and continuity for other types of functions

Limits and continuity are defined analogously for functions of 3 (and n) variables:

If f is a function with domain D ⊆ Rn, then the limit of f(x)as x approaches a is defined to be

L = limx→a

f(x)

if for every ε > 0, there is a δ > 0 such that |f(x) − L| < ε whenever x ∈ D and0 < |x− a| < δ.

A function f of n real variables is said to be continuous at a if f(a) = limx→a f(x).

Now limits and continuity look just like for functions of 1 real variable.

Example 75. Show that lim(x,y)→(0,0)xy

x2+y2 does not exist.

Example 76. Show that lim(x,y)→(0,0)xy4

x2+y8 does not exist.

22

§15.3: Partial Derivatives. Partial derivative of a function with respectto x

If f is a function of two variables x and y, we can fix y and think of f as a functiononly of x: that is g(x) = f(x, b). If the function g has a derivative, we can think ofthis as a derivative of f as well, in a certain sense.

The partial derivative of f with respect to x at (a, b) is

fx(a, b) = g′(a) where g(x) = f(x, b).

Or equivalently:

fx(a, b) = limh→0

f(a+ h, b)− f(a, b)

h

Partial derivative of a function with respect to y

Define similarly the partial derivative of f with respect to y at (a, b) to be

fy(a, b) = limh→0

f(a, b+ h)− f(a, b)

h= g′(b)

where g(y) = f(a, y).

So more generally, we can think of the partial derivatives as the functions definedby

fx(x, y) = limh→0

f(x+ h, y)− f(x, y)

h

fy(x, y) = limh→0

f(x, y + h)− f(x, y)

h

Calculating partial derivatives

Other ways to denote the partial derivatives of a function z = f(x, y) include:

fx(x, y) = fx =∂f

∂x=

∂

∂xf(x, y) = f1 = D1f = Dxf

fy(x, y) = fy =∂f

∂y=

∂

∂yf(x, y) = f2 = D2f = Dyf

Calculating partial derivatives:

• To find fx, treat y as a constant and differentiate with respect to x.• To find fy, treat x as a constant and differentiate with respect to y.

Example 77. Let f(x, y) = x3y4 + 3x2y2 − 2x+ 3y + 1. Calculate fx and fy.

Example 78. Let f(x, y) = exy sin(x+ y)− ln(x2y3). Calculate fx and fy.

23

Visualizing partial derivatives Partial derivatives give us tangents along thecurves we get taking a fixed trace.

Alternatively, they say how fast f is changing as we vary x or y only.

Example 79. Interpret the meaning of fx(1, 1) and fy(1, 1) where f(x, y) = x2+y2.

Implicit differentiation

Sometimes a function f(x, y) is defined implicitly by an equation in x, y, z. Inthis case, often the best way to calculate partial derivatives is through implicitdifferentiation.

Example 80. Use implicit differentiation to find ∂z/∂x and ∂z/∂y where

y3z + x2 ln y = xz

Example 81. Use implicit differentiation to find ∂z/∂x and ∂z/∂y where

ez = x+ y + xyz

Partial derivatives of functions of many variables

We define partial derivatives for functions of more than 2 variables analogously.For example, if u = f(x1, . . . , xn) is a function of n variables, then we can writeany of

∂u

∂xi=

∂f

∂xi= fxi = fi = Dif

to mean

limh→0

f(x1, . . . , xi−1, xi + h, xi+1, . . . , xn)− f(x1, . . . , xi, . . . , xn)

h.

We can calculate fxiby treating every other variable as a constant and differenti-

ating with respect to xi.

Iterating partial derivatives

We can also iterate partial derivatives. If we want to take the partial derivative offx with respect to x, we might write:

(fx)x = fxx = f11 =∂

∂x

(∂f

∂x

)=∂2f

∂x2=∂2z

∂x2

or if we wish to take the partial derivative of fx with respect to y we might write:

(fx)y = fxy = f12 =∂

∂y

(∂f

∂x

)=

∂2f

∂y∂x=

∂2z

∂y∂x

Example 82. Find all second partial derivatives of

f(x, y) = sin(xy) cos(xy)

24

An important theorem (due to Clairaut) states: If fxy and fyx are both definedand continuous on a disk D containing (a, b), then fxy(a, b) = fyx(a, b).

We often use equations including partial derivatives, known as partial differentialequations. For example, PDEs can be used to express laws or properties in physics.

One well known PDE is Laplace’s equation:

∂2u

∂x2+∂2u

∂y2= 0

A solution to this equation is called a harmonic function.

Another famous PDE is the wave equation

∂2u

∂t2= a2

∂2u

∂x2

where u is a function of x and t, and a is a constant.

Example 83. Show that u(x, t) = cos(3x+ 2t) satisfies the wave equation.

Example 84. Is either of the functions u = yx2 − xy2 or u = ex sin y − ey sinx asolution to Laplace’s equation?

§15.4: Tangent Planes and Linear Approximations. Tangent plane

Consider a surface S with equation z = f(x, y), where f has continuous first partialderivatives. Let C1 be the trace of the surface in the plane x = a and let C2 be thetrace of the surface in the plane y = b.

Then if T1 and T2 are the tangent lines to the curves at the point P (x0, y0, z0), thetangent plane to the surface S at the point P is the plane containing both tangentlines T1 and T2.

Equation of tangent plane

Actually, the tangent plane contains the tangent line along any curve C lying onthe surface S and passing through P .

The equation of the tangent plane to the surface z = f(x, y) at the point P (x0, y0, z0)is given by:

z − z0 = fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0).

Example 85. Find the tangent plane to the surface z = x2 − 2y2 at the point(1,−1,−1).

Example 86. Find the tangent plane to the surface z = sin(x3 + y2) at the point(−1, 1, 0).

Linear approximations

25

We can use the tangent plane to approximate a function near a point. If z =ax+ by + c is the equation of the tangent plane to a surface at P (x0, y0, z0), thenthe function

L(x, y) = f(x0, y0) + fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0)

is called the linearization of f at (x0, y0).

We say that

f(x, y) ≈ f(x0, y0) + fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0)

is the linear/tangent plane approximation of f at (x0, y0).

Example 87. Find the linearization L(x, y) of the function f(x, y) = 1xy at (2, 2).

Example 88. Linearly approximate the function f(x, y) = x2−y2x2+y2 near (1,−1).

Increments

If x changes from a to a + ∆x and y changes from b to b + ∆y, then define theincrement of z to be

∆z = f(a+ ∆x, b+ ∆y)− f(a, b).

In other words, we consider how z changes as both independent variables change,not just one.

Differentiability

If z = f(x, y) then f is differentiable at (a, b) if ∆z can be written in the form

∆z = fx(a, b)∆x+ fy(a, b)∆y + ε1∆x+ ε2∆y

and ε1, ε2 → 0 as (∆x,∆y)→ (0, 0).

A function is differentiable at a point (a, b) when the tangent plane approximationis very good near (a, b).

A better way to determine differentiability is: If fx and fy both exist near (a, b)and are continuous at (a, b), then f is differentiable at (a, b).

The total differential

If z = f(x, y) is differentiable then we define two new independent variables dx anddy (essentially representing ∆x and ∆y).

The differential dz, or the total differential, is

dz = fx(x, y)dx+ fy(x, y)dy =∂z

∂xdx+

∂z

∂ydy

26

So we can think of our linear approximation from earlier as stating

f(x, y) ≈ f(a, b) + dz

Example 89. Show that z = 2x cos y − y sin(x − y) is differentiable and find thetotal differential.

Example 90. Where is z = xy ln(xy) differentiable? Find the differential of thefunction.

Functions of many variables

If u = f(x1, . . . , xn), then the increment of u is

∆u = f(x1 + ∆x1, . . . , xn + ∆xn)− f(x1, . . . , xn).

The differential du is defined in terms of the differentials dx1, . . . , dxn by

du =∂u

∂x1dx1 + · · ·+ ∂u

∂xndxn.

§15.5: The Chain Rule. Chain rule, version 1 Let z = f(x, y) be a differen-tiable function of x and y, and let x = g(t) and y = h(t) be differentiable functionsof t.

Then z is a differentiable function of t and

dz

dt=∂f

∂x

dx

dt+∂f

∂y

dy

dt

Example 91. If z = cos(x− y), where x = 2 ln t and y = t3, find dzdt .

Example 92. If z =√x2 + y2, where x = cos t and y = sin t, find dz

dt .

Chain rule, version 2 Let z = f(x, y) be a differentiable function of x and y andlet x = g(s, t) and y = h(s, t) be differentiable functions of s and t.

Then z is a function of s and t, and

∂z

∂s=∂z

∂x

∂x

∂s+∂z

∂y

∂y

∂s

and∂z

∂t=∂z

∂x

∂x

∂t+∂z

∂y

∂y

∂t

Example 93. If z = x2y + y2x, where x = t and y = 1− t, then find ∂z∂s and ∂z

∂t .

Example 94. If z = exe−y, where x = s/t and y = t/s, then find ∂z∂s and ∂z

∂t .

27

Types of variables Recall that for a function z = f(x, y), we called x and yindependent variables and z the dependent variable.

Here, since x = g(s, t) and y = h(s, t), the real independent variables are s and t.We call x and y intermediate variables, and z is still the dependent variable.

We can draw tree diagrams using all of our variables to help us remember how touse the chain rule. (This will be especially helpful for functions of more than 2variables.)

Chain rule, general version

Let u be a differentiable function of n variables x1, . . . , xn and let each xj be adifferentiable function of m variables t1, . . . , tm.

Then u is a function of t1, . . . , tm and for each i = 1, . . .m

∂u

∂ti=

∂u

∂x1

∂x1∂ti

+ · · ·+ ∂u

∂xn

∂xn∂ti

.

Example 95. If f(x, y, z) = exy − sin(yz), where x = 3t2 − 2su, y = s3 + u3, andz = st, then find ∂f/∂t.

Implicitly defined function

Before we dealt with equations which define functions implicitly. The chain rulegives us a more formal way to understand implicit differentiation.

Consider the case of just to two variables. If some equation F (x, y) = 0 defines yimplicitly as a function of x, then the chain rule says that:

∂F

∂x

dx

dx+∂F

∂y

dy

dx= 0.

Implicit differentiation Rearranging we have

dy

dx= −

∂F∂x∂F∂y

= −FxFy.

Theorem 0.1 (Implicit Function Theorem). If F is defined around (a, b) andF (a, b) = 0, Fy(a, b) 6= 0, and both Fx and Fy are continuous, then F (x, y) = 0does in fact define y as a function of x (near (a, b) at least).

Example 96. Find dy/dx, where√xy = 2− x2y.

Implicit differentiation, more variables

28

Suppose that the equation F (x, y, z) = 0 defines z implicitly as a function of x andy. Then

∂z

∂x= −

∂F∂x∂F∂z

= −FxFz

and

∂z

∂y= −

∂F∂y

∂F∂z

= −FyFz.

Again, the Implicit Function Theorem gives the conditions for when this is a validprocess.

Example 97. Find ∂z/∂x and ∂z/∂y, where x2 + y2 + z2 = xyz.

§15.6: Directional Derivatives and the Gradient Vector. §15.6: DirectionalDerivatives and the Gradient Vector

Directional derivatives

Partial derivatives tell us the rate of change of z as we move in a specific direction(either the direction of x or the direction of y). We can think of the direction ofmovement through the point as the unit vector i or j.

The directional derivative of f at (x0, y0) in the direction of a unit vector u = 〈a, b〉is defined to be

Duf(x0, y0) = limh→0

f(x0 + ha, y0 + hb)− f(x0, y0)

h

Computing directional derivatives

In the case that f is a differentiable function of x and y, then the directionalderivative of f in the direction of the unit vector u = 〈a, b〉 becomes

Duf(x, y) = fx(x, y)a+ fy(x, y)b.

Example 98. Find the directional derivative of f(x, y) = sin(x + y) − sin(x − y)

at (π2 ,−π2 ) in the direction of 〈

√22 ,−

√22 〉.

Gradient

If f is a function of two variables, the gradient of f is the vector function ∇f definedto be

∇f(x, y) = 〈fx(x, y), fy(x, y)〉.

Now note that we can rewrite the directional derivative in the direction of u to beDuf = ∇f · u.

29

Thought of another way, Du is the scalar projection of ∇f onto u.

Example 99. If f(x, y) = x4y + x2 cos y, then compute ∇f and find the rate ofchange of f at (1, 0) in the direction of 〈2, 3〉.

Example 100. If f(x, y) = 2√x+ y + xy, then compute ∇f and find the rate of

change of f at (1, 2) in the direction of 〈−1,−2〉.

Functions of three variables

We can define directional derivatives for functions of 3 (or more) variables analo-gously:

The directional derivative of f at x = (x0, y0, z0) in the direction of a unit vectoru = 〈a, b, c〉 is

Duf(x0, y0, z0) = limh→0

f(x0 + ha, y0 + hb, z0 + hc)− f(x0, y0, z0)

h

Or in terms of vectors

Duf(x) = limh→0

f(x + hu)− f(x)

h

Gradient and computing directional derivatives If f is a function of threevariables, the gradient of f is

∇f(x, y, z) = 〈fx(x, y, z), fy(x, y, z), fz(x, y, z)〉.

Once again we can rewrite the directional derivative as

Duf(x, y, z) = ∇f(x, y, z) · u

Example 101. Find the directional derivative of f(x, y, z) = ex+y+z − ln(xyz) atthe point P (1, 1, 1) and in the direction of 〈−1,−1,−1〉.

Maximizing rate of change

Recall that for two vectors a and b, the value of |a · b| will be maximized when aand b have the same direction.

Theorem 0.2. If f is a differentiable function (of 2 or 3 variables), then themaximum value of of Duf(x) is equal to |∇f(x)|; this maximum value occurs whenu has the same direction as ∇f(x).

Example 102. If f(x, y) = x2e−y, in what direction is f changing fastest at thepoint P (1, 0), and what is the maximum rate of change?

Example 103. If f(x, y, z) = 1x2−xyz−z3 , in what direction is f changing fastest

at the point (1, 0,−1) and what is the maximum rate of change of f?

30

Tangent planes and normal lines

Suppose that S is a surface with equation F (x, y, z) = k (in other words, it is alevel surface of a function of three variables), and let P (x0, y0, z0) be an arbitrarypoint on S. Let X be a curve on the surface S which passes through the point P(and is parametrized by r s.t. r(t0) = 〈x0, y0, z0〉).

The gradient vector at P , which is ∇F (x0, y0, z0) is perpendicular to the tangentvector r′(t0), no matter what r was.

Tangent planes and normal lines

The tangent plane to the level surface F (x, y, z) = k at a point P (x0, y0, z0) is

Fx(x0, y0, z0)(x− x0) + Fy(x0, y0, z0)(y − y0) + Fz(x0, y0, z0)(z − z0) = 0

The normal line to S at P is the line passing through P and with direction∇F (x0, y0, z0). Symmetric equations for this line are

x− x0Fx(x0, y0, z0)

=y − y0

Fy(x0, y0, z0)=

z − z0Fz(x0, y0, z0)

Note that in the special case that our level surface is the graph of a functionf(x, y) = z, then this tangent plane equation gives rise to the earlier one we learned.

Example 104. Find equations for the tangent plane and normal line to the surfacexyz3 − x2z2 − 2x = 10 at the point (2, 0,−3).

Example 105. Find equations for the tangent plane and normal line to the surfaceex − ey = xye2z at the point (0, 0, 1).

§15.7: Maximum and Minimum Values. Minimums of a function

A function f(x, y) is said to have a local maximum at (a, b) if there is a disk centeredat (a, b) such that for all (x, y) in the disk we have f(x, y) ≤ f(a, b).

In other words, the value of f(a, b) is at least as large as the value of f(x, y), aslong as we only look at (x, y) which are very close to (a, b).

In this case, the number f(a, b) is the local maximum value.

A function f(x, y) has a global maximum at (a, b) if f(x, y) ≤ f(a, b) for all (x, y)(no matter where we look).

Maximums of a function

A function f(x, y) has a local minimum at (a, b) if f(x, y) ≥ f(a, b) for all (x, y)near (a, b).

A function f(x, y) has a global minimum at (a, b) if f(x, y) ≥ f(a, b) for all (x, y).

31

We call the number f(a, b) the local (global) minimum value if f has a local (global)minimum at (a, b).

Critical points

Like in calculus of a single variable, maximums and minimums are related to deriva-tives.

If f has a local max or min at (a, b) and the fx(a, b) and fy(a, b) exist, then fx(a, b) =fy(a, b) = 0.

In general, a critical point/stationary point is any point (a, b) such that fx(a, b) = 0(or does not exist) and fy(a, b) = 0 (or does not exist).

But while every local max/min is at a critical point, it is not the case that allcritical points are local maximums or minimums!

Second derivative test

The second derivative tests give us a way to check this for sure.

If fx(a, b) = fy(a, b) = 0 and all the second partial derivatives of f are continuousaround (a, b), then let

D = D(a, b) = fxx(a, b)fyy(a, b)− fxy(a, b)2.

Then:

• If D > 0 and fxx(a, b) > 0, then f(a, b) is a local min.• If D > 0 and fxx(a, b) < 0, then f(a, b) is a local max.• If D < 0 then f(a, b) is neither a local max nor min; we call (a, b) a saddle

point.

(The graph of f will cross its tangent plane at a saddle point.)

Example 106. Find any local maximums, local minimums, and saddle points off(x, y) = 9− 2x+ 4y − x2 − 4y2.

Example 107. Find any local maximums, local minimums, and saddle points of

f(x, y) = e4y−x2−y2 .

Example 108. Find any local maximums, local minimums, and saddle points off(x, y) = y2 − x2.

Finding global maximums and minimums

A boundary point of a set X is a point which has distance 0 from X (there arepoints of X arbitrarily close, that is).

A closed set is a set containing all of its boundary points.

32

A bounded set is a set contained within some disk. Equivalently, a set X is boundedif it is contained within a box: there is a d such that for all (x, y) ∈ X, |x| < d and|y| < d.

Finding global maximums and minimums

Theorem 0.3 (Extreme value theorem). If f is continuous on a closed and boundedset D in R2, then f attains an absolute maximum (and absolute minimum) in D.

The global max/min must occur at either a critical point of f in D, or on theboundary of D.

So to find the global maximum, for instance:

(1) Evaluate f at all critical points of f in D.(2) Find the extreme values of f on the boundary of D.(3) The largest of the values found above is the global maximum.

Example 109. Find the global maximum and minimum values of f(x, y) = x2 −2xy + 2y on the rectangle D = {(x, y) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2}.Example 110. Find the global maximum and minimum values of f(x, y) = x4 +y4 − 4xy on the rectangle D = {(x, y) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2}.Example 111. Find the global maximum and minimum values of f(x, y) = 2x3+y4

on D = {(x, y) | x2 + y2 ≤ 1}.

§15.8: Lagrange Multipliers. Functions subject to a constraint

In the last example from the previous section, we needed to maximize (and mini-mize) the value of f(x, y) = 2x3 + y4 on the circle x2 + y2 = 1. In other words, weare maximizing the function f , but subject to the constraint g(x, y) = x2 + y2 = 1.

How do you maximize a function f(x, y, z) subject to a constraint g(x, y, z) = k?

Functions subject to a constraint

The key observation is: if the maximum value of f , subject to the constraintg(x, y, z) = k, is f(x0, y0, z0) = c then the level surface f(x, y, z) = c is tangent tothe level surface g(x, y, z) = k and so their gradient vectors are parallel.

So if f has an extreme value, subject to the constraint g(x, y, z) = k, at P (x0, y0, z0)then there is a number λ, called the Lagrange multiplier, such that

∇f(x0, y0, z0) = λ∇g(x0, y0, z0).

Method of Lagrange multipliers

So if we want to maximize and minimize a function f subject to the constraintg(x, y, z) = k (when a max/min do exist and ∇g 6= 0):

33

(1) Find all points (x, y, z) and values of λ such that ∇f(x, y, z) = λ∇g(x, y, z)and g(x, y, z) = k.

(2) Evaluate and compare f at all of these points.

Example 112. Find the extreme values of f(x, y) = x2+2y2 on the circle x2+y2 =1.

Example 113. Find the closest and farthest points on the sphere x2 + y2 + z2 = 1from the point (4, 2,−4).

A function subject to two simultaneous constraints

Suppose we are trying to maximize f(x, y, z) subject to the two constraints g(x, y, z) =k1 and h(x, y, z) = k2.

If f has an extreme value at P (x0, y0, z0) (subject to the constraints) then thereare numbers λ and µ (again called the Lagrange multipliers) such that

∇f(x0, y0, z0) = λ∇g(x0, y0, z0) + µ∇h(x0, y0, z0).

Example 114. Find the maximum value of the function f(x, y, z) = x + 2y + 3zon the curve of intersection of the plane x− y+ z = 1 and the cylinder x2 + y2 = 1.

Example 115. Find the maximum value of the function f(x, y, z) = x+ 2y on thecurve of intersection of the plane x− y + z = 1 and the cylinder y2 + z2 = 4.

Chapter 16: Multiple Integrals

§16.1: Double Integrals over Rectangles. Double Riemann sums

Let R = [a, b] × [c, d] and let S = {(x, y, z) | 0 ≤ z ≤ f(x, y)}, where f is non-negative. Then the volume V of the solid is approximated by:

V ≈m∑i=1

n∑j=1

f(x∗ij , y∗ij)∆A

once we divide R into small rectangles of area ∆A and pick sample points (x∗ij , y∗ij)

in each sub-rectangle.

This is called a double Riemann sum.

Double integrals and volume

The double integral of f over the rectangle R is∫∫R

f(x, y)dA = limn.m→∞

m∑i=1

n∑j=1

f(x∗ij , y∗ij)∆A

A function f is integrable if this limit exists.

34

For a function f which is non-negative, the actual volume of S is equal to thisdouble integral.

Example 116. Estimate the volume of the solid above the square R = [0, 2]× [0, 2]and below the surface z = 16−x2−2y2 by dividing the rectangle into 4 equal squaresand using upper right corners as test points.

Example 117. Repeat the previous example with m = n = 4.

Midpoint rule

The Midpoint Rule for double integrals is just a special type of double Riemannsums. It says that ∫∫

R

f(x, y)dA ≈m∑i=1

n∑j=1

f(xi, yj)∆A

where xi and yj are the midpoints of the subdivisions of R.

Example 118. Estimate the volume of the solid between z = x + 2y2 and abovethe rectangle R = [0, 2]× [0, 4] using the Midpoint Rule and m = n = 2. Comparethis to what you get using the lower right corners for sample points.

Average value of a function

Recall that the average value of a function of one variable can be calculated by:

fave =1

b− a

∫ b

a

f(x)dx

Similarly, the average value of a function of two variables can be calculated as:

fave =1

A(R)

∫∫R

f(x, y)dA

where A(R) is the area of the rectangle R.

Properties of double integrals

When we actually get to calculating integrals, we have several nice properties:∫∫R

[f(x, y) + g(x, y)]dA =

∫∫R

f(x, y)dA+

∫∫R

g(x, y)dA

∫∫R

cf(x, y)dA = c

∫∫R

f(x, y)dA

If f(x, y) ≥ g(x, y) for all x, y then∫∫R

f(x, y)dA ≥∫∫R

g(x, y)dA

35

§16.2: Iterated Integrals. Iterated integrals

We can use the fundamental theorem of calculus to help with integration.

To calculate an integrated integral over the rectangle R = [a, b]× [c, d], we use thenotation ∫ b

a

∫ d

c

f(x, y)dydx =

∫ b

a

[∫ d

c

f(x, y)dy

]dx

where A(x) =∫ dcf(x, y)dy means to integrate f as a function of y (treating x as a

constant) and∫ baA(x)d(x) is self-explanatory.

Iterated integrals and double integrals

Theorem 0.4 (Fubini’s theorem). If f is continuous on the rectangle [a, b]× [c, d],then ∫∫

R

f(x, y)dA =

∫ b

a

∫ d

c

f(x, y)dydx =

∫ d

c

∫ b

a

f(x, y)dxdy

In particular, integration order does not matter.

A corollary of this is:∫∫R

g(x)h(y)dA =

∫ b

a

g(x)dx

∫ d

c

h(y)dy

where R = [a, b]× [c, d].

Example 119. Calculate the double integral∫∫R

(1 + 4xy)dA

where R = [0, 1]× [1, 3].

Example 120. Calculate the iterated integral∫ 3

1

∫ 5

1

ln y

xydydx


x(y + x2)4dA

where R = [0, 1]× [0, 1].


xy2

x2 + 1dA

where R = [0, 1]× [−3, 3].

36

Example 123. Find the average value of f(x, y) = x2y over the rectangle R =[−1, 1]× [0, 5].

§16.3: Double Integrals over General Regions. Defining double integralsover other regions

If D is a bounded region, we can imagine D being contained in some rectangularregion R.

Then define a function F by

F (x, y) =

{f(x, y) if (x, y) is in D0 otherwise

Then define ∫∫D

f(x, y)dA =

∫∫R

F (x, y)dA

Types of bounded regions to consider

A region is of type I if it has the form

D = {(x, y) | a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}for two continuous functions g1 and g2.

A region is of type II if it has the form

D = {(x, y) | c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)}for two continuous functions h1 and h2.

Calculating double integrals on regions of type I If f is continuous on a typeI region

D = {(x, y) | a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}then ∫∫

D

f(x, y)dA =

∫ b

a

∫ g2(x)

g1(x)

f(x, y)dydx

Calculating double integrals on regions of type II If f is continuous on atype II region

D = {(x, y) | c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)}then ∫∫

D

f(x, y)dA =

∫ d

c

∫ h2(y)

h1(y)

f(x, y)dxdy

Example 124. Evaluate∫∫D

(x + 2y)dA where D is the region bounded by the

parabolas y = 2x2 and y = 1 + x2.

37


y2 dA where D = {(x, y) | −1 ≤ y ≤ 1,−y − 2 ≤ x ≤

y}.

Example 126. Find the volume of the solid lying under the paraboloid z = x2+y2

and above the region D in the xy-plane bounded by the line y = 2x and the parabolay = x2.

Example 127. Find the volume of the solid bounded by the cylinder x2 + y2 = 1and the planes y = z, x = 0, z = 0 in the first octant.

Additional properties of integrals If D = D1 ∪D2 and D1 and D2 overlap atmost on their boundaries, then∫∫

D

f(x, y)dA =

∫∫D1

f(x, y)dA+

∫∫D2

f(x, y)dA

Another straightforward property is that:∫∫D

1dA = A(D)

Finally, if m ≤ f(x, y) ≤M for all (x, y) ∈ D then

m ·A(D) ≤∫∫D

f(x, y)dA ≤M ·A(D)


xy dA where D = {(x, y) | x ≥ 0, y ≥ 0, 1 ≤ x2 + y2 ≤

4}.

§16.4: Double Integrals in Polar Coordinates. Converting to polar coor-dinates

Some regions are easier to think about in polar coordinates rather than in rectan-gular coordinates.

Recall the rules for converting between rectangular coordinates and polar coordi-nates:

r2 = x2 + y2

x = r cos θ

y = r sin θ

Integrating over polar rectangles

38

A polar rectangle has the form

R = {(r, θ) | a ≤ r ≤ b, α ≤ θ ≤ β}.

To integrate using polar coordinates (note the important difference from the previ-ous section): ∫∫

D

f(x, y)dA =

∫ β

α

∫ b

a

f(r cos θ, r sin θ)r drdθ

When integrating using polar coordinates, remember your trig identities!


xy dA where D = {(x, y) | x ≥ 0, y ≥ 0, 1 ≤ x2 + y2 ≤

4}.

Example 130. Evaluate∫∫R

(3x+ 4y2)dA where R is the region in the upper half-

plane bounded by the circles x2 + y2 = 1 and x2 + y2 = 4.

A more general region might have the form D = {(r, θ) | α ≤ θ ≤ β, h1(θ) ≤ r ≤h2(θ)}.

To integrate over a region D as above:∫∫D

f(x, y)dA =

∫ β

α

∫ h2(θ)

h1(θ)

f(r cos θ, r sin θ)r drdθ

Example 131. Find the volume of the solid that lies under the paraboloid z =x2 + y2 above the xy-plane and inside the cylinder x2 + y2 = 2x.

Example 132. What is the area enclosed by one loop of the four-leaved roser = cos 2θ?

§16.6: Triple Integrals. Triple Riemann sums

Let f(x, y, z) be defined on the box B = [a, b]× [c, d]× [r, s].

Subdivide B into l ·m · n sub-boxes of volume ∆V = ∆x∆y∆z.

The triple Riemann sum is

l∑i=1

m∑j=1

n∑k=1

f(x∗ijk, y∗ijk, z

∗ijk)∆V

where each (x∗ijk, y∗ijk, z

∗ijk) is a sample point from a sub-box.

Triple integrals

39

The triple integral of f on B is∫∫∫B

f(x, y, z)dV = liml,m.n→∞

l∑i=1

m∑j=1

n∑k=1

f(x∗ijk, y∗ijk, z

∗ijk)∆V

when this limit exists.

Theorem 0.5 (Fubini’s theorem for triple integrals).∫∫∫B

f(x, y, z)dV =

∫ s

r

∫ d

c

∫ b

a

f(x, y, z) dx dy dz

More general regions

We can also extend these techniques to more general regions.

Type 1: E = {(x, y, z) | (x, y) ∈ D,u1(x, y) ≤ z ≤ u2(x, y)} where D is theprojection of E onto the xy-plane.

Type 2: E = {(x, y, z) | (y, z) ∈ D,u1(y, z) ≤ x ≤ u2(y, z)} where D is theprojection of E onto the xy-plane.

Type 3: E = {(x, y, z) | (x, z) ∈ D,u1(x, z) ≤ y ≤ u2(x, z)} where D is theprojection of E onto the xy-plane.

Of course, now we might also have to think about the region D in some plane anddecide what type of region it is, as well.

Example 133. Evaluate∫∫∫E

2z dV where E is the solid tetrahedron bounded by

the four planes x = 0, y = 0, z = 0 and x+ 2y+ z = 2. Is there more than one waywe can do this?


√x2 + z2 dV where E is the region bounded by the

paraboloid y = x2 + z2 and the plane y = 4.


6xy dV where E lies under the plane z = 1 + x + y

and above the region in the xy-plane bounded by the curves y =√x, y = 0 and

x = 1.

§16.7: Triple Integrals in Cylindrical Coordinates. Using cylindrical co-ordinates

In cylindrical coordinates, a point P is represented by the ordered triple (r, θ, z)where

x = r cos θ

y = r sin θ

z = z

40

Or to convert back:

r2 = x2 + y2

tan θ =y

xz = z

Triple integrals in cylindrical coordinates

Suppose that

E = {(x, y, z) | (x, y) ∈ D,u1(x, y) ≤ z ≤ u2(x, y)}

where D is given in polar coordinates by

D = {(r, θ | α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ)}.

If f is continuous on E, then∫∫∫E

f(x, y, z)dV

=

∫ β

α

∫ h2(θ)

h1(θ)

∫ u2(r cos θ,r sin θ)

u1(r cos θ,r sin θ)

f(r cos θ, r sin θ, z)r dz dr dθ

Example 136. Evaluate∫ 2

−2

∫ √4−x2

−√4−x2

∫ 2

√x2+y2

(x2 + y2) dz dy dx

Example 137. Find the volume of the solid that is enclosed by the cone z =√x2 + y2 and the sphere x2 + y2 + z2 = 4.


√x2 + y2dV where E is the region that lies inside

the cylinder x2 + y2 = 16 and between the planes z = −5 and z = 4.

§16.8: Triple Integrals in Spherical Coordinates. Using spherical coordi-nates The spherical coordinates (ρ, θ, φ) of a point P in space can be found usingthe conversions:

x = ρ sinφ cos θ

y = ρ sinφ sin θ

z = ρ cosφ

Note that

ρ2 = x2 + y2 + z2

Integrating over spherical “rectangular boxes”

41

If E is the spherical wedge

E = {(ρ, θ, φ) | a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d}then ∫∫∫

E

f(x, y, z)dV

=

∫ d

c

∫ β

α

∫ b

a

f(ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ)ρ2 sinφ dρ dθ dφ

Example 139. Evaluate∫∫∫B

e(x2+y2+z2)3/2dV whereB is the unit ballB = {(x, y, z) |

x2 + y2 + z2 ≤ 1}.


z2 dV where E is the solid hemisphere x2 +y2 +z2 ≤

9, y ≥ 0.

Chapter 17 - Vector Calculus

§17.1: Vector Fields. Vector fields Let D be a set in R2. A vector field onR2 is a function F that assigns to each point (x, y) in D a 2-dimensional vectorF(x, y) = 〈u, v〉.

We can write F in terms of its component functions P and Q:

F(x, y) = P (x, y)i +Q(x, y)j

We also call P and Q scalar fields.

Vector fields in R3

Let E be a set in R3. A vector field on R3 is a function F that assigns to each point(x, y, z) in E a 3-dimensional vector F(x, y, z) = 〈u, v, w〉.

Say that a vector field F is continuous if its component functions are continuous.

For convenience, we also write x = 〈x, y, z〉 and write F(x).

Example 141. F(x, y) = xi− yj defines a vector field. Sketch some of the vectorsF(x, y).

Example 142. Sketch the vector field on R3 given by F(x, y, z) = zk.

Example 143. Sketch the vector field on R2 given by F(x, y) = 〈sin(x+ y), x〉.

Examples of vector fields

A velocity field is a vector field V where each V(x, y, z) represents a velocity vector.

42

A gravitational field is a vector field F where each vector F(x, y, z) is a gravitationalforce.

Similarly we can define force fields and electric fields.

Recall that ∇f(x, y) is a vector for each (x, y). Thus, ∇f(x, y) is in fact a vectorfield, called a gradient vector field.

Example 144. Find the gradient vector field of f(x, y) = x2y − y3.

Example 145. Find the gradient vector field of f(x, y, z) =√x2 + y2 + z2.

Conservative vector fields

A vector field F is called a conservative vector field if it is the gradient of a scalarfunction, that is if F = ∇f for some function f .

We call f the potential function for F.

§17.2: Line Integrals. Line integrals with respect to arc length

How do we integrate a function along a curve instead of an interval?

If f is defined on a smooth curve C given by r(t) = x(t)i + y(t)j for a ≤ t ≤ b, thenthe line integral of f (with respect to arc length) along C is∫

C

f(x, y)ds = limn→∞

n∑i=1

f(x∗i , y∗i )∆si

when this limit exists.

∆si is the length of a subarc from when we divide the curve C up.

More about line integrals

We can compute this using the following:∫C

f(x, y)ds =

∫ b

a

f(x(t), y(t))

√(dx

dt

)2

+

(dy

dt

)2

dt

As when we learned about arc length, the parametrization used does not matter,as long as it is nice.

If C is a piecewise-smooth curve, then we can break line integrals into multipleparts ∫

C

f(x, y)ds =

∫C1

f(x, y)ds+ . . .+

∫Cn

f(x, y)ds.

43

Example 146. Evaluate∫C

(2+x2y)ds, where C is the upper half of the unit circle

x2 + y2 = 1.


2xds where C consists of the arc C1 of the parabola

y = x2 from (0, 0) to (1, 1) followed by the vertical line segment C2 from (1, 1) to(1, 2).

Line integrals with respect to x and y The line integrals of f along C withrespect to x and y are∫

C

f(x, y)dx = limn→∞

n∑i=1

f(x∗i , y∗i )∆xi =

∫ b

a

f(x(t), y(t))x′(t)dt

∫C

f(x, y)dy = limn→∞

n∑i=1

f(x∗i , y∗i )∆yi =

∫ b

a

f(x(t), y(t))y′(t)dt

Recall that a line segment connecting points r0 and r1 can be parametrized byr(t) = (1− t)r0 + tr1 for 0 ≤ t ≤ 1.

Example 148. Evaluate∫Cxeydx where C is the arc of the curve x = ey from

(1, 0) to (e, 1).

Example 149. Evaluate∫Cy2dx+xdy where C is the line segment from (−5,−3)

to (0, 2).

Orientations

A parametrization of a curve C carries with it an orientation, or direction, corre-sponding to the direction you move along C as t increases.

We can write −C to mean the same curve as C with the opposite orientation.

Then ∫−C

f(x, y)dx = −∫C

f(x, y)dx∫−C

f(x, y)dy = −∫C

f(x, y)dy∫−C

f(x, y)ds =

∫C

f(x, y)ds

Line integrals and functions of 3 variables

If r(t) = x(t)i + y(t)j + z(t)k for a ≤ t ≤ b traces the smooth curve C, then the lineintegral of f along C (with respect to arc length) is∫

C

f(x, y, z)ds =

∫ b

a

f(x(t), y(t), z(t))

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt

44

We can also define line integrals along C with respect to x, y, z.

More about line integrals

If we wanted to condense our line integrals to a more precise notation:∫C

f(x, y, z)ds =

∫ b

a

f(r(t)) · |r′(t)|dt

Note that if the function f is the constant 1 everywhere, a line integral will give usthe length of the curve C.

Example 150. Evaluate∫Cy sin z ds where C is the circular helix given by the

equations x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π.

Example 151. Evaluate∫Cx2y√z dz where C is the curve with parametrization

x = t3, y = t, z = t2 for 0 ≤ t ≤ 1.

Line integrals of vector fields

Let F be a continuous vector field defined on a smooth curve C given by a vectorfunction r(t) for a ≤ t ≤ b.

The line integral of F along C is∫C

F · dr =

∫ b

a

F(r(t)) · r′(t)dt =

∫C

F ·Tds

where T is the unit tangent vector.

Example 152. Find the work done by the force field F(x, y) = x2i−xyj in movinga particle along the quarter circle r(t) = cos ti + sin tj for 0 ≤ t ≤ π/2.


F · dr where F(x, y, z) = xyi + yzj + zxk and C is the

twisted cubic given by r(t) = 〈t, t2, t3〉 for 0 ≤ t ≤ 1.

Scalars and vectors

Note that there is a connection between line integrals of scalar and vector fields.∫C

F · dr =

∫C

Pdx+Qdy +Rdz

where F = P i +Qj +Rk.

§17.3: The Fundamental Theorem for Line Integrals. Fundamental the-orem for line integrals Recall the fundamental theorem of calculus.

Let C be a smooth curve given by the vector function r(t), a ≤ t ≤ b.

45

Theorem 0.6. If f is a differentiable function of 2 or 3 variables whose gradientvector ∇f is continuous on C, then∫

C

∇f · dr = f(r(b))− f(r(a))

Example 154. Find∫C

F · dr where F = ∇f , f(x, y) = 2x2 + y2 − xy, and C isthe curve from (−1, 1) to (3, 2).

Example 155. Find∫C

F · dr where F(x, y) = 〈2x, 3y2〉 and C is the quarter ofthe unit circle in the first quadrant, from (1, 0) to (0, 1). Note that F = ∇f , wheref(x, y) = x2 + y3.

Independence of path

If F is a continuous vector field with domain D, we say that∫C

F ·dr is independent

of path if∫C1

F · dr =∫C2

F · dr for any two paths C1 and C2 in D with the same

initial and terminal points.

In general, line integrals of vector fields will not be independent of path.

Line integrals of conservative vector fields are always independent of path.

Closed curves

A curve is closed if its terminal point and initial point coincide (not the same as aset being closed).

Theorem 0.7.∫C

F · dr is independent of path in D if and only if∫C

F · dr = 0for every closed path C in D.

Conservative vector fields

A region is called open if all of its points are interior points (not very close to theboundary). The complement of an open set is a closed set.

A region is called connected if every pair of points in D is joined by a path lying inD.

Theorem 0.8. If F is a vector field that is continuous on an open connected regionD and

∫C

F · dr is independent of path in D, then F is a conservative vector fieldon D; in other words, there is a function f such that F = ∇f .

Determining when F is conservative

Recall from partial derivatives that fxy = fyx when both exist and are continuous.

Theorem 0.9. If F(x, y) = P (x, y)i+Q(x, y)j is a conservative vector field, whereP and Q have continuous first-order partial derivatives on a domain F , then wehave

∂P

∂y=∂Q

∂x

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Simply connected regions

A simple curve is a curve that never intersects itself (besides the endpoints).

A simply-connected region is a connected region such that any region enclosed bya simple curve is a part of D.

In other words, a simply-connected region is one where every loop can be shrunkdown to a point, and remain inside the region.

Theorem 0.10. Let F = P i+Qj be a vector field on an open simply-connected re-gion F . If P and Q have continuous first-order derivatives and Py = Qx everywherein D, then F is conservative.

Example 156. Decide whether or not F(x, y) = (x−y)i+ (x−2)j is conservative.

Example 157. Determine whether or not the vector field F(x, y) = (3 + 2xy)i +(x2 − 3y2)j is conservative.

Example 158. Find a function f such that F = ∇f , where F(x, y) = (3 + 2xy)i +(x2 − 3y2)j. Evaluate

∫C

F · dr where C is the curve r(t) = et sin ti + et cos tj

Example 159. If F(x, y, z) = y2i+(2xy+e3z)j+3ye3zk, find a function such that∇f = F.

Example 160. Find a function f such that F = ∇f , where F(x, y, z) = yzi +xzj+(xy+2z)k, then compute

∫C

F ·dr where C is the line segment from (1, 0,−2)to (4, 6, 3).

Example 161. Find a function f such that F = ∇f , where F(x, y, z) = sin y i +(x cos y + cos z)j − (y sin z)k, then compute

∫C

F · dr where C is given by r(t) =sin t i + tj + 2tk for 0 ≤ t ≤ π/2.

§17.4: Green’s Theorem. Notation for Green’s Theorem

We use the convention that the positive orientation of a simple closed curve Cmeans a single counterclockwise traversal of C.

The positively orientated boundary curve of D can also be denoted by ∂D.

We also use∮

to mean the line integral along the positive orientation of a closedcurve.

Green’s Theorem

Theorem 0.11. Let C be a positively oriented, piecewise-smooth, simple closedcurve in the plane and let D be the region bounded by C. If P and Q havecontinuous partial derivatives on an open region that contains D, then∫

C

Pdx+Qdy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dA

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Example 162. Evaluate∫Cx4 dx+xy dy where C is the triangular curve consisting

of the line segments from (0, 0) to (1, 0), then to (0, 1), then back to (0, 0).

Example 163. Evaluate∮C

(3y − esin x)dx + (7x +√y4 + 1)dy, where C is the

circle x2 + y2 = 9.

An Application of Green’s Theorem

We can calculate the area of a region D by

A =

∮C

xdy = −∮C

ydx =1

2

∮C

xdy − ydx

Example 164. Find the area enclosed by the ellipse x2

a2 + y2

b2 = 1.

Example 165. Evaluate∮Cy2dx + 3xydy where C is the boundary of the semi-

annular region D in the upper half plane between the circles x2 + y2 = 1 andx2 + y2 = 4.

Example 166. Use Green’s Theorem to evaluate∫C

F·dr where F(x, y) = 〈y cosx−xy sinx, xy + x cosx〉 and C is the triangle from (0, 0) to (0, 4) to (2, 0) to (0, 0).

Example 167. Use Green’s Theorem to evaluate∫C

F·dr where F (x, y) = 〈y2 sinx, x2−2y cosx〉 and C is the triangle from (0, 0) to (1, 0) to (0, 1) to (0, 0).

§17.5: Curl and Divergence. Del operator Let F = P i +Qj +Rk is a vectorfield on R3 such that the partial derivatives of P,Q and R exist.

Define a vector differential operator ∇ (read “del”) as

∇ = i∂

∂x+ j

∂

∂y+ k

∂

∂z

Note that this gives the same meaning to ∇f as we would expect.

Since ∇ and F are both vectors, we can define ∇× F and ∇ · F.

Curl and divergence

The curl of F is the vector field

curl F = ∇× F =

(∂R

∂y− ∂Q

∂z

)i +

(∂P

∂z− ∂R

∂x

)j +

(∂Q

∂x− ∂P

∂y

)k

The divergence of F is a the function

div F = ∇ · F =∂P

∂x+∂Q

∂y+∂R

∂z

Some useful results about curl and divergence

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Theorem 0.12. If F is a vector field defined on all of R3 whose component func-tions have continuous partial derivatives and curl F = 0, then F is a conservativevector field.

Theorem 0.13. If F = P i +Qj +Rk is a vector field on R3 and P,Q and R havecontinuous second-order partial derivatives then

div curl F = 0

Example 168. Determine whether or not F(x, y, z) = y2z3i + 2xyz3j + 3xy2z2kis conservative. If it is, find an f such that F = ∇f .

Example 169. Determine whether or not F(x, y, z) = 2xy cos(z)i + x2 cos(z)j −x2y sin(z)k is conservative. If it is, find an f such that F = ∇f .

Example 170. Compute div F where F(x, y, z) = xzi + xyzj− y2k.

Example 171. Consider F(x, y, z) = xzi + xyzj − y2k. Can F be the curl ofanother function?

Example 172. Determine whether or not F(x, y, z) = y2exi + 2yexj + (z + 1)ezkis conservative. If it is, find an f such that F = ∇f . Evaluate

∫C

F · dr where

r(t) = ti + t3j + t5k for 0 ≤ t ≤ 1.

Example 173. Determine whether or not F(x, y, z) = 2y2i − 2xj − 2z2k is con-servative. If it is, find an f such that F = ∇f . Evaluate

∫C

F · dr where r(t) =(1− t)i + tj− 3tk for 0 ≤ t ≤ 1.

Example 174. Determine whether or not F(x, y, z) = i + sin zj + y cos zk is con-servative. If it is, find an f such that F = ∇f .

Green’s Theorem restated in terms of curl and divergence

Theorem 0.14. ∮C

F · dr =

∫∫D

(curl F) · k dA

Theorem 0.15. ∮C

F · n ds =

∫∫D

div F(x, y) dA

§17.6: Parametric Surfaces and Their Areas. Parametric surfaces

Let r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k be a vector-valued function defined on aregion D in the uv-plane.

The set S of points (x, y, z) ∈ R3 such that x = x(u, v), y = y(u, v), and z = z(u, v)for some (u, v) ∈ D is called a parametric surface.

We call x = x(u, v), y = y(u, v), and z = z(u, v) the parametric equations of S.

Parametrizing surfaces

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Parametrizing a surface S means finding x(u, v), y(u, v), z(u, v).

One special type of surface is a surface of revolution:

Let S be the surface obtained by the rotation of the curve y = f(x) for a ≤ x ≤ babout the x-axis, where f(x) ≥ 0.

S is parametrized by:

x = x y = f(x) cos θ z = f(x) sin θ

Example 175. Find a parametrization of the plane containing the points (1, 1, 1),(2,−1, 2) and (3, 0, 1).

Example 176. Find a parametric representation of the sphere x2 + y2 + z2 = 4.

Example 177. Parametrize the elliptic paraboloid z = x2 + 2y2.

Example 178. Find a parametrization of the surface generated by rotation thecurve y = sinx for 0 ≤ x ≤ 2π about the x-axis.

Example 179. Find a parametrization of the part of the cylinder y2 + z2 = 16lying between the planes x = 0 and x = 5.

Example 180. Parametrize the surface which is the top half of the cone z2 =4x2 + 4y2.

Tangent Planes

Let S be the surface traced out by the function

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k.

What is the equation for the tangent plane to S at a point P0 = r(u0, v0)?

A plane is determined by a normal vector and a point. Or equivalently, a plane isdetermined by two non-parallel vectors contained within the plane and a point.

Keeping u fixed as u = u0 we can find one tangent vector in this plane:

rv =∂x

∂v(u0, v0)i +

∂y

∂v(u0, v0)j +

∂z

∂v(u0, v0)k

Keeping v fixed as v = v0 we can find one tangent vector in this plane:

ru =∂x

∂u(u0, v0)i +

∂y

∂u(u0, v0)j +

∂z

∂u(u0, v0)k

Then ru × rv gives a normal vector to the plane.

Of course this only makes sense if ru × rv 6= 0. So call S smooth if ru × rv 6= 0.

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Example 181. Find the tangent plane to the surface with parametric equationsx = u2, y = v2 and z = u+ 2v at the point (1, 1, 3).

Example 182. Find the tangent plane to the surface given by r(u, v) = cosui +sinu cos vj + sin vk at u = v = π

3

Surface Area

Suppose S is parametrized by (and covered exactly once by)

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k (u, v) ∈ D

Then the surface area of S is

A(S) =

∫∫D

|ru × rv|dA

where

ru =∂x

∂u(u0, v0)i +

∂y

∂u(u0, v0)j +

∂z

∂u(u0, v0)k

and

rv =∂x

∂v(u0, v0)i +

∂y

∂v(u0, v0)j +

∂z

∂v(u0, v0)k

Surface area of the graph of a function A special type of surface is one whichis the graph of a function. These can be parametrized by

x = x y = y z = f(x, y)

The surface area of the graph of f above a region D is

A(S) =

∫∫D

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dA

Example 183. Find the surface area of a sphere of radius 2.

Example 184. Find the surface area of the part of the paraboloid z = x2 + y2

that lies under the plane z = 9.

Example 185. Find the surface area of the part of the plane 3x+ 2y+ z = 6 thatlies in the first octant.

Example 186. Find the surface area of the part of the sphere x2 + y2 + z2 = 2that lies between the planes z = 1 and z = 0.

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§17.7: Surface Integrals. Definition of surface integrals Let S be a surfaceparametrized by the vector equation

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k (u, v) ∈ D

where D is a rectangle.

The surface integral of f over the surface S is defined to be∫∫S

f(x, y, z) dS = limm,n→∞

m∑i=1

n∑j=1

f(P ∗ij)∆Sij

where each P ∗ij is a point in a “patch” of S and ∆Sij is the surface area of that“patch”.

Computing surface integrals To compute surface integrals, use∫∫S

f(x, y, z) dS =

∫∫D

f(r(u, v))|ru × rv| dA

where

ru =∂x

∂u(u0, v0)i +

∂y

∂u(u0, v0)j +

∂z

∂u(u0, v0)k

rv =∂x

∂v(u0, v0)i +

∂y

∂v(u0, v0)j +

∂z

∂v(u0, v0)k

This works where D is any region, not just a rectangle.

Surfaces which are graphs of functions

If S is the graph of a function g(x, y), we can parametrize S using

x = x y = y z = g(x, y)

Then a surface integral over S becomes∫∫S

f(x, y, z)dS =

∫∫D

f(x, y, g(x, y))

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dA

It may also be convenient to think of S as the graph of g in a slightly differentway: either x = g(y, z) or y = g(x, z). These will give slightly different forms forcalculating surface integrals.

Example 187. Evaluate∫∫S

x2 dS where S is the unit sphere x2 + y2 + z2 = 1.

Example 188. Evaluate∫∫Sy dS where S is the surface z = x+ y2 for 0 ≤ x ≤ 1,

0 ≤ y ≤ 2.


√1 + x2 + y2 dS where S is parametrized by r(u, v) =

〈u cos v, u sin v, v〉 for 0 ≤ u ≤ 1 and 0 ≤ v ≤ π.

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(xz+y) dS where S is the part of the plane x+2y+z = 2

that lies in the first octant.

Oriented surfaces

Let S be a surface which has a tangent plane at every point (x, y, z) on S (excludingthe boundaries, perhaps).

Then there are two unit normal vectors one can pick at each point.

We say that S is oriented if it is possible to pick n at each point (x, y, z) in a waysuch that n varies continuously over S.

The choice of n is called an orientation.

Picking orientations

If S is the graph of a function g, then we can pick an orientation by

n =− ∂g∂x i− ∂g

∂y j + k√(∂g∂x

)2+(∂g∂y

)2+ 1

If S is a smooth orientable surface then it has the natural orientation

n =ru × rv|ru × rv|

If S is a closed surface (the boundary of a solid region), we make the conventionthat the positive orientation points outwards.

Surface integrals of vector fields (flux) If F is a continuous vector field definedon an oriented surface S with unit normal vector n then the surface integral of Fover S is ∫∫

S

F · dS =

∫∫S

F · n dS

This is also called the flux of F across S.

Computing flux

When S is parametrized by r(u, v), then this becomes∫∫S

F · dS =

∫∫D

F · (ru × rv) dA

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When S is the graph of a function, this becomes:∫∫S

F · dS =

∫∫D

(−P ∂g

∂x−Q∂g

∂y+R

)dA


F · dS where F(x, y, z) = yi + xj + zk and S is the

boundary of the solid region E enclosed by the paraboloid z = 1− x2 − y2 and theplane z = 0.

Example 192. Find the flux of the vector field F(x, y, z) = zi+yj+xk across theunit sphere x2 + y2 + z2 = 1.

Example 193. Find the flux of the vector field F(x, y, z) = yj−zk where S consistsof the paraboloid y = x2 + z2 for 0 ≤ y ≤ 1 and the disk x2 + z2 ≤ 1, y = 1.

Example 194. Find the flux of F(x, y, z) = x2i+y2j+z2k where S is the boundary

of the solid half-cylinder 0 ≤ z ≤√

1− y2, 0 ≤ x ≤ 2.


F·dS where F(x, y, z) = xi+yj+2zk and S is the part

of the paraboloid z = 4− x2 − y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1and has upward orientation.

§17.8: Stokes’ Theorem. Stokes’ Theorem

Theorem 0.16. Let S be an oriented piecewise-smooth surface that is boundedby a simple, closed, piecewise-smooth boundary curve C with positive orientation.Let F be a vector field whose components have continuous partial derivatives onan open region in R3 containing S. Then∫

C

F · dr =

∫∫S

curl F · dS

Green’s Theorem can be seen as a special case of Stokes’ Theorem.


F · dr where F(x, y, z) = −y2i + xj + z2k and C is

the curve of intersection of the plane y + z = 2 and the cylinder x2 + y2 = 1 (C isoriented to be counterclockwise as viewed from above).

Example 197. Compute∫∫S

curl F · dS where F(x, y, z) = xzi + yzj + xyk and S

is the part of the sphere x2 + y2 + z2 = 4 that lies inside the cylinder x2 + y2 = 1and above the xy-plane (oriented upwards).


curl F ·dS where F(x, y, z) = exyi+exzj+x2zk and S

is the half of the ellipsoid 4x2 + y2 + 4z2 = 4 that lies on the right of the xz-plane,oriented in the direction of the positive y-axis.


F ·dr where F(x, y, z) = e−xi+exk+ezk and C is theboundary of the part of the plane 2x+ y+ 2z = 2 in the first octant. C is orientedcounterclockwise as viewed from above.

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curl F ·dS where F(x, y, z) = x2z2i+y2z2j+xyzk and

S is the part of the paraboloid z = x2 + y2 that lies inside the cylinder x2 + y2 = 4,oriented upwards.

§17.9: The Divergence Theorem. The Divergence Theorem

Theorem 0.17. Let E be a simple solid region with boundary surface S given withpositive (outward) orientation. Let F be a vector field whose component functionshave continuous partial derivatives on an open region that contains E. Then∫∫

S

F · dS =

∫∫∫E

div F dV

Example 201. Find the flux of the vector field F(x, y, z) = zi + yj + xk over theunit sphere x2 + y2 + z2 = 1.


F · dS where F(x, y, z) = xyi + (y2 + exz2

)j + sin(xy)k

and S is the surface of the region E bounded by the parabolic cylinder z = 1− x2and the planes z = 0, y = 0, and y + z = 2.


F · dS where F(x, y, z) = xzi− 2yj + 3xk and S is the

sphere x2 + y2 + z2 = 4, with outward orientation.

Example 204. Calculate the flux of F across S where F(x, y, z) = xzi+3yzj+z2kand S is the surface of solid bounded by the cylinder x2 + y2 = 1 and the planesz = y + 1 and z = y − 1.

Example 205. Calculate the flux of F across S where F(x, y, z) = x2yzi+x2yzj+xyz2k and S is the surface of the box bounded by the coordinate planes and theplanes x = 3, y = 2, and z = 1.

Calculus III Final Review Notes

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