Calculus Vector Principia Mathematica. Lynne Ryan Associate Professor Mathematics Blue Ridge Community College

Transcription

1 Calculus Vector Principia Mathematica Lynne Ryan Associate Professor Mathematics Blue Ridge Community College

2 Defining a vector Vectors in the plane A scalar is a quantity that can be represented by a single real number (such as length or area). A vector (denoted v) is a quantity that has both magnitude (denoted v ) and direction (denoted θ). An example of a vector quantity would be force, such as a force of 4 N directed at a 45 above the horizontal. Vectors can be sketched in the coordinate plane as directed line segments. A vector denoted v =< v 1, v 2 > can be represented with its tail at the origin (0, 0), and head at the point (v 1, v 2 ). A vector v =< v 1, v 2 > has x component v 1 and y component v 2. The relationship between components, magnitude, and direction angle is shown below: v 1 = v cosθ v 2 = v sinθ V v v2 2 = v 2 tan θ = v 2 v 1 Example: What are the magnitude and direction of the vector v =< 3, 4 >? 1

3 Example: What are the components of a vector with magnitude 25 m and direction θ = 2π 3? Vectors can be relocated in space. For a vector with tail at the point (a 1, a 2 ) and head at the point (b 1, b 2 ) v =< b 1 a 1, b 2 a 2 > A vector is really a set of directed line segments; all directed line segments with the same magnitude and direction are considered to be the same vector, regardless of where their heads and tails are located. 2

4 Defining a vector Vectors in three or more dimensions Vectors in three dimensional space can be defined as ordered triples of components; the vector v =< v 1, v 2, v 3 > has x component v 1 y component v 2 z component v 3 It may also be denoted < v x, v y, v z >. We can generalize the idea of a vector in n-dimensional space as an ordered n-tuple: The v 1... v n are the components of v. v =< v 1, v 2,..., v n > Vectors may be located anywhere in space. A vector with tail at (a 1, a 2,..., a n ) and head at (b 1, b 2,..., b n ) has components v =< b 1 a 1, b 2 a 2,..., b n a n > Example: How would you denote a vector with tail at (3, 2, 1) and head at (1, 1, 4)? For vectors in space, the magnitude is still obtained from the Pythagorean Theorem: v 2 = v v v 2 3 or, in general, n v 2 = v1 2 + v vn 2 = vi 2 i=1 3

5 It takes more than one angle to fix the direction of a vector v =< v 1, v 2, v 3 > in space. Direction cosines can be used. Let Then α be the angle between v and the positive x axis β be the angle between v and the positive y axis γ be the angle between v and the positive z axis cos α = v 1 v cos β = v 2 v (The angles themselves are obtained by taking inverse cosines.) cos γ = v 3 v Example: Find the magnitude and direction cosines of a vector with tail at (3, 2, 1) and head at (1, 1, 4). What are the angles between the vector and the x, y, and z axes? The distance formula can be written in terms of vectors. If v is the vector with tail at (a 1, a 2,..., a n ) and head at (b 1, b 2,..., b n ), then the distance between head and tail is simply the magnitude of the vector v =< b 1 a 1, b 2 a 2,..., b n a n >=< v 1, v 2,..., v n >: d = v = v v v 2 n Example: What is the distance between the points (1, 4, 5) and ( 1, 2, 5)? 4

6 Vector addition and scalar multiplication Using components Let u =< u 1, u 2,..., u n > and v =< v 1, v 2,..., v n > be vectors and let c be a scalar. We define vector addition: u + v =< u 1 + v 1, u 2 + v 2,..., u n + v n > and scalar multiplication: cv =< cv 1, cv 2,..., cv n > Example: For u =< 1, 4, 5 >, v =< 1, 5, 2 >, c = 4, find cu + v Properties of vector addition: * u + v = v + u (commutative property of addition) * u + (v + w) = (u + v) + w (associative property of addition) * u + 0 = 0 + u = u (additive identity; the identity element is the zero vector, a vector with u i = 0 for all i) * u + u = u + u = 0 (additive inverse) Properties of scalar multiplication: * c(u + v) = cu + cv (distributive property; scalar over vector sum) * (c + d)u = cu + du (distributive property; vector over scalar sum) * c(du) = (cd)u (associative; scalars with vector) * 1u = u (multiplicative identity for scalar multiplication) 5

7 Proof: Prove that for vectors u and v, u + v = v + u Suppose that we have two vectors in n dimensional space: u = < u 1, u 2,..., u n > v = < v 1, v 2,..., v n > By the definition of vector addition, u + v =< u 1 + v 1, u 2 + v 2,..., u n + v n > Since the components themselves are scalars, by the commutative property of addition u + v = < u 1 + v 1, u 2 + v 2,..., u n + v n > = < v 1 + u 1, +v 2 + u 2,..., v n + u n > = v + u The set of all vectors in the plane (v =< v 1, v 2 >) with accompanying scalars and the operations of vector addition and scalar multiplication defined on them form a vector space, R 2. In general, for any n, the set of all vectors in the form v =< v 1, v 2,..., v n > with accompanying scalars form a vector space R n under the operations of vector addition and scalar multiplication. 6

8 Vector addition and scalar multiplication Graphical interpretation Multiplying a vector by 1 has the effect of reversing the direction of the vector. Multiplying a vector by a constant c scales the vector to c times the original length cv = c v Vectors can be added graphically by either the parallelogram method, or the tail to head method. An expression in the form cu+dv (combining operations of scalar multiplication and vector addition) is a linear combination of u and v. Example: For the vectors u =< 1, 2 > and v =< 2, 3 >, find 2u + v by sketching (use the parallelogram method). 7

9 Unit vectors and direction Any vector of length (magnitude) 1 is a unit vector. Given a vector v, you can construct a unit vector u in the same direction as v by normalizing v: Example: Normalize v =< 1, 1, 5 >. u = 1 v v A unit vector can be used to represent the direction of a given vector - normalizing v gives us a way to borrow its direction, without borrowing its magnitude. To find a vector w with a given magnitude, and in the direction of a given vector v, * Find the unit vector in the direction of v. * Multiply the unit vector by the given magnitude, to scale it to the correct length. Example: Find a vector w in the direction of v =< 2, 3 >, but with w =5. 8

10 In the plane, the vectors i =< 1, 0 > and j =< 0, 1 > are the standard unit vectors. In space, the standard unit vectors are i =< 1, 0, 0 >, j =< 0, 1, 0 >, andk =< 0, 0, 1 >. Any vector can be written as a linear combination of the standard unit vectors; for example < 2, 3, 4 >= 2< 1, 0, 0 > 3 < 0, 1, 0 > +4 < 0, 0, 1 >= 2i 3j +4k The vectors i and j are referred to as the standard basis for R 2 (they span and are independent). i, j, andk are the standard basis for R 3, and so on. The idea of basis is covered in detail in Linear Algebra. Example: Express < 2, 0, 4 > as a linear combination of the standard basis vectors. Example: Calculate (3i 2k) 3(i + j +5k) 9

11 Application Forces Vectors can be used to represent forces - a force has both a magnitude and a direction. The resultant of a collection of forces is the net force. Example: Find the resultant force: the vector sum of < 1, 4 > and < 2, 5 >. What are the magnitude and direction of the resultant? Sketch. Example: An example of a tension problem is presented in the lecture. Take the time to work through the steps, and sketch and write yourself some notes as you go! 10

12 The dot product Defining the dot product For u =< u 1, u 2,..., u n > and v =< v 1, v 2,..., v n > in R n we define the dot product n u v := u 1 v 1 + u 2 v u n v n = u i v i i=1 Note the dot product is a scalar. This function is a type of inner product. Example: For u =< 1, 2, 3 >, v =< 4, 1, 0 >, find u v. Properties of the dot product: * u v = v u * (u + v) w = u w + v w * cu v = c(u v) = u cv * u u 0 and u u = 0 if any only if u = 0 Proof: The first property: Suppose u and v in R n. Then n n u v = u i v i = v i u i = v u i=1 i=1 since u i and v i are real numbers, and the commutative property of multiplication holds. Example: Prove the second property: 11

13 Proof: The fourth property: Prove u u 0 and u u = 0 if any only if u = 0 [First, prove that u u 0] u u = = n u i u i i=1 n (u i ) 2 i=1 Since (u i ) 2 0 for all u i, we must have n i=1 (u i ) 2 0. So, u u 0. [Now, the if and only if] Suppose u u = 0. Then n i=1 (u i ) 2 = 0. So each u i must equal zero, since if any u i were nonzero, (u i ) 2 would add a positive nonzero value to the sum. Since each u i = 0, u = 0. [The other direction] Suppose u = 0. Then u i = 0 for each i, and n i=1 (u i ) 2 = 0. So u u = 0. Dot product and magnitude are related by the following formula: Exercise: Prove it! v 2 = v v 12

14 The dot product The angle between two vectors If u and v are both unit vectors, then u v = cos θ If u and v are not both unit vectors, then we need to normalize: ( ) ( ) u v = cos θ u v This can be rearranged as u v = u v cosθ The proof (for vectors in the plane) comes from the Law of Cosines. Please take a few minutes to work through the derivation - sketch and write it down! 13

15 The relationship cos θ = u v u v is useful for calculating the angle between two vectors (and can be used to define the idea of angle between generalized vectors in any vector space). Example: Find the angle between the vectors u =< 1, 2 > and v =< 3, 4 >. Example: Find the angle between the vectors u = 3i 2j + k and v = j + 5k. 14

16 The dot product Orthogonal vectors Two vectors u and v are orthogonal (perpendicular) if and only u v = 0. Example: Are the vectors u =< 1, 2, 4 > and v =< 1, 2, 4 > orthogonal? Example: Find a vector orthogonal to u =< 1, 3 >. How many such vectors are there, and what do they look like relative to u? Example: Find a vector orthogonal to u =< 1, 3, 2 >. How many such vectors are there, and what do they look like? 15

17 An orthogonal set is a collection of vectors that is pairwise orthogonal (the dot product of any pair of vectors in the set is zero). Example: The set of vectors {< 1, 2, 1 >, < 4, 2, 0 >, < 2, 4, 10 >} is an orthogonal set. The vectors in any pairing are perpendicular - dot all possible combinations to see: A set of vectors is orthonormal if it is orthogonal and all the vectors in the set are normalized (have magnitude equal to 1)(are unit vectors). The set of vectors in the previous example is not orthonormal - the vectors are not unit vectors. However, you can construct an orthonormal set from an orthogonal set by normalizing each of the vectors: 16

18 The dot product Vector projections and components Recall that a vector can be broken down into horizontal and vertical components: v =< v 1, v 2 >= v 1 i + v 2 j The v 1 and v 2 are scalars, and indicate the length of each component; the i and j are vectors indicating the horizontal and vertical directions. In some applications, it may be more convenient to break a vector down into perpendicular components that are not horizontal and vertical. The vector projection of v onto u is denoted proj u v and is shown below. It is also called the vector component of v along u. The length (magnitude) of proj u v is denoted comp u v comp u v = proj u v comp u v = v cosθ ( ) u v comp u v = u The direction of proj u v is that of u. However, we need to normalize u in order to borrow its direction: Direction of proj u v = u u 17

19 Putting it all together, or proj u v = (magnitude)(direction) ( ) u = (comp u v) u ( ) ( ) u v u = u u proj u v = ( ) u v u 2 u Example: Let v =< 1, 4 > and u =< 3, 1 >. Find proj u v, and proj v u. Sketch both. The vector component of v orthogonal to u is denoted orth u v and is shown below: 18

20 v is the vector sum of orth u v and proj u v: v = proj u v + orth u v so orth u v = v proj u v Example: In the previous example, we found that for v =< 1, 4 > and u =< 3, 1 >, proj u v =< 3 10, 1 10 >. Use this to find orth uv. Example: Sketch and work the last example of a box on a hill: 19

21 Application Work When a constant force F is used to move an object through a distance d, and the force is entirely in the same direction as the object, we have W = Fd In this version, familiar from Physics courses, force is represented as a scalar (and that s OK, since the force and the distance line up), but this is really just a special case. When the force is directed at an angle from the line of motion, the above no longer holds. It is necessary to think of force as a vector F, with magnitude and direction. Only the component of F in the direction of the motion does work. We can also think of distance as a vector, so that we can describe work as W = (comp d F)( d ) W = F d cosθ W = F d (Work is the dot product of the force vector and the distance vector.) Example: What is the work done by a force of 30 N directed 70 above the horizontal, moving a box up an incline of 25? 20

22 Example: A force of (30i 20j + 100k)N is applied to move an object from a location at (1, 1, 1)m to a location at (10, 5, 1)m. What is the work done by the force? 21

23 Theorems Cauchy-Schwarz inequality, Triangle inequality, Pythagorean theorem This section is a collection of three important theorems involving dot product and magnitude. Please work through the proofs! Cauchy-Schwarz inequality For any two vectors u and v in a vector space, u v u v Example: Verify the Cauchy-Schwarz inequality for u =< 1, 4, 3 > and v =< 1, 0, 2 >. Proof: Work through the proof: Note that this form of the proof relies on the idea of angle- this proof (which is the simple one) is OK for vectors in R n, where we can verify the dot product formula through geometry. The C-S inequality does hold for vectors in general vector spaces, but you have to be a little careful or you end up going in circles on the idea of angle- a different proof approach is preferred. 22

24 Triangle inequality For any two vectors u and v in a vector space u + v u + v Proof: Work through the proof: The proof relies on the Cauchy-Schwarz inequality. When the C-S inequality is established for general vector spaces, this also establishes the Triangle inequality for general vector spaces. 23

25 Pythagorean theorem If u and v are orthogonal, then u + v 2 = u 2 + v 2 Proof: Prove it! Start it like the proof of the Triangle inequality... but what do you know about the dot product of orthogonal vectors? Use that, and the proof is over with quickly! 24

26 The cross product Defining the cross product The cross product of two vectors u and v in space is defined by where θ is the angle between u and v. u v = ( u v sinθ)n n is a unit vector orthogonal to both u and v, and indicates the direction of the cross product. The magnitude is given by the ( u v sin θ) part; i.e. the cross product is a multiple of n that has this length. It is important to distinguish between dot and cross products - the dot product results in a scalar, the cross product results in a vector. There are two possible choices for the unit direction vector n; the right hand rule is used to determine direction. Example: Compute i j. 25

27 Example: For an arbitrary vector v, what is v v? Example: Let u and v be vectors which lie in the xy plane. u = 10 and θ u = 75 (the angle u makes with the x-axis). v = 20 and θ v = 15 (the angle v makes with the x-axis). What are the magnitude and direction of u v? The cross product is only defined for vectors in space (R 3 ). Vectors in the plane are a subset of vectors in space, and can be crossed, but we should think of them as having a third (zero) component: v = v 1 i + v 2 j = v 1 i + v 2 j + 0k 26

28 The cross product Computing the cross product The definition u v = ( u v sinθ)n is inconvenient for computing the cross product of arbitrary vectors in space. A list of what you would have to do appears on the slides. The point to this list is not that you should do it, but that you shouldn t! Instead, we ll derive a computational formula based on the components of u and v. Let u =< u 1, u 2, u 3 > and v =< v 1, v 2, v 3 > be vectors in R 3. The cross product can be computed by u v =< u 2 v 3 u 3 v 2, (u 1 v 3 u 3 v 1 ), u 1 v 2 u 2 v 1 > Example: Let u =< 1, 2, 3 >, v =< 5, 1, 2 >. Compute u v. Verify that u v is orthogonal to both u and v. We need to verify that the computational formula matches the definition, and gives the correct magnitude and direction. 27

29 Verify magnitude: u v sinθ = u v 1 cos 2 θ ( ) 2 u v = u v 1 u v = u v ( u v )2 (u v) 2 ( u v ) 2 = ( u v ) 2 (u v) 2 = (u u u 2 3)(v1 2 + v2 2 + v3) 2 (u 1 v 1 + u 2 v 2 + u 3 v 3 ) Verify direction (orthogonality): = (a good bit of algebra - distribute, regroup, factor) = (u 2 v 3 u 3 v 2 ) 2 + (u 1 v 3 u 3 v 1 ) 2 + (u 1 v 2 u 2 v 1 ) 2 = u v (u v) u = < u 2 v 3 u 3 v 2, (u 1 v 3 u 3 v 1 ), u 1 v 2 u 2 v 1 > < u 1, u 2, u 3 > = (u 2 v 3 u 3 v 2 )u 1 (u 1 v 3 u 3 v 1 )u 2 + (u 1 v 2 u 2 v 1 )u 3 = u 1 u 2 v 3 u 1 u 3 v 2 u 1 u 2 v 3 + u 2 u 3 v 1 + u 1 u 3 v 2 u 2 u 3 v 1 = 0 (Similar to verify (u v) v = 0.) The cross product can be used to find a vector which is orthogonal to two given vectors (since this is what it, by definition, produces). Example: Let u = 5i j + k, v = i 2j + 4k. Find a vector which is orthogonal to both u and v. Is this vector unique? 28

30 The cross product Using determinants to compute The cross product formula u v =< u 2 v 3 u 3 v 2, (u 1 v 3 u 3 v 1 ), u 1 v 2 u 2 v 1 > can be remembered using a determinant structure. [ ] a b For a 2x2 matrix: c d the determinant is denoted a b c d and is computed by a b c d = ad bc Example: Compute Computing the cross product: This is best explained by example. Suppose we wish to calculate u v for u = 3i 2j + k,v = i + 7j + 4k Start by arranging u and v in a 3x3 determinant structure, with i, j, and k forming the top row, and the components of u and v forming the second and third rows: i j k u 1 u 2 u 3 v 1 v 2 v 3 = i j k Set up a structure and fill it in (watch the demo): i j + k 29

31 And compute using the rule for 2x2 determinants: ( 8 7)i (12 + 1)j + (21 2)k 15i 13j + 19k (or < 15, 13, 19 >) In general u v = = i j k u 1 u 2 u 3 v 1 v 2 v 3 u 2 u 3 v 2 v 3 i u 1 u 3 v 1 v 3 j + u 1 u 2 v 1 v 2 k (Don t memorize that as a formula; learn the pattern of crossing out columns in turn!) Example: Use the determinant structure to compute u v for u =< 1, 1, 4 >, v =< 3, 0, 1 >. 30

32 The cross product The area of a parallelogram The area of a parallelogram with adjacent edges formed by the vectors u and v is given by A = u v v Example: Use u v to compute the area of a parallelogram with adjacent edges bounded by u =< 1, 1, 2 > and v =< 5, 4, 3 >. 31

33 Example: Compute the area of a parallelogram with vertices A = (5, 2, 0), B = (2, 6, 1), C = (2, 4, 7), D = (5, 0, 6). The area of a triangle with adjacent edges formed by the vectors u and v is given by A = u v 2 Example: Compute the area of a triangle with vertices A = (1, 1, 1), B = ( 1, 4, 7), C = (0, 2, 2). 32

34 The cross product Properties of the cross product For all of the following, u, v, and w are vectors in R 3, c scalar. Geometric properties of the cross product: * u v is orthogonal to both u and v. * u v = u v sinθ. * u v = 0 if and only if u = cv for some c. * u v gives the area of a parallelogram with adjacent sides u and v. Algebraic properties of the cross product: * v v = 0 * v 0 = 0 = 0 v * c(u v) = (cu) v = u (cv) * u (v + w) = (u v) + (u w) * u v = (v u) Proof: Prove the third property: c(u v) = (cu) v = u (cv) 33

35 The commutative property does not hold: u v v u (This is important; we re used to being able to multiply in any order, and we ve now met a multiplication-type operation where that is no longer the case.) Instead u v = (v u) (Same magnitudes, but opposite directions because of the right hand rule.) 34

36 The cross product The triple scalar product The triple scalar product of vectors u, v and w in space is defined by u (v w) Example: Compute the triple scalar product of the vectors u =< 1, 2, 2 > v =< 1, 4, 7 > w =< 0, 2, 2 > The triple scalar product can be computed using the determinant structure: u 1 u 2 u 3 u (v w) = v 1 v 2 v 3 w 1 w 2 w 3 v = u 2 v 3 1 w 2 w 3 u 2 v 1 v 3 w 1 w 3 + u 3 v 1 v 2 w 1 w 2 35

37 Example: Compute the triple scalar product of the vector (use the determinant structure): u =< 1, 2, 2 > v =< 1, 4, 7 > w =< 0, 2, 2 > A parallelpiped is a polyhedron whose faces are parallelograms. The triple scalar product can be used to compute the volume of a parallelpiped with edges u, v and w: V = u (v w) Example: Compute the volume of a parallelpiped with edges u = 5i 2j+k, v = i +j 4k, w = 3i + k. Properties of the triple scalar product: u (v w) = v (w u) = w (u v) u (v w) = (u v) w 36

38 Application Torque Torque is the name given to the twisting effect of applying a force to an object, causing it to pivot. It depends on 3 things: (1) the amount of force applied, (2) the direction of the force, and (3) the distance from the point of application of the force to the pivot point. The lever arm is the line along which the force is transmitted to the pivot. It is denoted by a vector r. When r and F are orthogonal, we have τ = r F * That τ is supposed to be boldface - torque is a vector. Can t seem to get Greek and bold at the same time. Example: What is the magnitude of the torque produced by applying a force of 3 lb to the pictured wrench? When the force is directed at an angle, the component which produces torque is the orthogonal component of F along r: orth r F. The magnitude of orth r F is F sinθ, and τ = r F sinθ Example: What is the magnitude of the torque produced by applying a force of 3 lb to the pictured wrench? 37

39 We define torque as a vector by the cross product: τ = r F The convention for direction comes from the right hand rule, and we have * Clockwise rotation = IN * Counterclockwise rotation = OUT If we further apply the convention to a right handed coordinate system, where i j = k determines the direction of the positive z axis, we have * Clockwise rotation = IN = unit direction vector ( k) * Counterclockwise rotation = OUT = unit direction vector (+k) Example: What is the torque produced by applying a force of 3 lb to the pictured wrench? Express the answer as a vector. 38

40 Expressing torque as a vector cross product makes it possible to work with forces and pivoting objects in 3D space: Suppose you have a lever with one end at the origin, and the other located at the point (20, 30, 10) cm. The lever is free to pivot about the origin. A force of < 5, 10, 4 > N is applied to the end of the lever, causing it to pivot. What is the torque produced by the force? Compute the cross product τ = r F: And interpret your answer (what do the magnitude and direction tell you?): 39

41 Vector algebra Summary of operations Let u and v be vectors, c scalar. We have defined * The norm (magnitude) of a vector [ v ] * Scalar multiplication [cv] * Vector addition [u + v] * Dot product of vectors [u v] * Cross product of vectors [u v] Expressions containing these operations need to make sense; for example, we can t add a scalar and a vector (or cross a scalar and a vector, or dot a scalar and a vector). Examples: For each of the following expressions, decide whether the expression is defined, and if so, whether the final result is a vector or a scalar. (u v)(3u v) (u v) ( 2v) (u v) ( 2v) As long as the expression is defined, we compute by building up in pieces. Example: Let u =< 1, 1, 3 >, v =< 3, 4, 2 >, w =< 0, 1, 7 >. Find (u v) + 3w 40

42 Lines and planes Lines Lines in 2D can be defined by a point and a slope. You should recall the various forms of equations of lines: * y y 0 = m(x x 0 ) (point-slope) * y = mx + b (slope-intercept) * Ax + By + C = 0 (general or standard) In 3D space, the idea of slope no longer makes sense. We can define a line in space using a point (to fix its location) and a vector (to indicate its direction). The equation of a line in space is given by where * P 0 is a given point on the line. * r 0 is the position vector pointing to P 0 r = r 0 + tv * v is a vector indicating the direction of the line, called the parallel vector for the line * t is an arbitrary scalar parameter. The quantity tv produces an infinite number of scalar multiples of the vector v, extending the line infinitely in both directions. * r, the vector sum of r 0 and tv, is the position vector pointing to any arbitrary point (x, y, z) on the line. r varies with t. 41

43 The equation or r = r 0 + tv < x, y, z >=< x 0, y 0, z 0 > +t < a, b, c > is the point-parallel form of the equation of a line. Example: Write the equation of a line passing through the point (1, 4, 5) with direction parallel to the vector < 1, 2, 7 >. Example: Give three points on the line Sketch. < x, y, z >=< 1, 1, 1 > +t < 2, 5, 3 > By rearranging the point-normal form and equating components, we arrive at the parametric equations for a line: x = x 0 + at y = y 0 + bt z = z 0 + ct 42

44 Example: Write the parametric equations for the line passing through (1, 5, 3) parallel to v =< 2, 0, 1 >. If we solve the parametric equations for t x x 0 = at t = x x 0 a y y 0 = bt t = y y 0 b z z 0 = ct t = z z 0 c and equate: x x 0 = y y 0 = z z 0 a b c we arrive at the symmetric equations for the line. Example: Write the symmetric equations for the line passing through (1, 5, 3) parallel to v =< 2, 0, 1 >. 43

45 Example: Write the parametric and symmetric equations for the line passing through the points ( 2, 4, 5) and (1, 0, 1). And, having done all this for lines in 3D space, we should note it works perfectly well for lines in the plane as well: * < x, y >=< x 0, y 0 > +t < a, b > (point-parallel) * x = x 0 + at y = y 0 + bt (parametric) * x x 0 a = y y 0 b (symmetric) 44

46 Lines and planes Intersection of lines We solve for the coordinates of points of intersection by equating expressions for the x, y, and z expressions of the parametric equations for the lines. The parametric equations are As an example, look at x = x 0 + at y = y 0 + bt z = z 0 + ct L 1 : x = 1 2t y = 1 + 3t z = 1 + 4t L 2 : x = 1 y = 2 t z = 3 2t And be careful, there s a catch. Equating the x expressions gives a solution of 1 2t = 1 t = 0, but substituting this solution into the equations for y and z yields an inconsistent system. This is not the solution. Be sure to use a different parameter for the second line: L 1 : x = 1 2t y = 1 + 3t z = 1 + 4t L 2 : x = 1 y = 2 s z = 3 2s We can now equate and solve (work through the solution here): 45

47 Example: Find the intersection of the lines L 1 : < x, y, z >=< 0, 1, 2 > +t < 1, 1, 4 > L 2 : < x, y, z >=< 1, 1, 0 > +t < 1, 1, 2 > or show that they do not intersect. Parallel lines lie in the same plane and do not intersect. They must have parallel direction vectors (i.e., v 1 and v 2 must be scalar multiples of each other). Skew lines do NOT lie in the same plane, and do not intersect. The direction vectors cannot be scalar multiples of each other. Example: We have shown that the lines do not intersect. Are they parallel or skew? L 1 : < x, y, z >=< 0, 1, 2 > +t < 1, 1, 4 > L 2 : < x, y, z >=< 1, 1, 0 > +t < 1, 1, 2 > 46

48 Example: The lines L 1 : L 2 : x 1 = y = z 5 x = y + 4 = z do not intersect (take the time to verify this for practice!). Are they parallel or skew? *The point to this question is to make sure you can start with any form of a line and get back to the other forms. What part of the symmetric equations shows the parallel vector? What part shows the point? 47

49 Lines and planes Planes A plane can be defined by specifying a point and a normal vector. The plane is the set of all (x, y, z) such that the vector from the fixed point to any (x, y, z) is orthogonal to the normal vector. The point-normal form of the equation of a plane is n (r r 0 ) = 0 Suppose we have a plane with normal n =< 1, 2, 5 >, and a point on the plane (3, 4, 6). n =< 1, 2, 5 > r 0 =< 3, 4, 6 > r =< x, y, z > then n (r r 0 ) = 0 becomes < 1, 2, 5 > < x 3, y 4, z 6 >= 0 *You should learn point-normal as < n 1, n 2, n 3 > < x x 0, y y 0, z z 0 >= 0 If you work through the dot product, you get the scalar form: n 1 (x x 0 ) + n 2 (y y 0 ) + n 3 (z z 0 ) = 0 In our example: 1(x 3) 2(y 4) + 5(z 6) = 0 48

50 Distributing out the scalar form gives the general form: Ax + By + Cz + D = 0 In our example: 1(x 3) 2(y 4) + 5(z 6) = 0 x 3 2y z 30 = 0 x 2y + 5z 25 = 0 Notice that the normal vector is still visible in Ax+By+Cz+D = 0: < n 1, n 2, n 3 >=< A, B, C >. Example: Write equations (point-normal, scalar, and general forms) for a plane passing through (5, 1, 2) and normal to the vector < 6, 2, 3 >. Example: The equation of a plane is given: 3x 2z = 4. What is a normal vector for this plane? Give the coordinates of three points on the plane. 49

51 The general form can be rearranged again: x 2y + 5z 25 = 0 5z = x + 2y + 25 z = 1 5 x y + 5 This expresses z as a multivariable function of x and y [z = f(x, y)], with domain the entire xy plane: < x <, < y <. This form is usually needed to plot using software. Planes can be hand sketched using intercepts: * to get the x-intercept, set y = 0, z = 0 * to get the y-intercept, set x = 0, z = 0 * to get the z-intercept, set x = 0, y = 0 The angle between planes is the (acute) angle between normal vectors. Compute from cos θ = n 1 n 2 n 1 n 2 Example: Find the angle between planes 3x 2y + z 5 = 0 and x + y + 3 = 0. 50

52 You can determine whether planes are parallel, orthogonal, or neither by comparing normal vectors. Examples: Determine whether the planes are parallel, orthogonal, or neither: 3x 2y + z = 5 and x + y z = 1 3x y + z = 7 and 2x + y z = 1 x 2y + 3z = 2 and 2x + 4y 6z = 1 51

53 Lines and planes Writing equations of lines and planes Examples: * Write the symmetric equations of a line through the point (1, 1, 2) that is parallel to the line with the parametric equations x = 3t, y = 5 2t, z = 4 + t * Write the general equation of the plane that contains the intersecting lines L 1 : x = 3t, y = 5 2t, z = 4 + t L 2 : x = 3 t, y = 7 + t, z = 5 + 2t There are too many possible relationships between lines and planes to learn each type of problem. Instead, we need to develop a general plan. The key to setting up these problems is to keep in mind that * Lines are defined in term of parallel vectors (lying on the line) * Planes are defined in terms of normal vectors. The Plan: * What are you given? * What do you want? * How are they related? * What s the equation, and what form is it in? 52

54 Work through the first example: Write the symmetric equations of a line through the point (1, 1, 2) that is parallel to the line with the parametric equations * What are you given? x = 3t, y = 5 2t, z = 4 + t * What do you want? * How are they related? * What s the equation, and what form is it in? 53

55 Work through the second example: Write the general equation of the plane that contains the intersecting lines L 1 : x = 3t, y = 5 2t, z = 4 + t L 2 : x = 3 t, y = 7 + t, z = 5 + 2t * What are you given? * What do you want? * How are they related? * What s the equation, and what form is it in? 54

56 Lines and planes Intersection of planes Any two planes which are not parallel will eventually intersect in a line which is common to both planes. To solve for the line of intersection, simultaneously solve the system of equations of the planes. Equations should be in general form (and then, move the constant over to the right of the equals). Example: Find the intersection of the planes (work through the solution): 3x 4y + z 10 = 0 x 2y z + 4 = 0 Three planes in space may intersect in a point, a line, or not at all. Solving three equations in three unknowns is tedious; continue on to Using SciLab to solve for the intersection of three planes for a crash course in matrix solving. 55

57 Lines and planes Distances in space The distance between two points a = (a 1, a 2, a 3 ) and b = (b 1, b 2, b 3 ) in space is the magnitude of the vector v =< b 1 a 1, b 2 a 2, b 3 a 3 > d = v = (b 1 a 1 ) 2 + (b 2 a 2 ) 2 + (b 3 a 3 ) 2 Example: What is the distance between the points (1, 4, 5) and (1, 2, 5)? The shortest distance from a point to a line is the orthogonal distance. However, we generally don t know the exact point on the line which is closest to the given point. On the other hand, it is easy to find an arbitrary point on the line. How do we relate the distance we can find to the distance we want? Set up some vectors: And do some algebra: Start with d = b sin θ, and multiply both sides by v : v d = b v sinθ The right hand side is the magnitude of the cross product of b and v, so and v d = b v d = b v v 56

58 So, to calculate the distance from a point to a line: * A point P and the equation of a line L need to be given. * Read the direction vector v from the equation of the line. * Find the coordinates of a point Q on the line. * Write the vector b from Q to P. * And compute d = b v v Example: What is the distance from the point (1, 4, 7) to the line x = 2 t, y = 3 + 2t, z = t Finding the distance from a point to a plane has the same problem - we generally don t know the exact point on the plane closest to the given point. As before, we can start by finding an arbitrary point on the plane. Then, since planes are defined by their normals, we can get the distance by projecting: comp n b = b n n 57

59 To calculate the distance from a point to a plane: * A point P and the equation of a plane Ax + By + Cz + D = 0 (or one of the other forms) need to be given. * Read the normal vector n from the equation of the plane. * Find the coordinates of a point Q on the plane. * Write the vector b from Q to P. * And compute d = b n n Example: What is the distance from the point (1, 4, 7) to the plane 3x 2y + z 1 = 0 58

60 Line to plane and plane to plane distances only make sense when the objects are parallel (if they aren t they ll eventually intersect). To calculate distance from a line to a plane, or a plane to a plane: * Verify that the figures are parallel. * Find an arbitrary point on the line (or first plane). * And go through the process of computing point-to-plane. Example: What is the distance between the planes 3x 2y + z 1 = 0 6x + 4y 2z 10 = 0 Distance from line to line is a little tricky - you can t pick arbitrary points on both lines. The trick is to recognize that non-intersecting lines lie in parallel planes. Figure out what the planes are, and restate as a plan-to-plane problem. Because the planes are parallel, they have the same normal vector n, orthogonal to both v 1 and v 2. Find n = v 1 v 2. To calculate the distance from a line to a line: * Equations of two lines L 1 and L 2 need to be given. * Read the direction vectors v 1 and v 2 from the equations for L 1 and L 2. Find a point P on L 1 and a point Q on L 2. * Find n = v 1 v 2 * Use n and point Q to write the equation of a plane. * And and find the distance from point P to the plane. 59

61 Example: What is the distance between the lines L1 : x = 1 + t, y = 2 + 3t, z = 4 t L2 : x = 2t, y = 3 + t, z = 3 + 4t Solution (in steps): What are v 1 and v 2? Find a point on each line (let t = 0). Find n = v 1 v 2. Use n and Q to write the equation of a plane containing L 2. Find the distance from the point P +(1, 2, 4) to the plane 13x 6y 5z +3 = 0. Point Q = (0, 3, 3) is a known point on the plane. 60

62 Vector valued functions Defining vector valued functions A line is an example of a vector valued function; a function which takes a scalar (t) as input, and returns a vector r(t) =< f(t), g(t), h(t) > or r(t) = f(t)i + g(t)j + h(t)k The components of the vector are all scalar functions of t. The graph of a vector valued function is a curve in space, where the points on the curve are traced out by the position vector < x, y, z >=< f(t), g(t), h(t) > The functions f(t), g(t), and h(t) are the component functions. We can write vector valued functions in parametric form: x = f(t) y = g(t) z = h(t) Example: What are the parametric equations for the vector valued function r(t) =< sint, t 2, t > The domain of a vector valued function is the set of values of t for which the function is defined. r(t) is defined wherever all three of its component functions are defined, so the domain of r(t) =< f(t), g(t), h(t) > is the intersection of the domains of f(t), g(t), and h(t). The range of a vector valued function is the set of all vectors < x, y, z > such that < x, y, z >= r(t) for some t in the domain. 61

63 Example: What are the domain and range of the vector valued function r(t) =< t 2, 3t + 1, t 2 > Example: Give a few points on the graph of Can you envision what it looks like? r(t) =< t 2, 3t + 1, t 2 > 62

64 Vector valued functions Plane curves A vector valued function with two components r(t) =< f(t), g(t) >= f(t)i + g(t)j and parametric equations traces out a curve in the xy plane: x = f(t), y = g(t) < x, y >=< f(t), g(t) > We can eliminate the parameter t and write an equation for the curve in terms of x and y. This is the Cartesian equation for the curve. Example: Eliminate the parameter and write the Cartesian equation for the curve r(t) =< 3t + 1, t 2 > What is the graph of this function? 63

65 Example: Eliminate the parameter and write the Cartesian equation for the curve What is the graph of this function? r(t) =< 3sin t, 2cos t > The curves both trace out the circle r(t) =< cos t, sint > s(t) =< cos t, sint > x 2 + y 2 = 1 but they provide more information than the Cartesian equation. Parametric curves have a direction in which the curve is traced out. To determine direction, pick a few values for t. For example, the curve r(t) =< t, t 2 > has parametric equations x = t, y = t 2 and draws y = x 2 in the direction indicated (make yourself a little margin sketch). Domain information is also part of the vector valued function. For example, the graph of r(t) =< t, 3t > is not the entire parabola y = 3x 2. The domain of r(t) =< t, 3t > is t 0, because of the radical. This is passed through to the values of x and y: since x = t, x 0, and since y = 3t but t 0, y 0 as well. And, if we start picking values for t, we ll never draw the left side of the parabola. 64

66 When we rewrite the parametric equations as the single equation y = 3x 2 we have to keep track of those restrictions; in particular, since y = f(x), we need to specify that the domain for this function of x must be x 0. This should both be included with the expression for the function: and be reflected in the graph. y = 3x 2, x 0 Example: Eliminate the parameter and write the Cartesian equation for What is the graph of this function? r(t) =< 2 t, 3t > 65

67 Vector valued functions Projections onto planes Seeing a 3D object projected onto a 2D plane can help you visualize the object. In particular, we look at projections of space curves onto the xy, xz, and yz planes. Examine the projections of the graph of r(t) =< cos t, sint, t > Make some sketches, and write the equations for the various projections. 66

68 Example: What are the projections of the graph of r(t) =< t, t 2, e t > onto the xy, yz, and xz planes? Write equations and sketch. 67

69 Calculus of vector valued functions Limits and continuity For a vector valued function r(t) =< f(t), g(t), h(t) > we define the limit as t approaches a of r(t) by lim t a r(t) =< lim t a f(t), limg(t), lim h(t) > t a t a Example: Find lim t 3 r(t) for r(t) =< t 3, t 1 t + 4, 9 t >. Example: Find lim t 2 r(t) for r(t) =< e t, t2 4 t + 2, 3 t2 >. 68

70 Example: Find lim t 1 r(t) for r(t) =< e t, ln(2 t), 3 t 1 >. A vector valued function r(t) =< f(t), g(t), h(t) > is continuous at a if and only if * r(a) exists * lim t a r(t) exists * lim t a r(t) = r(a) Example: Are the three previous examples continuous at their respective values for a? r(t) =< t 3, t t + 1 4, 9 t > (a = 3) r(t) =< e t, t2 4 t + 2, 3 t2 > (a = 2) r(t) =< e t, ln(2 t), 3 t 1 > (a = 1) 69

71 A vector valued function is continuous on an open interval (a, b) if it is continuous at every value for t in the interval. Left and right limits are defined by: lim t a lim t a r(t) =< lim f(t), lim g(t), lim h(t) > t a t a t a r(t) =< lim f(t), lim g(t), lim h(t) > + t a + t a + t a + A vector valued function is continuous on a closed interval [a, b] if it is continuous on the open interval (a, b) and if lim r(t) = r(a) t a+ lim r(t) = r(b) t b Example: On what intervals are the previous examples continuous? r(t) =< t 3, t t + 1 4, 9 t > (a = 3) r(t) =< e t, t2 4 t + 2, 3 t2 > (a = 2) r(t) =< e t, ln(2 t), 3 t 1 > (a = 1) 70

72 Calculus of vector valued functions Differentiation For a vector valued function r(t) =< f(t), g(t), h(t) >, we define the derivative of r(t), denoted r (t) or dr dt by After some vector algebra: r r(t + t) r(t) (t) = lim t 0 t r(t + t) r(t) t = 1 (< f(t + t), g(t + t), h(t + t) > < f(t), g(t), h(t) >) t = 1 < f(t + t) f(t), g(t + t) g(t), h(t + t) h(t) > t f(t + t) f(t) g(t + t) g(t) h(t + t) h(t) = <,, > t t t And some vector calculus: r (t) = r(t + t) r(t) lim t 0 t f(t + t) f(t) < = lim t 0 = < lim t 0 t f(t + t) f(t), lim t t 0 g(t + t) g(t) h(t + t) h(t),, > t t g(t + t) g(t), lim t t 0 h(t + t) h(t) t > We get r (t) =< f (t), g (t), h (t) > Example: Given r(t) =< e t, t t, 1 t 2 >, find r (t). 71

73 Example: Given r(t) =< sint, t 2, t >, find r (t). Then, find r(4) and r (4). The geometric interpretation of the derivative at t = a is a vector tangent to the curve traced by r(t) at the point defined by r(a). The derivative function takes r (t) at each value of t and uses it as the position to trace out another curve. A curve is smooth on an interval if its derivative is never zero on that interval. A curve has a cusp at t = a if r (a) = 0. Example: Sketch and work the example shown for r(t) =< t 3, t 2 > at t = 0. 72

74 Calculus of vector valued functions Tangents and tangent lines For a vector valued function r(t) and a given t = a, r (a) gives us a vector tangent to the curve at the point r(a). The unit tangent vector at t = a is obtained by normalizing r (a): T(a) = r (a) r (a) The unit tangent vector function gives the value of the unit tangent vector at any t: T(t) = r (t) r (t) Example: Find the unit tangent vector function for r(t) =< sint, cost, t 2 >. Example: Find a vector tangent to the curve traced by r(t) = (3t)i + (t 2 )j at t = 1. Then, find the unit tangent vector at t = 1. Sketch the curve, and indicate r(1), r (1) and T(1) on the sketch. 73

75 The equation of a line in space is given by < x, y, z >=< x 0, y 0, z 0 > +t < a, b, c > where (x 0, y 0, z 0 ) is a point on the line, and < a, b, c > is a parallel vector indicating the direction of the line. To write the equation of the tangent line at t = a, we get the point by evaluating r(a), and the direction vector by evaluating r (a). Example: Write the equation of the line tangent to the curve r(t) = (3t)i + (t 2 )j at t = 1. Convert to Cartesian form, and add a sketch of the tangent line to your previous sketch. Example: Write the equation of the line tangent to the curve r(t) =< e 2t, t 3, 5t > at the point (e 4, 8, 10). 74

76 Calculus of vector valued functions Integration For a vector valued function r(t) =< f(t), g(t), h(t) > we define the indefinite integral (antiderivative), denoted r(t)dt, by r(t)dt =< f(t) dt, g(t) dt, h(t)dt > Example: Given r(t) =< sint, 1 t, t3 + t >, find r(t)dt. For a vector valued function r(t) =< f(t), g(t), h(t) > we define the definite integral, denoted b r(t)dt, by the limit of the Riemann sum: a b a b a r(t) dt = lim n n r(t i) t i=1 = < lim n r(t) dt = < b a n i=1 f(t) dt, f(t i) t, lim n b a n i=1 g(t i) t, lim n b g(t) dt, h(t)dt > a n h(t i) t > i=1 75

77 We can extend the Fundamental Theorem of Calculus to vector valued functions: if r(t) is continuous on (a, b), then b a r(t)dt = R(b) R(a) where R(t) is any antiderivative of r(t): R (t) = r(t). Example: Given r(t) =< t 3, t 4, t 5 >, find 1 0 r(t)dt. 76

78 Calculus of vector valued functions Normal and binormal vectors The derivative of the unit tangent vector, T (t), always produces vectors which are normal to the curve r(t). Proof: Prove that for any vector valued function v(t), if v(t) = c (constant), then v (t) is orthogonal to v(t) for all t. Then, apply that result to T(t): since T(t) is by definition a unit vector with T(t) =1forallt, T(t) andt (t) are orthogonal for all t, andforanyt = a, T (a) gives a vector normal to the curve defined by r(t). The unit normal vector is obtained by normalizing T : N(t) = T (t) T (t) And, we can obtain a third vector orthogonal to both T and N, thebinormal vector: B(t) =T(t) N(t) 77

79 Example: Find T(t), N(t) andb(t) forr(t) =< sin(3t 2 ), cos(3t 2 ), 0 >. Example: Sketch the graph of r(t) =< sin(3t 2 ), cos(3t 2 ), 0 >. Place the vectors T(0) and N(0) on the graph. What is the direction of B(0) (in or out of the page)? 78

80 The normal plane at r(a) containes the vectors N and B, and is orthogonal to the vector T. The osculating plane at R(a) contains the vectors T and N, and is orthogonal to the vector B. The circle contained in this plane and tangential to the curve is the osculating circle and is the closest approximation to the curve near r(a) 79

81 Example: Find T, N and B for r(t) =<t 2, 2 3 t3,t> at the point (1, 2 3, 1). Example: point (1, 2 3, 1). Write the equations for the normal and osculating planes for r(t) =<t 2, 2 3 t3,t > at the 80

82 Calculus of vector valued functions Arc length Recall that the length of a curve can be approximated using lengths of line segments n L ( xi ) 2 + ( y i ) 2 If the curve is parameterized by x = f(t) and y = g(t), then i=1 x i f (t i ) t i y i g (t i ) t i and we have the Riemann sum L = = n ( xi ) 2 + ( y i ) 2 i=1 n [f (t i ) t i ] 2 + [g (t i ) t i ] 2 i=1 n [f (t i )] 2 + [g (t i )] 2 t i i=1 Taking the limit as n gives the length of the curve, or arc length, on the interval (a, b): L = L = b a b a [f (t)] 2 + [g (t)] 2 dt (dx ) 2 + dt For a space curve, an analogous argument gives L = L = L = b a b a b a (dx ) 2 + dt ( ) 2 dy + dt ( ) 2 dy dt dt ( ) 2 dz dt dt [f (t)] 2 + [g (t)] 2 + [h (t)] 2 dt (dx ) 2 + dt ( ) 2 dy + dt ( ) 2 dz dt dt If the curve is defined by r(t) =< f(t), g(t), h(t) >, the formula can be written compactly as L = b a r (t) dt 81

83 Example: Find the length of the curve given by r(t) =< e t cos t, e t sint >, 0 t π. The arc length function gives the length of a curve on (a, t) as t varies (a is fixed): s(t) = t a r (u) du Example: Find the arc length function for the curve r(t) =< 2 sin t, 5t, 2 cos t > t 0. 82

84 Differentiating the arc length function d dt s(t) = d dt t a r (u) du and applying the Fundamental Theorem of Calculus gives ds dt = r (t) The rate of change of the arc length function with respect to t is the same as the magnitude of the rate of change of the position vector. 83

85 Calculus of vector valued functions Curvature We can think of the unit tangent vector, T(t) = r (t) r, as providing information about (t) the direction of the curve at any t, without any additional magnitude information. Curvature is a measure of the tightness of the curve- how fast the direction is changing as you travel along length of curve. Curvature is defined as the norm of the rate of change of the unit tangent vector with respect to arc length: κ = dt ds The implication is that T (unit tangent) is expressed in terms of s (arc length), and not in terms of the usual parameter t. While it is possible to reparameterize T in terms of s, it is easier to modify the formula: dt = dt dt ds ( ds dt dt dt = dt ds ) ds dt (chain rule) (don t forget, s is scalar, and ds dt = r (t) ) T (t) = κ r (t) Emphasizing that κ is a function of t: (You ll see it both ways in various texts.) κ = T (t) r (t) κ(t) = T (t) r (t) 84

86 Example: Find curvature as a function of t for r(t) =< t 2, 2t, lnt >. Example: What are the values of the curvature at t = 1 and t = 4 for r(t) =< t 2, 2t, lnt >? 85

87 The radius of curvature is the reciprocal of the curvature value: ρ = 1 κ The radius of that of an inscribed circle tangent to the curve (so high κ value = small ρ value = tight curve, and vice versa). An alternate curvature formula (the derivation of this formula is at the end of the notes): κ(t) = r (t) r (t) r (t) 3 Example: Use the alternate formula to find κ(t) for r(t) =< t 2, 2t, lnt >. 86

88 The is also a curvature formula for plane curves in the form y = f(x): (derivation at end of notes) κ(x) = f (x) [1 + (f (x)) 2 ] 3/2 Example: Find the curvature and radius of curvature of the graph of y = cos2x at x = 0. Sketch the graph and the inscribed circle. 87

89 Derivation of Start with T(t) = r (t) r (t) and solve for r (t): Substitute r (t) = ds dt : r (t) = κ(t) = r (t) r (t) r (t) 3 r (t) = r (t) T(t) ( ) ds T(t) dt Differentiate both sides with respect to t. This involves the scalar/vector function product rule on the right: d dt r (t) = d ( ) ds dt [ T(t)] dt r (t) = d ( ) ( ) ds ds d dt [ ]T(t) + dt dt dt [T(t)] ( ) ( ) d r 2 s ds (t) = dt 2 T(t) + T (t) dt Cross r (t) with both sides: r (t) r (t) = r (t) [( ) d 2 s dt 2 T(t) + ( ) ] ds T (t) dt Apply properties of cross product (distributes over vector addition; scalars can move to the front): ( ) ( ) d r (t) r 2 s ds (t) = dt 2 r (t) T(t) + r (t) T (t) dt Substitute in r (t) = r (t) T(t): ( d r (t) r 2 s (t) = dt 2 ) r (t) T(t) T(t) + ( ) ds r (t) T(t) T (t) dt Note that T(t) T(t) = 0 (property of cross product; vector crossed with itself): ( ) ds r (t) r (t) = 0 + r (t) T(t) T (t) dt Substitute in r (t) = ds dt and combine: r (t) r (t) = ( ) 2 ds T(t) T (t) dt Take magnitudes of both sides: ( ) 2 ds r (t) r (t) = T(t) T (t) dt ( ) 2 ds = dt T(t) T (t) 88

90 We have T(t) T (t) = T(t) T (t) sinθ. We have also proven that since T(t) = 1 (constant), T(t) and T (t) are orthogonal and therefore sinθ = sin(π/2) = 1. So ( ) 2 r (t) r ds (t) = dt T(t) T (t) But T(t) = 1: ( ) 2 r (t) r ds (t) = dt T (t) Substitute back ds dt = r (t) (and yes, we did have to work with ds dt so the original substitution was necessary): for a while to differentiate, Solve for T (t) : Divide both sides by r (t) : r (t) r (t) = r (t) 2 T (t) T (t) = r (t) r (t) r (t) 2 T (t) r (t) = r (t) r (t) r (t) 3 The expression on the left was our first formula for κ(t). So κ(t) = r (t) r (t) r (t) 3 Because it involves the cross product, this formula only makes sense for vectors in R 3 (and R 2 as a subset of R 3 with 0k). This is still pretty useful, as we are unlikely to be sketching any four dimensional curves. It should be noted that the idea of curvature (and the original formula) could be defined in any dimension. 89

91 Derivation of for plane curves in the form y = f(x): κ(x) = f (x) [1 + (f (x)) 2 ] 3/2 If y = f(x), then a parameterization of the curve is x = t, y = f(t), giving us the vector function r(t) =< t, f(t), 0 > Note that and apply the previous formula: r (t) =< 1, f (t), 0 > r (t) =< 0, f (t), 0 > r (t) r (t) = i j k 1 f (t) 0 0 f (t) 0 =< 0, 0, f (t) > Then and r (t) = 1 + (f (t)) 2 κ(t) = r (t) r (t) = (f (t)) 2 = f (t) f (t) ( 1 + (f (t)) 2 ) 3 Since t = x, κ(x) = f (x) [1 + (f (x)) 2 ] 3/2 90

92 Application: Differential equations and initial value problems A differential equation is an equation involving various order derivatives of a function (the solution is a function that satisfies the equation). Simple differential equations of the form y = f (x), y = f (x) etc. can be solved by antidifferentiation, and we can do something analogous with vector valued functions. Example: Given r (t) =< t 2, sec 2 t, 1 >, what is r(t)? An initial value problem is a differential equation with additional information about the function and/or its derivatives that allows you to solve for constants of integration. For example, to solve y = x 2 + sinx y(0) = 1 we antidifferentiate y = 1 3 x3 cos x + C and use y(0) = 1 to solve for the constant: y(0) = cos 0 + C = C = 1 C = 2 For a vector valued function... y = 1 3 x3 cos x

93 Example: Given r (t) =< t 2, sec 2 t, 1 >, r(0) =< 1, 2, 5 >, what is r(t)? Example: Given r (t) =< 2t, 3t 2 + 1, cos t >, r ( π ) =< 0, 0, 1 >, r(0) =< 1, 1, 1 >, what is r(t)? 2 92

94 Application: Position, velocity, acceleration If t is interpreted as a time parameter, the vector valued function r(t) traces out a curve over time, and at t = a, r(a) pointstotheposition of the object moving along that curve. The derivative r (t) gives the rate of change of position with respect to time; i.e. the velocity. v(t) =r (t). Differentiating again gives rate of change of velocity; i.e. the acceleration. a(t) =v (t) =r (t). Example: What are the velocity and acceleration functions for an object whose position is given by r(t) =<te t,t 2, tan t>? Example: What are the velocity and position functions for an object whose acceleration is given by a(t) =< 2t, 3t 2, 4t 3 > with v(0) =< 1, 1, 0 >, r(0) =< 0, 0, 10 >? 93

95 In 2D and 3D motion, velocity and acceleration are vector valued functions, having both magnitude and direction. The speed of a moving object is the magnitude of the velocity, denoted v or v. The direction of the velocity vector (which indicates the direction of motion) is the unit tangent vector T(t) = r (t) r (t) = v(t) v(t) and velocity can be expressed as a product of magnitude and direction: ( ) v(t) v(t) = v(t) = v T v(t) Example: What is the speed function for the object whose path is described by r(t) =<te t,t 2, tan t>? Howfastisitmovingatt =1s(r in meters)? Express v(1) in the form v = vt. In 1D motion, an object moving at a constant speed has zero acceleration. Is the same true for 2D or 3D motion? 94

96 Application: Tangential and normal components of acceleration ( Unit tangent T = v v ) ( ) and unit normal N = T T vectors provide information about the direction of motion. We know that velocity is tangential to the curve, and can be expressed as v = vt where v = v We are interested in how acceleration is related to these vectors. Since v = v T and a = v, we obtain by differentiating: d dt v = d dt [ v T] ( ) d = dt [ v ] T + v d dt [T] ( ) d = dt [ v ] T + v T ( ) ( d T ) = dt [ v ] T + v T T ( ) d = dt [ v ] T +( v T T ) T ( ) d a = dt [ v ] T +( v T ) N In this expression, the scalar coefficient of the unit tangent vector is the tangential component of acceleration: a T = d dt [ v ] The scalar coefficient of the unit normal vector is the normal component of acceleration: a N = v T The form of the expression tells us about the acceleration a = a T T + a N N Accleleration lies in a plane containing T and N (the osculating plane) The tangential component a T = dt d [ v ] is the rate of change of the speed. The normal component a N = v T provides information about changes in direction. The easiest way to visualize the effects of a N is in terms of centripetal force: F = ma N. 95

97 Computing a T and a N from their respective definitions is tedious (a T isn t too bad, but computing the vector T to get a N is inconvenient). Consider the relationship a = a T T + a N N Since T and N are orthogonal, we have a right triangle with a 2 =(a T ) 2 +(a N ) 2 We can further consider a T T as the vector projection of a onto T, anda T as the length of that projection. Working through the formulas we derived for the vector projection, we arrive at Using T = a T = proj T a = a T T v v,wehave a T T =proj T a a T = proj T a (or comp T a) = a T (since T =1) a T = a v v We can also consider a N N as the orthogonal projection of a onto T, anda N as the length of that projection. Working through the formulas we derived for the orthogonal projection, we arrive at a N N = orth T a Using T = v v,wehave a N = orth T a = a N = orth T a a T T a N = = a T (since T =1) a v v 96

98 So, after a great deal of explanation, we come up with two simple formulas for a T and a N : a T = a v v a N = a v v Alternately, compute a T and a, andgeta N from a N = a 2 (a T ) 2 Example: at t =1s? What are the tangential and normal components of acceleration for r(t) =<t 2,e t, ln t> m Important footnote to all this... Computing a T and a N turns out to be painless; however these are only the scalar components; i.e. magnitudes. In themselves, they tell us nothing about the direction of the accleration. To examine the directions that these magnitudes belong to, we must go through the process of computing the vectors T and N described in the section on Unit tangent and unit normal vectors. In particular, there is no nice, short way to compute the normal vector N. Once useful thing you can do to visualize tangential and normal components (without doing the vector calculations) is take a look in MVT s TNB Frames application (see the link from the web page). The TNB frames will show you the tangent, normal, and binormal vectors as you move along the curve. Combined with your calculations for v, a T and a N, you can get a feel both both the size AND the direction of the quantities involved. 97

99 Functions of several variables Functions of several variables Functions of several variables, or multivariable functions have the form for example The variables x 1,x 2,..., x n are all independent. f(x) =f(x 1,x 2,..., x n ) f(x 1,x 2,..., x n )=x x2 2 x 3 +sinx 4 For the most part, we ll be looking at functions of two variables, expressed as z = f(x, y) (e.g. z = x 2 y +siny) wherex and y are the independent variables, and z is dependent on x and y. The graph of z = f(x, y) is the set of all ordered triplets that satsify the relation defined by z = f(x, y). These triplets form a surface in space. The domain of a multivariable function f(x) =f(x 1,x 2,..., x n ) is the set of all values x 1,x 2,..., x n for which it is defined. For example, the domain of f(w, x, y, z) = w + x + y + z is the set of all values for w, x, y, and z which do not produce a negative quantity under the square root: {(w, x, y, z) w + x + y + z 0} Examples: Give the domains of f(x, y) =ln(xy) g(x, y, z) = 1 x 2 + y 2 + z 2 h(x, y) = 2 x 4 x2 y 2 98

100 In the case of a function of two variables, the domain will be a region in the xy plane. For example, we found the domain of f(x, y) =ln(xy) tobe{(x, y) xy > 0}. For this to hold, we must have x and y both positive (1st quadrant), or x and y both negative (3rd quadrant). Example: Sketch the domain of h(x, y) = 2 x 4 x2 y 2. The graph of a function of three variables is four dimensional (and tough to sketch). The domains of these functions are expressed in terms of three variables, and can be described as regions of 3D space. We found the domain of g(x, y, z) = 1 x 2 + y 2 + z 2 of space except the origin. to be {(x, y, z) (x, y, z) (0, 0, 0)}. Thisisall Example: Find and describe the domain of f(x, y, z) =ln(z sin(y)). Projections of surfaces onto various planes may not be that informative. Take a moment and sketch the example of the projections of f(x, y) =x 2 + y 2 : 99

101 Projecting a function of two variables onto the xy plane in particular shows the region that is the domain of the function. You can see (sort of, owing to the limitations of graphing software and some bad behavior around asymptotes) that the projection of h(x, y) = 2 x is the circular 4 x2 y2 region you found for the domain in a previous example. Operations on multivariable functions are analogous to operations on single variable functions: *(cf)(x, y) = cf(x, y) (scalar multiple) *(f ± g)(x, y) =f(x, y) ± g(x, y) (sum or difference) *(fg)(x, y) =f(x, y)g(x, y) (product) *( f y) g )(x, y) =f(x, g(x, y) (quotient) Domains of the resulting functions are the intersections of the domains of f and g, with g nonzero in the case of the quotient. We can define the composition of a single variable function with a multivariable function. Suppose g(x) is a function of one variable, and h(x, y) is a function of two variables. The composition (g h)(x, y) =g(h(x, y)) makes sense, and produces a function of two variables f(x, y) =(g h)(x, y) Example: What is the composition of g(x) = 1 x with h(x, y) =x 2 + y 2? 100

102 Functions of several variables Intersections of surfaces and traces Intersections of surfaces are curves in space, and we should be able to write a set of parametric equations that describe this curve. Example: Let z = f 1 (x, y) =8 x 2 y 2 z = f 2 (x, y) =x 2 + y 2 The intersection of these surfaces appears to give a circle. Work through the example screens where we solve for the intersection of the surfaces: Example: Find parametric equations for the intersection of the surfaces z =sin(x) andz = x + y 101

103 The trace of a surface in a plane is the intersection of that surface with the plane. Traces can be expressesed in parametric form (as above), or, more commonly, in Cartesian form (specify the plane that the equation lies in). For example, the trace of the surface z = f(x, y) =x 2 +4y 2 in the plane z = 2 is the ellipse 2=x 2 +4y 2 (the curve lies in the plane z =2). Example: What is the trace of the surface z = x 2 +4y 2 in the plane x =2? Example: What are the traces of the surface z =sinx +cosy in the xz, yz, and xy planes? 102

104 (The algebra on this one is kind of involved, so here s some more room. Just keep working...) The traces of a surface in the xy, xz, and yz planes can be used to visualize the surface. A 3D surface would be hand sketched by sketching its traces. Example: What are the traces in the xz, yz, and xy planes of the surface z =9 x 2 y 2? Use the traces to hand sketch the surface. 103

105 Example: You ve found the traces of the surface z =sinx +cosy in the xz, yz, and xy planes. Can you build up a picture of the surface and sketch it? 104

106 Calculus of multivariable functions Limits, part 1: the intuitive approach In single variable Calculus, we start by introducing the idea of a limit intuitively (through graphs and tables), before turning to the formal defnition. We re going to do the same thing here. We think of lim x a f(x) =L in terms of approaching - as x values get close to a, function values get close to L. And we think of contiunuity in terms of unbroken - can be drawn without lifting your pencil. Recall that most of the functions we work with (polynomial, rational, radical, trigonometric, logarithmic, exponential) are continuous on their domains. And the first technique we learn for evaluating limits is direct substitution- we expect that as long as we re in the domain of the function, that lim x a f(x) =f(a). So, we go through the same process with functions of two variables(or more, but as usual, we ll stick with two because we can graph the surfaces). Keep in mind you can approach a point (a, b) from all directions in the xy plane. Informal definition of limit: We say that the limit of f(x, y) as(x, y) approaches (a, b) isl and write lim f(x, y) =L (x,y) (a,b) if the values of f(x, y) can be made arbitrarily close to L by choosing points (x, y) sufficiently close to (a, b). A function is continuous at (a, b) if and only if lim f(x, y) =f(a, b) (x,y) (a,b) A function is continuous on an open region R if it continuous at every point in R. We expect all the familiar functions to be continuous on their domains, and for compositions g(h(x, y)) of continuous g(x) andh(x, y) to be continuous. If we know that the function we re interested in is continuous on its domain, and (a, b) is in the domain of the function, the limit may be evaluated by direct substitution: Example: Find lim (x,y) (1,2) x 2 y + xy 2. lim f(x, y) =f(a, b) (x,y) (a,b) 105

107 Example: Find lim (x,y) (1,2) x 2 y +2x 3xy +6. Example: Find lim (x,y) ( 1,2) x 2 y +2x 3xy +6. Examples: Where are the functions below continuous? f(x, y) =ln(xy) h(x, y) = 2 x 4 x2 y 2 106

108 Calculus of multivariable functions Limits, part 2: using paths to show a limit does not exist For a single variable function, one way to show a limit does NOT exist is to consider the limit from the left and from the right; if these values do not agree, then the limit does not exist. For example, for the function { x 2 x<1 f(x) = x +3 x 1 we know that lim x 1 f(x) does not exist, because and lim x 1 lim f(x) = lim x 1 x 1 x2 =1 f(x) = lim + +(x +3)=4 x 1 When we do this, we are taking two different paths to the value a = 1, from the left and from the right. We can extend this approach to functions of two variables - if we are interested in lim f(x, y) (x,y) (a,b) we can look at that limit by approaching (a, b) along various paths in the (x, y) plane. Unlike the single variable case, though, there are an infinite number of paths passing through a given (a, b), so we can never show a limit exists by considering paths - there s no way to consider them all! If we suspect the limit doesn t exist, though, we only need to find two that disagree. Example: What is lim 2x 2 3y 2 (x,y) (0,0) x 2 + y 2 along the path y = 0 (i.e., the x-axis)? Note that limiting values to this path puts you on a curve that is the trace of the function in the xz plane. To solve, simply substitute y = 0 in the expression, and find the limit. 107

109 2x Example: What is lim 2 3y 2 (x,y) (0,0) x 2 + y 2 along the path x = 0 (i.e., the y-axis)? Note that limiting values to this path puts you on a curve that is the trace of the function in the yz plane. To solve, simply substitute x = 0 in the expression, and find the limit. Example: What does that tell you about lim (x,y) (0,0) 2x 2 3y 2 x 2 + y 2 overall? Example: Show that lim (x,y) (0,0) xy x 2 + y 2 does not exist by considering paths. 108

110 Instead of looking at specific paths one at a time (y =0,x =0,y = x, y =2x, andsoon,ifwe re considering a limit going to the origin), you can shorten things by checking all paths of a certain type at once. For example, all lines through the origin (except x =0)havetheformy = kx. Making the substitution y = kx and examining the limit as x 0 will give us an expression that mayormaynotdependonthevalueofk. If it depends on k, we know the limit does not exist. Example: Find lim (x,y) (0,0) x 2 x 2 + y 2 along the paths y = kx. What can you conclude? Example: Find lim (x,y) (0,0) xy 2 x 2 + y 4 along the paths y = kx. What can you conclude? 109

111 Calculus of multivariable functions Limits, part 3: the delta-epsilon definition We have the informal definition of a limit: We say that the limit of f(x, y) as(x, y) approaches (a, b) isl and write lim f(x, y) =L (x,y) (a,b) if the values of f(x, y) can be made arbitrarily close to L by choosing points (x, y) sufficiently close to (a, b). In this section, we are going to look at the formal definition of a limit, and define exactly what we mean by sufficiently close to. We start by defining a δ-neighborhood about a point (a, b) in the xy plane: a δ-neighborhood about (a, b) is a disk centered at (a, b) with radius δ>0. We can can have open or closed δ- neighborhoods, depending on whether we include the boundary of the disk. Open disk: {(x, y) (x a) 2 +(y b) 2 <δ} Closed disk: {(x, y) (x a) 2 +(y b) 2 δ} Recall the definition of the limit for a single variable function: Let f be a function defined on an open interval containing a (except possibly at a), and let L be a real number. The statement lim f(x) =L x a means that given any ɛ>0, there exists a δ>0 such that if 0 < x a <δ, then f(x) L <ɛ We can extend this to the formal δ-ɛ defintion of the limit for a function of two variables: Let f be a function of two variables defined on an open disk centered at (a, b) (except, possibly, at (a, b) itself),andletl be a real number. We say that the limit of f(x, y) as (x, y) approaches (a, b) isl and write lim f(x, y) =L (x,y) (a,b) if, given any ɛ>0, there exists δ>0 such that if 0 < (x a) 2 +(y b) 2 <δ then f(x, y) L <ɛ Examples are posted as separate.pdf s. 110

112 General notation: We ve been looking at these concepts in terms of two variable functions, with the note that they can be extended to functions of three or more variables by analogy. Since we re being formal here, I d like to provide the general definition of the limit for a function of n variables using proper notation. The easiest way to do this is to make use of vectors (or ordered n-tuples). Let x and a be vectors in R n : x =< x 1,x 2,..., x n > We define an open δ-neighborhood about a by And the limit a =< a 1,a 2,..., a n > {x x a <δ} Let f(x) =f(< x 1,x 2,..., x n >) be a function of n variables defined on an open interval containing a =< a 1,a 2,..., a n > (except possibly at a), and let L be a real number. The statement lim f(x) =L x a means that given any ɛ>0, there exists a δ>0 such that if 0 < x a <δ, then f(x) L <ɛ Finally, recall discussing continuity on an open region. Now that we ve defined open negihborhood, we can formally define open region. Apoint(a, b) in a region R is an interior point if there exists a δ-neighborhood about (a, b) that lies entirely within R. A point (a, b) in a region R is a boundary point if every δ-neighborhood about (a, b) contains points both inside and outside R. If every point in R is an interior point, it s an open region (it doesn t contain its own boundary). If R contains all its boundary points, it s a closed region. 111

113 Calculus of multivariable functions Limits, part 4: using the squeeze theorem to prove the limit exists Squeeze Theorem: If in a neighborhood of (a, b), then g(x, y) f(x, y) h(x, y) (assuming these limits exist). lim g(x, y) lim (x,y) (a,b) f(x, y) lim (x,y) (a,b) h(x, y) (x,y) (a,b) If lim (x,y) (a,b) g(x, y) and lim (x,y) (a,b) h(x, y) existandareequal, then lim (x,y) (a,b) f(x, y) exists, and lim f(x, y) = lim (x,y) (a,b) g(x, y) = lim (x,y) (a,b) h(x, y) (x,y) (a,b) Example: Use the squeeze theorem to find 3x 2 y 2 lim (x,y) (0,0) x 2 + y 2 * First, check some preliminaries - what does direct substitution give you? Can you reduce the expression? What does the graph look like? * Now, set up the squeeze: * Next, pass the limit through: 112

114 * And conclude: Example: Use the squeeze theorem to find y 4 sin 2 y lim (x,y) (0,0) x 2 + y 2 113

115 Hint: for functions which are not nonnegative for all (x, y), try squeezing the absolute value of the function. This works when you suspect the limit is 0. The absolute value gives the distance from the function to 0, and if that turns out to be 0, you know the function itself must be approaching 0. Example: Use the squeeze theorem to show the limit is zero: lim (x,y) (0,0) 2xy 2 x 4 + y 2 114

116 Calculus of multivariable functions Limits, part 5: general strategy In the past few sections, we ve illustrated various techniques for showing whether a limit does/does not exist. It s up to you to decide what technique to use - a typical limit problem reads Find the limit, if it exists, or show that it doesn t. Strategy. Which way you proceed depends on your intuition about whether the limit exists or not. * Look at domain, try direct substitution. You may get a conclusive answer and not have to go through the rest of the process! * If you get an indeterminate form (usually 0 0 ), see if there s any algebra you can do to reduce. * Now, if that hasn t gotten you anywhere, you need to decide. Exists, or doesn t exist. If at all possible, GRAPH! Inspection will tell you if anything weird is going on. * If you think the limit doesn t exist, prove it by trying paths. * If you think the limit does exist, prove it by trying to squeeze. The worst part of doing limit problems is deciding which way to go... and that s been simplified considerably with graphing software. If you aren t in a position to graph and inspect, the usual approach is to try paths first, and if all the paths seem to be heading to the same place, change tactics and try for a squeeze. Example 1: Find the limit, if it exists, or show that it doesn t: lim (x,y) (0,0) 6x 2 y 2 x 4 + y 4 * Substitute: * Algebra: 115

117 * Graph: * Paths: Example 2: Find the limit, if it exists, or show that it doesn t: lim (x,y) (0,0) x 3 y 2 x 4 + y 2 * Substitute: * Algebra: * Graph: 116

118 * Squeeze: Now that we have a stragtegy for examining limits, we can discuss the continuity of piecewise functions, suchas 6x 2 y 2 f(x, y) = x 4 + y 4 (x, y) = (0, 0) 1 (x, y) =(0, 0) When considering piecewise functions, we can apply the usual rules on the open intervals on which the pieces are defined. The main question is What happens at the spots the pieces switch? (In this case, what happens at (0, 0)?) Example: Discuss the continuity of 6x 2 y 2 f(x, y) = x 4 + y 4 (x, y) = (0, 0) 1 (x, y) =(0, 0) 117

119 Calculus of multivariable functions Partial differentiation We define partial derivatives of functions of two or more variables by considering what happens if we hold one of the variables constant. Given a point (a, b), and a function f(x, y), we define the partial derivative of f with respect to x at the point (a, b) by f x (a, b) = lim h 0 f(a + h, b) f(a, b) h Similarly, we define the partial derivative of f with respect to y at the point (a, b) by The definition of f x (a, b) isequivalentto f y (a, b) = lim k 0 f(a, b + k) f(a, b) k Hold y = b constant in the function f(x, y); i.e., intersect the function with the plane y = b. f(x, y) becomes f(x, b) =g(x), a curve in that plane. Now, differentiate with respect to x, and evaluate at a. f x (a, b) =g (a) - the slope of the tangent to the curve in that plane. The definition of f y (a, b) isequivalentto Hold x = a constant in the function f(x, y); i.e., intersect the function with the plane x = a. f(x, y) becomes f(a, y) =h(y), a curve in that plane. Now, differentiate with respect to y, and evaluate at b. f y (a, b) =h (b) - the slope of the tangent to the curve in that plane. 118

120 We get from the partial derivative at a point to the partial derivative function in exactly the same way as the one variable case- within a fixed plane (say y = b) the values of the slope vary as we move along the trace. So, we could talk about f x (x, b) as a function of x. Then, if we allow the planes y = b to vary as well, the slope will vary depending on which plane we re in. So we can define f x (x, y) as a function of two variables (and the same reasoning holds for f y (x, y)): f x (x, y) = lim h 0 f(x + h, y) f(x, y) h f y (x, y) = lim k 0 f(x, y + k) f(x, y) k Since the partial derivative is defined by holding one variable constant and differentiating with respect to the other value, it follows that all the differentiation rules for functions of one variable apply here. Before working examples, it will help to introduce the partial differentation operators, x and y. These act in the same way that d dx acts on a single variable function; they indicate differentiate this (with respect to x or with respect to y, as indicated). So, f x (x, y) =f x = f x = f(x, y) x f y (x, y) =f y = f y = f(x, y) y Example: Find f x (x, y) andf y (x, y) forf(x, y) =x 4 + x 2 y 2 y sin x. 119

121 Example: Find f x (x, y) andf y (x, y) forf(x, y) = x + ey x y. Example: Find f x (x, y) andf y (x, y) forf(x, y) =ln(2x +3y + xy). 120

122 To find a partial derivative at a point (a, b), differentiate and evaluate at (a, b). And keep in mind what that gets you- f x (a, b) gives the slope of the line tangent to the surface at (a, b), in the plane y = b; f y (a, b) gives the slope of the line tangent to the surface at (a, b), in the plane x = a. Example: Find f x (1, 2) and f y (1, 2) for f(x, y) =x y. Hint: think about this one a bit - there are two different differentiation rules applying, depending on who is constant. If you re holding y constant, the appropriate rule is the power rule (x n ). If you re holding x constant, the appropriate rule is the exponential rule (a x,orinthiscase,a y ). We can define partial derivatives for functions of more than two variables- differentiate with respect to one variable while holding the others constant. For example, for a function f(x, y, z), f z (x, y, z) = lim h 0 f(x, y, z + h) f(x, y, z) h (hold x and y constant while differentiating with respect to x). Example: Find f x (x, y, z), f y (x, y, z), and f z (x, y, z) forf(x, y, z) =e 2x+y+z2. 121

123 We can define higher order derivatives. There are four second partial derivatives: * f xx =(f x ) x : Take the function f x and differentiate it with respect to x. * f xy =(f x ) y : Take the function f x and differentiate it with respect to y. * f yx =(f y ) x : Take the function f y and differentiate it with respect to x. * f yy =(f y ) y : Take the function f y and differentiate it with respect to y. f xy and f yx are referred to as mixed partial derivatives. There s a posted live example for higher order derivatives. Notation note: if you re using the differentiation operators, pay attention to the difference in ordering of the x s and y s (you ll notice it in the mixed partials): * f xx = x (f x)= x ( ) f x * f xy = y (f x)= y ( ) f x * f yx = x (f y)= x ( ) f y ) * f yy = y (f y)= y ( f y = 2 f x 2 = 2 f y x = 2 f x y = 2 f y 2 Clairaut s Theorem (equality of mixed partials): Suppose f is defined on a disk D that contains the point (a, b). If the functions f xy and f yx are continuous at (a, b), then f xy (a, b) =f yx (a, b) Higher higher order partial deriviatives, e.g f xxx or f xyx defined analogously. (Example posted) 122

124 Calculus of multivariable functions Differentials Review of differentials for single variable functions The idea behind differentials goes back to the secant line vs. tangent line problem. Initially, we noted that slope of a secant line between two points (a, f(a)) and (b, f(b)) on a curve is given by m sec = f(a + h) f(a) h = f(b) f(a) b a and the slope of a line tangent to the curve at (a, f(a)) is given by m tan = lim h 0 f(a + h) f(a) h = f (a) One of the first explorations of differentiation (before learning the various rules), was to compute the slopes of various secant lines, and notice that as Δx = h grew smaller and smaller, the slope values approached the value of the slope of the tangent. In other words, For small values of Δx = h, the slope of the secant line between (a, f(a)) and (b, f(b)) provides a decent approximation to the slope of the tangent line at (a, f(a)). Now that we ve got some fairly easy ways to compute f (x) for any function f(x), we can turn this idea around: For small values of Δx = h, the slope of the tangent line at (a, f(a)) provides a decent approximation to the slope of the secant line between (a, f(a)) and (b, f(b)). In addition, A line tangent to a curve provides a decent approximation to the curve itself near the point of tangency. 123

125 In functions of a single variable, we manipulate the notation to obtain dy dx = f (x) dy = f (x)dx The quantities dy and dx on their own are strange little things they are infinitesimally small quantities called differentials. The most useful way to interpret them in the current context is that they represent the change in y and the change in x along the tangent line of the function. If we let dx =Δx (a change in the horizontal displacement), from a fixed point x = a, weget Δy dy = f (x)dx Δx represents a change in x (usually relative to some fixed point x = a. Δy represents the corresponding change in the function value: Δy = f(a +Δx) f(a). Δx and Δy are called increments of x and y. dy dx gives the slope of the tangent (again, generally at a fixed x = a). The individual symbols dx and dy are called differentials, and are related by dy = f (x)dx (or dy = f (a)dx at x = a). When we let dx =Δx, thenδy dy. 124

126 The chain rule The chain rule - part 1 The chain rule for functions of one variable describes the relationship between the derivatives of functions in a composition. Suppose y = f(x) and x = g(t). Then y(g(t)) is a function of t, and Example: If y = x 2 and x = sint, what is dy dt? y (t) = f (g(t))g (t) dy dt = dy dx dx dt In the previous example, we could also have composed before differentiating: so f(t) = f(x(t)) = f(sint) = (sint) 2 f (t) = 2(sin t) 1 (cos t) = 2sin t cost That s still using the chain rule, just without explicitly writing out the steps (you re using the chain rule every time you differentiate a composition and think derivative of the outside times derivative of the inside ). There are several versions of the chain rule for multivariable functions. To figure out a chain rule for a function that makes sense for that function, you need to identify the independent variables (which are they and how many are there?) the intermediate variables (which are they and how many are there?) the dependent variable (ultimately, what are you differentiating?) 125

127 Case 1: f = f(x, y), x = x(t) and y = y(t) This is best explained by example... Example: Suppose f(x, y) = x 2 y 3 Here, x and y appear to be independent variables, and z = f(x, y) is the dependent variable. Because x and y are independent of each other, it makes sense to talk about the rate of change of z with respect to x, or z with respect to y. These give us the partial derivatives f x (x, y) = f x Find z x = f x(x, y) and z y = f y(x, y): or z x, and f y(x, y) = f y or z y Now, what if x and y themselves are dependent on another variable, say t? Let x = x(t) = sint y = y(t) = lnt Here, t is the independent variable. Since both x and y are functions of a single variable, we can differentiate them with respect to t (notice these aren t partial derivatives!): Find dx dt and dy dt : All together, we have z = f(x, y) = x 2 y 3, x = x(t) = sint, and y = y(t) = lnt. Perform the composition z(t) = f(x(t), y(t)): 126

128 How many variables is z ultimately a function of, and what derivative makes sense? Find that derivative: Now, see how all the pieces fit together. We have x = sin t and y = lnt dx = cost and dy dt dt = 1 t z x = 2x and z y = 3y2 dz dt = 2(sint)(cost) 3(lnt)2 ( 1 t ) dz dt = 2x dx dt 3y2 dy dt The chain rule for is given by z = f = f(x, y), dz dt = z dx x dt + z y dz dt = z dx x dt + z y dy dt x = x(t) and y = y(t) (or df dt = f x dx dt + f y dy, depending on how you prefer to notate). dt z or f is the dependent variable, t is the independent variable, and x and y are the intermediate variables. z is a multivariable function of its intermediate variables, and ultimately a single variable function of the independent variable. dy dt A proof of this chain rule is attached. 127

129 Example: Use the chain rule to find dz dt for z = f(x, y) = 4exy, with x = 5t 3 and y = e t. We can generalize this case of the chain rule to multivariable functions of n variables, where the n variables are still all functions of a single variable (say t), and ultimately, the function depends on t: f = f(x 1, x 2,..., x n ) with x 1 = x 1 (t), x 2 = x 2 (t),..., x n = x n (t) df dt = f dx 1 x 1 dt + f dx 2 f dx n x 2 dt x n dt Example: Find dw dt for w = f(x, y, z) = x2 yz + yz 3, x = e t, y = 2t, z = cos t. 128

130 The chain rule Implicit differentiation The chain rule can be used for differentiating implicitly. Before we go there, take a look at Case 1a z = f = f(x, y), y = y(x) This is a variation on case 1: Suppose you have a function of two variables, z = f(x, y). Now, instead of x and y both dependent on another parameter t, suppose that x is independent, but y is dependent on x. Soz = f(x, y) is really z = f(x, y(x)), and is at the bottom level a function of one variable x. It now makes sense to talk about x z and z, but also to talk about dz y dx. Applying the chain rule as in case 1: dz dx = z x dx dx + z y dy dx dz dx = z x + z y dy dx You ll also frequently see it written in the following way: f = f x + f y y One context where this occurs is in the study of differential equations in the form y = f(x, y) such asy = x sin y or y = x 2 + y 2 where y (implicitly) depends on x. Given this kind of setup for y, it s possible to find y and higher derivatives; note y = f(x, y) y = f (x, y) =f x + f y y so y = f x + f y f Example: For y = x sin y, compute y using y = f x + f y f. 129

131 Implicit differentiation We could have accomplished the same thing using implicit differentiation, but this can be a good bit easier for messy expressions. In fact, the chain rule is going to give us a nice alternative to implicit differentiation. The first thing you want to do here is make sure you recall how to differntiate implicitly. If we have an equation in x and y, wherey is understood to be an implicit function of x, we use the chain rule (and product and quotient rules as needed) to find dx dy. Example: Use implicit differentiation to find dy dx for x2 x sin y = y 3. The chain rule with partial derivatives gives us an alternate way to find dx dy. Suppose you have an expression F (x, y) =0 where x is independent, and y is dependent on x, as in case 1a. Differentiate both sides, using the chain rule to find df dx : And solve for dy dx : d d F (x, y) = dx dx 0 F x + F dy y dx =0 F x + F dy y dx =0 F dy y dx = F x F dy dx = x = F y F x F y So, a convenient formula for finding dx dy,wheny is implicitly defined as a function of x, is dy dx = F x F y when the expression relating x and y is in the form F (x, y) =0 130

132 Example: Use the above formula to find dy dx for x2 x sin y = y 3. dx dy = F x F y F y =0. isn t going to work under all conditions; the obvious one is that we have a problem if The Implicit Function Theorem gives the hypotheses under which this formula holds: If F (x, y) is defined on a disk containing (a, b), F (a, b) =0,F x and F y are continuous on the disk, and F y (a, b) 0,thenF (x, y) = 0 implicitly defines y as a function of x near (a, b), and the differentiation formula holds. 1 A similar derivation gives us some formulas for implicit partial differentiation. Suppose we have an expression relating x, y, andz, wherez is implicitly a multivariable function of x and y. Ifwe rearrange the expression into the form F (x, y, z) =0,wehave z x = F x F z and z y = F y F z Example: Suppose z is implicitly a function of x and y, andxyz 3x 2 =sinz. Find z x and z y A proof of the Implicit Function Theorem is beyond the scope of this class; see for example Rudin, Principles of Mathematical Analysis, McGraw-Hill.

133 The chain rule The chain rule, part 2 We can construct more cases of the chain rule for functions of more than one variable that are of functions of more than one variable, say f = f(x, y), x = x(s, t), y = y(s, t). If we substitute in, we have that f = f(x(s, t),y(s, t); i.e. f is still a function of two variables, s and t. In this case, we would call s and t the independent variables, x and y the intermediate variables, and f (or z) the dependent variable. Instead of memorizing a bunch of variations, the trick is to ask yourself for any given problem, what derivatives make sense? Drawing a chain rule diagram can be helpful. Example: Let z = f(x, y) =x 2 + y 3, x =sin(t + s), y =cos(t s) What is z ultimately a function of? What derivatives of z does it now make sense to compute? What do you think the correct chain rules would be? Make a diagram. Compute the partials with respect to s and t of z: 132

134 Example: Let w = f(x, y, z) =x 2 + xyz, x =sin(t + s), y =2t, z =3s What is w ultimately a function of? What derivatives of w does it now make sense to compute? What do you think the correct chain rules would be? Make a diagram. Compute the partials with respect to s and t of w 133

135 Applications of partial differentiation Tangent lines in the planes x = a and y = b All of this bit is really just piecing together stuff you already know. Recall... f x (a, b) gives the slope of the line tangent to the graph of f at (a, b, f(a, b)) in the plane y = b. This line is in the direction of (parallel to) the x axis. f y (a, b) gives the slope of the line tangent to the graph of f at (a, b, f(a, b)) in the plane x = a. This line is in the direction of (parallel to) the y axis. To write the equation of a line in 3D space, we need a point on the line (which we have; it s (a, b, f(a, b))), and a vector which is parallel to the line. The question is, how to convert the values of f x and f y (slopes) into vectors? In 2D, we can go from slope to vector easily, by considering slope as a ratio of m 1, sketching a triangle with those proportions, and writing as a vector. A line with slope m has direction vector < 1,m>. We can still do this in the particular 3D cases under consideration, because in each case, we are holding one dimension constant in a plane. 134

136 The slope f x (a, b) is the slope of the trace in the plane y = b. Expressed in 3D, the rise/run are for z and x, while y is held constant. A direction vector for a tangent line in the plane y = b with this slope is given by < 1, 0,f x (a, b) > The slope f y (a, b) is the slope of the trace in the plane x = a. Expressed in 3D, the rise/run are for z and y, while x is held constant. A direction vector for a tangent line in the plane x = a with this slope is given by < 0, 1,f y (a, b) > Remeber that the point-parallel form of a line is given by <x,y,z >=< x 0,y 0,z 0 > +t <v 1,v 2,v 3 > We now have sufficient information to write the equations of tangent lines passing through (a, b) and parallel to the x and y axes. Point: <x 0,y 0,z 0 > = <a,b,f(a, b) > Vector:either<v 1,v 2,v 3 > = < 1, 0,f x (a, b) > (parallel to x axis) or <v 1,v 2,v 3 > = < 0, 1,f y (a, b) > (parallel to y axis). 135

137 Example: Write the equation of the line which is parallel to the x axis and tangent to the surface f(x, y) =x 2 3y + e y at the point (2, 0,f(2, 0)). Example: Write the equation of the line which is parallel to the y axis and tangent to the surface f(x, y) =x 2 3y + e y at the point (2, 0,f(2, 0)). 136

138 Applications of partial differentiation Tangent plane and normal line We have seen that that vectors < 1, 0,f x (a, b) > and < 0, 1,f y (a, b) > are tangential to the surface f(x, y) at(a, b, f(a, b)) in the direction of the x and y axes, respectively. The plane which passes through the point (a, b) and contains these vectors is the tangent plane to the function f(x, y) atthepoint(a, b). (It also contains all other vectors/lines which are tangent to the plane at this point, but we only need two vectors to define the plane). It is simple to derive the equation for the tangent plane at (a, b, f(a, b)): We ve got a point, we need a normal, and the normal is obtained by crossing the vectors which lie in the plane. Do it. < 1, 0,f x > < 0, 1,f y >=? And now, use the point-normal equation for a plane to write the equation of the tangent plane: 137

139 Example: Write the equation of the plane tangent to the surface f(x, y) =x 2 3y + e y at the point (2, 0,f(2, 0)). While we have a normal vector lying around, we might as well write the equation of the normal line passing through the surface at (a, b, f(a, b)). This line has the vector we just found as its direction vector, and its equation is: <x,y,z >=<a,b,f(a, b) > +t < f x (a, b), f y (a, b), 1 > or <x,y,z >=<a,b,f(a, b) > +t <f x (a, b),f y (a, b), 1 > 138

140 Example: Write the equation of the plane tangent to the surface and the normal line passing through the surface for z = 4 x 2 2y 2 at the point (1, 1, 1). The tangent plane is a linear function. It is the linear function that most closely resembles f(x, y) at the point (a, b). At that point, the plane and the function have the same (1) function value, and (2) partial derivatives with respect to x and y. As we move away from the point (a, b), the plane and the function no longer coincide, but in a small neighborhood around (a, b), the tangent plane provides a good approximation to the function. For these reasons, the equation of the tangent plane is also called the linearization of f(x, y) at(a, b), or the linear approximation to f(x, y) at (a, b). We can then say: * z = f(a, b)+f x (a, b)(x a)+f y (a, b)(y b) (equation of tangent plane) * L(x, y) =f(a, b)+f x (a, b)(x a)+f y (a, b)(y b) (linearization) * f(x, y) f(a, b)+f x (a, b)(x a)+f y (a, b)(y b) (linear approximation) They all say the same thing, and are interchangeable. Example: Find the linearization of f(x, y) =xe 2y +siny at the point (0, 0, 0). 139

141 Example: Use the linear approximation of f(x, y) =xe 2y +siny at the point (0, 0, 0) to compute the value of f(.05,.03). Compare the approximate value to the exact 1 value. This would be the point of having linear approximations - I have a value that estimates the function value... and required no calculation other than adding two numbers. We have seen the tangent plane before in the context of differentials (so now, we ve derived where that equation came from). Go back and look at that again. And note, when we say that the tangent plane provides an approximation to the function at (a, b), we are assuming that the function is differentiable (as defined in that section) at (a, b) - otherwise, all bets are off. 1 exact here would mean however many digits your calculator gives you 140

142 Applications of partial differentiation Directional derivatives Recall that we can specify direction (in the plane or in space) in terms of a unit vector. In the xy plane, in the direction of the x axis could be expressed as in the direction of the vector i =< 1, 0 >, and in the direction of the y axis could be expressed as in the direction of the vector j =< 0, 1 > When we compute the partial derivative f x for f(x, y), we get a function that, when evaluated at a fixed point, gives us the rate of change in f as we move parallel to the x-axis (i.e. the slope of the tangent line along the surface parallel to the x-axis). We can think of this as the rate of change in the direction of the vector < 1, 0 >. An equivalent way to write the formula for f x would be f x (x, y) = lim h 0 f(< x,y>+h <1, 0 >) f(< x,y>) h Similarly, f y gives us the rate of change in the direction of < 0, 1 > (parallel to the y axis): f y (x, y) = lim h 0 f(< x,y>+h <0, 1 >) f(< x,y>) h Suppose we wish to consider the rate of change in the function f along some other unit vector u =< u 1,u 2 >. Geometrically, for f = f(x, y), you can think of this as asking how does the function change as we walk out on a line which is not necessarily parallel to the x-axis or y-axis, and whose direction is given by a unit vector u. We refer to this quantity as the derivative of f in the direction of u, or the directional derivative of f along u, and define it by D u f = lim h 0 f(x + hu 1,y+ hu 2 ) f(x, y) h or f(< x,y>+hu) f(< x,y>) lim h 0 h (As written, this expression is the vector valued function; we could then evaluate the derivative at a specific point.) 141

143 As you d expect, we don t generally want to calculate the directional derivative from the definition. It turns out that it can be computed very conveniently: D u f =< f x,f y > u = f x (a, b)u 1 + f y (a, b)u 2 (atthepoint(a, b, f(a, b))) = f x (x, y)u 1 + f y (x, y)u 2 (as a function of x and y) (derivation attached) If the specified direction vector is not a unit vector, be sure to normalize it. Example: What is the directional derivative of the function f(x, y) =x 2 y 3xy 2 +5 in the direction of v =< 1, 2 > at the point ( 1, 3,f( 1, 3))? Compute f x and f y and evaluate at ( 1, 3) (express as a vector <f x,f y >): Normalize v =< 1, 2 > to obtain a unit vector u = v v : And compute D u f =< f x,f y > u: What you just found: the slope of the line tangent to the surface f(x, y) atthepoint(a, b), in the direction of the vector u. 142

144 A vector in the direction of this line is <u 1,u 2,D u f>. This allows us to write the equation of the tangent line using the point and direction vector. <x,y,z >=<a,b,f(a, b) > +t <u 1,u 2,D u f> Example: For the function f(x, y) =x 2 y 3xy 2 + 5, write the equation of the line tangent to the surface at ( 1, 3,f( 1, 3)) in the direction of v =< 1, 2 >. We can also specify direction as an angle θ, measured from the positive x axis. In this case, the unit vector u is obtained from u =< cos θ, sin θ> 143

145 Example: Find the directional derivative of f(x, y) =ln(x 2 + y 2 ) at the point (3, 4,f(3, 4)), in the direction indicated by angle θ = π 3. Write the equation of the line tangent to the curve at that point and in that direction. 144

146 Applications of partial differentiation The gradient vector Recall that the directional derivative D u f for a function f(x, y) iscalculatedby D u f =< f x,f y > <u 1,u 2 > The quantity <f x,f y > (a vector formed from the first partial derivatives of f) gets its own notation (among other things, it ll make writing formulas for functions of more than two variables a bit more compact). We call it the gradient vector. The gradient vector of a function f(x, y) is denoted by f, and is defined f =< f x,f y > Depending on the context, this can be a vector valued function: or, evaluated at a point (a, b), a single vector: f(x, y) =<f x (x, y),f y (x, y) > f(a, b) =<f x (a, b),f y (a, b) > For a function of three variables, f(x, y, z), we have and so on. f =< f x,f y,f z > Example: For f(x, y) =x 3 e y +2y 2, find f(x, y) and f(1, 0). What does f give us? First off, another way to compactly express the formula for calculating D u f: D u f = f u But there are some additional properties associated with the gradient vector... Note that the gradient vector (at a point) is a 2D vector associated with a 3D surface. We can move it around in space as needed, but it s convenient to visualize it as a direction vector in the xy plane, or in a plane parallel to the xy plane, in the same way that we think of u when looking at directional derivatives. 145

147 Consider the following: For any direction u, the derivative in that direction is calculated by D u f = f u By the definition of the dot product, this becomes D u f = f u cos θ where θ is the angle between f and u. Since u is by definition a unit vector, it has magnitude 1, and D u f = f cos θ *Whatθwould give the maximum value for D u f? * What does this mean? The gradient vector is associated with the maximum increase in the surface: * v = f gives the direction of maximum increase (where the surface has the steepest positive slope). * u = f is the unit vector in that direction. f * The slope in this direction is given by D u f = f u = f D u f = f f f = f 2 f The value of the directional derivative in the direction of the gradient vector is the 146 same as the norm of the gradient vector.

148 * Looking in the opposite direction, letting u = f gives the direction of maximum decrease. * And the slope in that direction is D u f = f Example: In what direction does the function f(x, y) =x 2 y sin(2y) + 5 increase most rapidly, from the point (1,π,f(1,π))? Decrease most rapidly from that point? What are the rates of increase and decrease (slopes)? Recall that the level curves of a function z = f(x, y) are obtained by setting f(x, y) =k, k constant. They represent the traces of the surface in the planes z = k. As such, they can be thought of as curves along with the rate of increase of the function is 0. How do gradient vectors relate to level curves? Derive it: * For a expression f(x, y) =k, whatis dx dy? Think in terms of implicit differentiation using partials of F (x, y) =f(x, y) k. * So, at a fixed point (a, b), what is the slope of the tangent? * Translate that slope into a tangent vector: 147

149 * What is the gradient of f at (a, b)? * What do you notice about how they are related? (Try dotting them.) * And conclude: Example: For the function z = f(x, y) =x 2 +4y 2, sketch the level curve corresponding to k =4,and find (and sketch) normal vectors (to the level curve) at various points along the curve. 148

150 Applications of partial differentiation Notation for higher dimensions We generally work with f(x, y) as the basic example for defining concepts such as derivatives, partial derivatives, the gradient, an so on - easy to write,, easy to visualize. It s easiest to extend these concepts by analogy - if I say that we define f(x, y) as f(x, y) =<f x (x, y),f y (x, y) > and that analogously for a function f(x, y, z), we have f(x, y, z) =<f x (x, y, z),f y (x, y, z),f z (x, y, z) > you ll be able to tell me immediately that for a function of four variables, f(w, x, y, z), f(w, x, y, z) =<f w (w, x, y, z),f x (w, x, y, z),f y (w, x, y, z),f z (w, x, y, z) > The concept here is I make a vector out of the partial derivatives, and you can figure out how that works no matter how many variables we ve got. The only drawback is that the notation isn t very general - I d write yet another formula for a function of five variables (and eventually I d start to run out of letters!). First off, if I m going to be working with a bunch of variables (or a non-specific number of variables), it s better to work with subscripts rather than working through the alphabet - (x, y) becomes (x 1,x 2 ), (x, y, z) becomes (x 1,x 2,x 3 ). A function such as becomes f(x, y, z) =sin(xy) z 2 f(x 1,x 2,x 3 )=sin(x 1 x 2 ) x 2 3 This allows me to talk about a function of n variables in general as f(x 1,x 2,..., x n ) Let s look at this in the context of partial derivatives: if I were talking about a function f(x, y, z), I d say that to get the partial with respect to x, holdy and z constant, and allow x to vary. To get the partial with respect to y, holdx and y constant, and allow y to vary, and so on. Under the subscripted notation, f(x 1,x 2,x 3 ), I d say to get the partial with respect to the first variable, x 1,holdx 2 and x 3 constant, and allow x 1 to vary, and so on. Doesn t seem like much of an improvement... but it allows me to describe a partial derivative easily for a function of n variables in a general sense: For a function f(x 1,x 2,..., x n ), to obtain the partial derivative with respect to the ith variable, x i, hold all the other variables constant and allow x i to vary. Note we still aren t up to the notation, but we re getting there! 149

151 We need vectors - there s not a lot of distinction between a point (x 1,x 2,..., x n ) and the vector <x 1,x 2,..., x n > pointing to that point. Using vector notation, I can say let x =< x 1,x 2,..., x n >, and start referring to the function f(x 1,x 2,..., x n ) as simply Things suddenly got a lot more compact. I could denote the partial derivative of f with respect to (for example) the second variable as f f x2 (x) or f(x) or x 2 x 2 and the partial with respect to the i th variable in general as f f xi (x) or f(x) or x i x i f(x) Example: Let x =< x 1,x 2,...x 7 >, f(x) =x 1 x 5 +4x 3 cos(x 7 ). What are x 7 f(x) and x 2 f(x)? We still don t have quite enough notation to write the definitions of these things. Just one more thing though, and we ll be ready. We have seen i and j used to denote the vectors < 1, 0 > and < 0, 1 > in R 2, i =< 1, 0, 0 >, j =< 0, 1, 0 > and k =< 0, 0, 1 > in R 3. We re running into the same issue as with the components in higher dimensions - we re going to run out of letters, so we turn to subscripting: Let the vectors e i represent the elementary basis vectors for R n e 1 =< 1, 0, 0,..., 0 > e 2 =< 0, 1, 0,..., 0 > e n =< 0, 0, 0,..., 1 > e.g in R 3,wehavei = e 1, j = e 2, k = e 3. The point is we can have as many of these as we need for whatever dimension n we re working in. We have a subscripted list of vectors... whose individual components are subscripted. To refer to the third component of the second basis vector, I d use (e 2 )

152 The basis vectors can be described as vectors such that (e i ) i = 1 (the ith component of the ith vector is one), and (e i ) j =0,j i (all the other components are zero). We now have the pieces in place to write the definitions of partial derivatives, the gradient, and directional derivatives in any number of variables in a compact form. Partial derivatives: For we let f = f(x 1,x 2,..., x n ) x =< x 1,x 2,..., x n > We can then write, using vector notation f xi = lim h 0 f(x + he i ) f(x) h Gradient vector: For we let f = f(x 1,x 2,..., x n ) x =< x 1,x 2,..., x n > We can then write f(x) =<f x1,f x2,..., f xn > Directional derivative: For f = f(x 1,x 2,..., x n ), u =< u 1,u 2,..., u n > we let x =< x 1,x 2,..., x n > We can then write D u f = lim h 0 f(x + hu) f(x) h And we would compute the directional derivative using D u f = f u The additional notation is for the purpose of having a good formal definition of all these quantities in a general sense. From a practical standpoint, there s nothing new here - keep doing everything exactly the way you d expect to do it! * To differentiate with respect to a variable, hold all the other variables constant. * To write the gradient vector, form a vector from the partial derivatives. * To find a directional derivative, dot the gradient with the unit direction vector. 151

153 Example: Find the directional derivative of f(x, y, z) =xyz 3 at the point (1, 2, 1,f(1, 2, 1)) in the direction of u =< 3 29, 2 29, 4 29 >. 152

154 Optimization Maxima and minima of functions of one variable (a review) Start by recalling how to locate local (relative) maxima and minima for functions of one variable: Obtain the critical values of the function by finding all x where f (x) = 0 or f (x) does not exist. These critical values are potentially values at which the maximum or minimum function values are located, but some sort of test needs to be performed to determine if there s a maximum, a minimum, or neither at each. In other words, everywhere f has a max or a min, we know f = 0 (or does not exist), but the converse doesn t hold - not all values where f = 0 (or DNE) are necessarily extrema. That s why the testing. One test, the first derivative test involves checking for intervals of increasing/decreasing around the critical values. But there s another test, the second derivative test that allows you to immediately determine if a critical value leads to a maximum or minimum. The sign of the second derivative indicates whether the graph is concave up or concave down, which allows you to make the determination. This is the test you want to recall, as we ll be doing something analogous for surfaces. To obtain the max and min values themselves, you need to substitute back into the original function. 153

155 Example: Find the local extrema (maxima and minima) of the function f(x) = 2x 3 + 3x 2 36x + 6 Solution: Step 1: Find the critical values by setting f (x) = 0 and solving. Step 2: Examine the sign of f (x) at each critical value. Interpret the concavity to identify the locations of maxima and minima. Step 3: Find the value of f at each critical value. Identify maximum and minimum values. 154

156 Absolute extrema A key result of single variable Calculus is that a continuous function on a closed interval has an absolute maximum and absolute minimum value. To find absolute extrema on a given interval: Find all critical values that fall within the interval. Test each critical value, and the endpoints of the interval. In this case, the test is to simply evaluate f(x) at each of those values; the largest is the absolute maximum, smallest the absolute minimum. Example: Find the absolute extrema of the function on the interval [ 5, 5]. Solution: f(x) = 2x 3 + 3x 2 36x + 6 Step 1: Find the critical values (as before) by setting f (x) = 0 and solving. Step 2: Note whether the critical values fall within the interval. Evaluate f at critical values and endpoints, and identify maximum and minimum points. Look for these ideas: As we go through the topic in multivariable Calculus, you ll see them all extended by analogy. The horizontal tangent line translates to a horizontal tangent plane. f = 0 translates to f x = 0 and f y = 0. And, there is a version of the second derivative test, for determining whether your critical value is a max or a min. Absolute extrema will be extended to optimization under constraints. 155

157 Optimization Local extrema of functions of two variables Definition: A function of two variables has a local maximum [local minimum] at (a, b) if f(x, y) f(a, b) [f(x, y) f(a, b)] Theorem: Proof: for all (x, y) in some disk with center (a, b). If f has a local maximum or minimum and the first order partial derivatives exist there, then f x (a, b) = 0 and f y (a, b) = 0 Let g(x) = f(x, b) [hold y = b fixed and allow x to vary; you re now looking at the single variable function which is a slice of f(x, y) in the plane y = b]. If f has a local maximum [or minimum] at (a, b), then g must have a local maximum [or minimum] at a. Since we re now looking at a single variable function, we can apply the [previously established in single variable Calculus] result that g (a) = 0. Since g (a) = f x (a, b), we must have f x (a, b) = 0. The proof for f y (a, b) is analogous [hold x = a fixed and consider h(y) = f(a, y)]. This result is analogous to the one for single variable functions. If the function has a max or min, and the (partial) derivatives are known to exist (meaning we won t worry about the case analogous to f DNE), then both partials must be zero. This implies that setting the partials equal to zero and simultaneously solving for (x, y) pairs will give potential local extrema, which can then be tested. Critical values of f are values (a, b) such that f x (a, b) = 0 (or DNE) and f y (a, b) = 0 (or DNE). Critical values are obtained by finding the partials, setting equal to zero, and solving simultaneously. 156

158 Example: Find the critical values of f(x, y) = x 3 y + 12x 2 8y We still need to test the location (a, b) = (2, 4) to determine whether the graph has a maximum or minimum (or possibly neither) at that value. In single variable Calculus, we can determine this by checking the concavity - the value of the second derivative. This test extends to multivariable Calculus in the following way... Second derivatives test If the second partials (f xx, f xy, f yx, f yy ) are continuous on a disk centered at (a, b), where (a, b) is a critical value of f(x, y), compute the quantity and test according to the following rules: D = f xx (a, b)f yy (a, b) [f xy (a, b)] 2 If sign of D is and also sign of f xx (a, b) then f(a, b) is a: positive (D > 0) positive (f xx > 0) local minimum positive (D > 0) negative (f xx < 0) local maximum negative(d < 0) (doesn t matter) saddle point zero(d = 0) (doesn t matter) test fails *A derivation of how the second partials test gives results is generally omitted from Calculus texts at this point - it involves a Taylor polynomial expansion in two variables. I think the first time I saw one of those was in a partial differential equations course, a good bit further down the line. If you re curious, COW (Calculus on the Web) has an explanation/derivation. 157

159 The Hessian For those of you with some background in Linear Algebra, the quantity D = f xx (a, b)f yy (a, b) [f xy (a, b)] 2 can be expressed as the determinant of a matrix of the partials: D = f xx f xy f yx Note that when the second partials are continuous, we have f xy = f yx by Clairault s Theorem, so f xy f yx is the same as the (f xy ) 2 in the above formula for D. [ ] fxx f The matrix xy is referred to as the Hessian of f, so f yx f yy f yx f yy [ ] fxx f Hess(f) = xy f yy and D = det(hess(f)) 158

160 Example (continued): Obtain expressions for the second partials of the function f(x, y) = x 3 y + 12x 2 8y Then, evaluate the second partials at the critical value of (2, 4), and compute the value of D. Finally, apply the test, and determine whether ( 2, 4, f( 2, 4)) is a local maximum, minimum, or saddle point. You need the values of f xx (2, 4) and D. What s a saddle point? You should recall from single variable Calculus that when f (x) = 0, the graph has a horizontal tangent. This may indicate a maximum or minimum, but it s possible for the graph to have a horizontal tangent without it being either. The graph of x 3 is a typical example; there is a horizontal tangent at x = 0, and the graph crosses its tangent, rather than having a max or min there. A saddle point is the 3D version of this. The graph of f(x, y) has a horizontal tangent plane at (a, b), but the graph crosses its tangent plane. 159

161 Example: Find all local maxima, minima and saddle points of f(x, y) = x 4 + y 4 4xy + 1 Start by getting f x and f y, setting to 0, and solving for critical values. Then, get the second partials, and test each critical value. You might find it helpful to organize in a table. f xx = f xy = f yy = (a,b)[cv] f xx (a,b) f xy (a,b) f yy (a,b) D = f xx ( 1, 1) (0,0) (1,1) f xy f xy f yy Then, interpret the results, and get function values [z = f(a, b)] for each of the critical values: 160

162 Optimization Constrained optimization and Lagrange multipliers Constrained optimization is what it sounds like - the problem of finding a maximum or minimum value (optimization), subject to some other restrictions or constraints. Example: Suppose we have a rectangle inscribed within a given ellipse, say x2 9 + y2 4 to find the dimensions of the rectangle that have the maximum area. = 1. We would like A =?? 2y A =?? 2y 2x 2x The function that we wish to maximize, called the objective function, is the area function: f(x, y) = (2x)(2y) = 4xy The constraint is that x and y are related through the ellipse: We let g(x, y) be the constraint function x y2 4 = 1 x y2 4 1 = 0 g(x, y) = x2 9 + y2 4 1 Now, this particular example can be solved through single variable Calculus, and I ll suggest that you try that as a refresher. It ll also let us check the solution later on after we use the method of Lagrange multipliers. Solve x2 9 + y2 4 = 1 for y (use the positive square root). Substitute that into the area function, creating a function of one variable, x. Use the usual single variable Calculus process for finding a max or min value - differentiate, set equal to 0 and solve for x, and verify/test that you in fact have a maximum. 161

163 The solution is posted as a separate file, and linked in the lecture - you should get x = 3 2 [and can easily obtain y from there]. Now, back to Lagrange multipliers... If we look at the area as a function of two variables, we have a 3D surface. And just looking at f(x, y) = 4xy, we d see that that function by itself doesn t have a maximum value - as x and y get larger, the area grows without bound - so the constraint is needed to make sense of the problem. If we consider all possible values that the area could take on, e.g. an area of 1, 1.5, 2, , and so on, so f(x, y) = 4xy = 1 f(x, y) = 4xy = 1.5 f(x, y) = 4xy = 2 f(x, y) = 4xy = etc., etc., then we are looking at things in the form f(x, y) = 4xy = k i.e., the level curves of the function. Here is a contour plot showing some of the level curves of f(x, y): If we superimpose the constraint ellipse on the level curves, we see first of all that only some of the level curves intersect with the constraint, and therefore contain allowable values for x and y, and we can discard any that don t: 162

164 Now, it would help if I noted the heights that those level curves are at, which are the prospective values for area, because you d see that they are increasing as you move out from the center. So what we re really looking for is the level curve that just barely intersects the constraint equation - that s the one where the area will be maximal. Just barely intersects would be the same as tangent to. k = 12 k = 8 k = 5 k = 3 k = 1.5 k =.5 It looks like the maximal area that still intersects the constraint is when k = 12, so when 4xy = 12. Set 4xy = 12, so y = 3 x and sub into x2 9 + y2 4 = 1 : x (3/x)2 4 x 2 = x 2 = 1 4x = 36x 2 4x 4 36x = 0 (2x 2 9)(2x 2 9) = 0 2x 2 9 = 0 x =

165 Now, we need to formalize that approach... Method of Lagrange multipliers The trick here is to use the gradient vectors, and recall the useful result that the gradient vector at any point is always orthogonal to the level curve at that point. So it s orthogonal to a vector tangent to the level curve in the plane of the curve. If you want the curves to be tangent to each other, they need to have the parallel tangent vectors...which means they need to have parallel gradient vectors. And parallel for vectors means the same vector, or scalar multiples of each other. So we need to find (x 0, y 0 ) that solve f(x, y) = λ g(x, y) and also satisfy the constraint equation g(x, y) = 0. Lagrange s Theorem Proof: Let f and g have continuous first partial derivatives such that f has an extremum at a point (x 0, y 0 ) on the smooth constraint curve g(x, y) = 0. If g(x 0, y 0 ) 0, then there is a real number λ such that f(x 0, y 0 ) = λ g(x 0, y 0 ) Let r(t) be a parameterization of the smooth curve given by g(x, y) = 0; r(t) = x(t)i + y(t)j with r (t) 0, where x and y are continuous functions of t on an open interval I. [Note that you don t have to find the parameterization to work the problem, but asserting that one exists is needed for the proof. The smooth part is where r (t) 0 comes in.] Define a function h(t) = f(x(t), y(t)). Since (x 0, y 0 ) is an extreme value of f, we know that h(t 0 ) = f(x(t 0 ), y(t 0 )) = f(x 0, y 0 ) must be an extreme value of h. So h (t 0 ) = 0. [h(t) is a single variable function, and single variable Calculus tells us its derivative is zero wherever it has a max or min value.] Since h(t) is a composite function, h(t) = f(x(t), y(t)), we apply the chain rule to get or dh dt = f dx x dt + f dy y dt h (t) = f x (x(t), y(t))x (t) + f y (x(t), y(t))y (t) = f(x(t), y(t)) r (t) Since h (t 0 ) = 0, f(x 0, y 0 ) r (t 0 ) = 0, and f(x 0, y 0 ) is orthogonal to r (t 0 ). 164

166 Also, we have already established that the gradient is orthogonal to the level curve of a function at a point: g(x 0, y 0 ) is orthogonal to the level curve of g passing through that point. Since r(t) is the curve g(x, y), r (t) must be tangential to that curve, and so g(x 0, y 0 ) is also orthogonal to r (t 0 ). Therefore, f(x 0, y 0 ) and g(x 0, y 0 ) must be parallel to (i.e. scalar multiples of) each other, and so there exists some λ such that Back to the example... f(x 0, y 0 ) = λ g(x 0, y 0 ) Use the method of Lagrange multipliers to maximize f(x, y) = 4xy subject to the constraint x y2 4 1 = 0 Step 1: Find f for the objective function f(x, y) = 4xy, and g for the constraint function g(x, y) = x2 9 + y2 4 1 Step 2: Set up the equation f = λ g, and split that into two equations { fx = λg x f y = λg y 165

167 Step 3: Solve the system of three equations and three unknowns - the two equations that come from f = λ g, and also the constraint g(x, y) = 0. Occasionally you may be lucky enough to have a system of linear equations, but frequently, these will be nonlinear, and you need to resort to creative algebra. 166

168 In general, how do we know that the solution is the extremum we want? If the problem calls for a max, and we get one solution, what s telling us that that solution isn t in fact a min? Recall that with previous optimization techniques, there s been a test at the end to determine whether you have a max or min (recently, we had a second derivative type test for two variable optimization). The method guarantees an extremum of some sort - so to test that your solution (x 0, y 0 ) produces a maximum f(x 0, y 0 ), verifying that f(x 0, y 0 ) f(x, y) for all other (x, y) that satisfy the constraint reduces to the problem of verifying f(x 0, y 0 ) f(x, y) for just any one (x, y). Pick any other point on the graph of the constraint equation, and plug it in to f: Since (x, y) = (0, 2) is a point on x2 9 + y2 4 = 1, compute f(0, 2) = 4(0)(2) = 0. Since f( 3, 2) f(0, 2), f( 3, 2) = 12 is a maximum. 2 2 Finally... The process is more streamlined than it looks at first read - keep in mind we solved the same problem 3 ways and derived the method in the process. Implementing it isn t all that tedious. Look for a live example. The method extends to three (and more) variables - in the three variable case, the level curves become level surfaces. You ll see a couple of these in the suggested problems. 167

169 Iterated, double, and triple integrals Iterated integrals Integrals in the forms d b c a f(x, y) dx dy or b d a c f(x, y) dy dx or b d f a c e f(x, y, z) dz dy dx are iterated integrals; e.g (3x 2 y + y 2 ) dx dy is an example of an iterated integral of a function of two variables. We would interpret as d b c a [ d b c a f(x, y) dx dy f(x, y) dx ] dy where b a f(x, y) dx would indicate that we integrate f(x, y) with respect to x; in other words, treat y as a constant and integrate as you normally would for a function of one variable (x). Using the Fundamental Theorem of Calculus: b a f(x, y) dx = F (x, y)] x=b x=a = F (b, y) F (a, y) where F is an antiderivative of f (with respect to x); i.e., F x (x, y) =f(x, y). We would then go on to the outer integral and integrate the previous result. As with partial differentiation, integrating with respect to a variable while holding the others constant is simply working with a function of one variable, which is painless - assuming you recall all your integration techniques from first year Calculus! You may wish to refresh your memory on basic techniques, u substitution, and integration by parts. Although there s nothing new here, it will take a while to walk through even a basic exampleyou re working with definite integrals, and those take a while to go through. More examples will be posted, showing different integration techniques. 168

170 Example: (xy e y ) dx dy Start with the inside: 2 (xy e y ) dx 1 You are integrating with respect to x, so treat as a single variable function of x, with y constant. Until you ve practiced a few, you may want to explicitly use the rules for sums/differences and constant multiples: I ve split into two integrals, and pulled out constant multiples. Since we are integrating with respect to x, bothy and e y are constants. Now, integrate: And evaluate, as by the Fundamental Theorem. Since you are integrating with respect to x, you are evaluating x from1to2(y is just hanging out being y for a while): You have now obtained: 169

171 That was just the inner integral. You now have a function of y. Returning to the original problem, sub in your result for the innter integral: And you can finish integrating: So the final answer is: Example: π z ( ) y sin x 1+y 2 dz dy dx Go for the inner integral first: 3 1 2z ( ) y sin x 1+y 2 dz Now, nasty as that looks, everything except the z is a constant - pull it all out: And get to work z ( ) y sin x 2y sin x 1+y 2 dz = 1+y zdz 170

172 Here s another page for you... check against the solution when you re done. Hint: at some point, you ll need to do a u-substitution. But all the integration is pretty straightforward; it s just that there s so much of it! 171

173 Iterated, double, and triple integrals More general iterated integrals We have termed integrals in the form d b c a f(x, y) dx dy (and similar) iterated integrals, and we interpret them in terms of integration of single variable functions: d b [ d ] b f(x, y) dx dy = f(x, y) dx dy where c a b a c a f(x, y) dx would indicate that we integrate f(x, y) with respect to x; in other words, treat y as a constant and integrate as you normally would for a function of one variable (x). So consider this expression a bit more: b a f(x, y) dx Since y is constant (as far as integrating with respect to x is concerned), there s no reason the upper and lower bounds on the integral can t have y s in them; for example 3+y 1 y (x 2 + y 2 ) dx The FTC still holds - get the antiderivative, and evaluate. Example: Find 3+y 1 y (x 2 + y 2 ) dx 172

174 That was a single integral. Suppose that was the inner integral of an iterated integral, let s say Making the substitution gives y 0 1 y (x 2 + y 2 ) dx dy ( 8 3 y3 +4y 2 +10y ) dy While we were considering x, thaty was constant. Now we re integrating with respect to y. Now, it s a function of y. So integrate it: In general, since any function of x is a constant with respect to y, iterated integrals of the form: make sense, and can be evaluated. b y2 =g 2 (x) a y 1 =g 1 (x) f(x, y) dy dx Similarly, since any function of y is constant with respect to x, iterated integrals of the form also make sense, and can be evaluated. d x2 =h 2 (y) c x 1 =h 1 (y) f(x, y) dx dy 173

175 Example: Find 1 x 1 x2 dy dx 0 0 Example: Find 2 2y y 2 0 3y 2 6y 3ydxdy 174

176 The rules for allowable bounds extend to three integrals in a row. Look at b g2 (x) h2 (x,y) a g 1 (x) h 1 (x,y) f(x, y, z) dz dy dx * The inner (dz) integral can have bounds with x s and y s (everything but z) * The middle (dy) can have bounds with x s (everything but y... and z is already gone) *Theouter(dx) must have constant bounds (everything but x... and y and z are already gone. Which doesn t leave anything but constants.) The same type of rule holds for other arrangements (dx dy dz, dy dz dx, etc.). 175

177 Iterated, double, and triple integrals Iterated integrals and area We started off by looking at iterated integrals in the forms d b c a f(x, y) dx dy and b d a c f(x, y) dy dx Ignore the function and think about just the bounds for a moment; in fact, consider the iterated integrals For d b c a 1 dx dy and d b c a 1 dx dy b d a c 1 dy dx the inner dx integral tells me I ll be evaluating from a low bound of x 1 = a to a high bound of x 2 = b. Theouterdy integral tells me I ll evaluating from a low bound of y 1 = c to y 2 = d. Inother words (or symbols), we re talking about a x b c y d This describes a rectangular region in the plane. Now, what would give? d c b a d b c a 1 dx dy 1 dx = x] b a = b a (b a) dy =(b a)y] d c =(b a)(d c) (b a) is the width of that rectangular region, and (d c) is the height. So... d b c a 1 dx dy gives the area of that rectangular region bounded by a x b, c y d. What about b d a c 1 dy dx That s still c y d and a x b - the same region. And, d c 1 dy = y] d c = d c gives the same area. b a (d c) dx =(d c)x] b a =(d c)(b a) 176

178 So d b c a 1 dx dy = b d a c 1 dy dx order doesn t matter, area is area. Be careful not to overgeneralize this, though - right now, we re talking about specifically f(x, y) = 1, and constant bounds on the integral. We will denote the area of a rectangular region R with a x b, c y d by the symbol da This is a special case of the double integral over a region R. d b b d da = dx dy = dy dx R c a R translates as The area of the rectangular region R can be computed by using iterated integrals. Example: Find R da where R is the rectangle 3 x 5, 2 y 1. a c Turning to more general iterated integrals, the bounds on b g2 (x) a g 1 (x) 1 dy dx can be expressed as a x b g 1 (x) y g 2 (x) This (typically - note to follow) describes a region R with the vertical lines x 1 = a and x 2 = b as the left and right bounds, and the functions y 1 = g 1 (x) andy 2 = g 2 (x) as the bottom and top bounds. Regions of this type are referred to as vertically simple. A vertical line drawn anywhere in the region will always hit the same function g 1 (x) at the bottom, and the same function g 2 (x) atthe top. 177

179 Note: the preceding gives the impression that you ll always see vertical walls on the left and right - that isn t true (although that s the picture you want to have in mind if you re thinking about a generic vertically simple region). The region shown below is 0 x 1 x y x with 1 x da = R 0 x dy dx It is still vertically simple, even though you don t have walls - it is still true that the bottom function is always g 1 (x) =x, and the the top function is always g 2 (x) = x,for0 x 1. 1 x da = R 0 x dy dx Similarly, the bounds on can be expressed as d x2 =h 2 (y) c x 1 =h 1 (y) 1 dx dy h 1 (y) x h 2 (y) c y d This (typically) describes a region R with the horizontal lines y 1 = c and y 2 = d as the bottom and top bounds, and the functions x 1 = h 1 (y) andx 2 = h 2 (y) as the left and right bounds. Regions of this type are referred to as horizontally simple. A horizontal line drawn anywhere in the region will always hit the same function h 1 (y) on the left, and the same function h 2 (y) onthe right. 178

180 The same note: as with the vertically simple regions, you won t necessarily see the walls (horizontal, in this case), but can still have a horizontally simple region; the key is that it can be expressed as h 1 (y) x h 2 (y) The region below is c y d y 2 x y 0 y 1 with 1 y da = R 0 y 2 dx dy Notice it s the same region I used as a previous example. Some regions are both horizontally and vertically simple (and some are neither). You may get some choice as to how you want to set the area integral up. In the lecture, you ll see a link to an additional example, using MVT to plot a horizontally simple region. Example: Sketch the region R whose area is given by 1 3 y da = R 0 y 2 dx dy Is the region vertically or horizontally simple? 179

181 Example: Sketch the region R whose area is given by 2 e x da = R 1 0 dy dx Is the region vertically or horizontally simple? Example: Sketch the region R bounded by the graphs of 2x 3y =0,x + y =5andy =0. Hint: Since you won t be able to decide whether it s vertically or horizontally simple until after you ve made the sketch, you should solve in whatever way you prefer to sketch those lines. You can always rearrange it afterwards. Now that you ve established what type of region it is, rearrange the expressions for the bounds in the correct form needed to set up the inner integral: You could probably get the low and high bounds for y by inspection, but make a habit of solving for them whenever they involve a point of intersection: 180

182 Summarize: write the inequalities that describe the region, and write the iterated integral that will give you the area of the region: And solve for the area of the region R by evaluating the integral: 181

183 Iterated, double, and triple integrals The double integral and volume (defining) We re ready to take a look at the general definition of the double integral R f(x, y) da over a region R. We have already established the interpretation for the special case of showing that it gives the area of the region R, and can be computed using iterated integrals: R da = b g2 (x) a g 1 (x) R 1 da dy dx = d h2 (y) c h 1 (y) dx dy We ve managed to avoid the Riemann sum part of the picture so far, by building on results from single variable Calculus (and assuming that you have of course remembered the Riemann sum definition of a single definite integral!). To properly define the double integral and discuss the geometry, we need to return to Riemann sums and limits. A quick refresher: Assume first that f is a continuous function on an interval [a, b], and for convenience, assume f(x) 0 for all x [a, b]. The (single) definite integral is defined by where we b n f(x) dx = lim f(x i )Δx a n i=1 * Consider the interval [a, b] and form a partition of [a, b] by dividing into n subintervals. In the Riemann sum, we are using n subintervals of equal width (we can write a more general definition where the partition does not need subintervals of equal width, but the above is a bit simpler to work with and good enough for our purposes). *LetΔx = b n a. *Chooseapointx i in each subinterval (for convenience, I m showing the right endpoint). * Evaluate the function at x i giving f(x i ). At this point, we note that on any subinterval, we can draw a rectangle with Δx as the width of the base, and f(x i ) as the height. * Compute the area of a rectangle: A = f(x i )Δx. 182

184 * Note that each subinterval has its own rectangle, and the sum of these n rectangles gives a total area that approximately equals the area under the curve: n i=1 f(x i )Δx * And pushing n to infinity smoothes out the rectangular area, and gives us the exact area under the graph of f. b n f(x) dx = lim f(x i )Δx a n i=1 The Fundamental Theorem of Calculus is the thing that tells us we don t have to use the definition, but that there is a connection between the definite integral (representing area), and the antiderivative. The FTC is a marvelous thing, that lets us avoid much painful algebra. We are not going to prove it again here, though - in theory, you ve already done that at some point in your life. Suppose f(x) is a continuous function on an interval [a, b]. Then b where F is any antiderivative of f. a f(x) dx = F (b) F (a) If f changes sign on [a, b], we know that the definite integral gives us net area (the difference between the area that lies above the x axis and that which lies below), and we adjust our area problem setups accordingly to break into separate regions. Onward to 3D: b a f(x) dx = A1 A2 Returning to multivariable Calculus, what do you expect the double integral f(x, y) da to give? Volume under the surface (assuming f(x, y) 0onR). R 183

185 Definition of the double integral: Suppose f(x, y) is a continuous function on a region R. For convenience, and for visualization purposes, let R be a rectangular region a x y, c y d, and assume f(x, y) 0 for all (x, y) R. Thedouble integral over R is defined by where... n f(x, y) da = lim f(x i,y i )ΔA R n i=1 * Consider the region R and form a partition by dividing [a, b] intoj subintervals of equal width, and [c, d] intok subintervals of equal width. (Again, this is a simplification of a more general partitioning we could set up.) This creates a grid of rectangles in the xy plane. There are n = j k rectangles total. *LetΔx = b j a,andδy = d k c (the length and width of each subinterval). The area of each rectangle is ΔA =Δx Δy. *Chooseapoint(x i,y i ) in each subinterval (for convenience, I m showing the center point). * Evaluate the function at (x i,y i ) giving f(x i,y i ). At this point, we note that on any rectangle, we can draw a rectangular prism with ΔA as the area of the base, and f(x i,y i ) as the height. * Compute the volume of a prism: V = f(x i,y i )ΔA. * Note that each rectangle has its own prism, and the sum of these n prisms gives a total volume that approximately equals the volume under the curve: n i=1 f(x i,y i )ΔA * And pushing n to infinity smoothes out the rectangular volume, and gives us the exact volume under the graph of f. n f(x, y) da = lim f(x i,y i )ΔA R n i=1 184

186 Things to consider: As noted before, the rectangles don t have to all be the same: the subintervals don t have to be of equal length and width (it s just convenient to simplify things a bit). We could write a more general version of the definition. The region R doesn t have to be rectangular. This will work perfectly well over any region in the xy plane. Keep those vertically and horizontally simple regions in mind. The definition doesn t tell you how to compute the double integral... just as the definition of the definite integral doesn t tell you how to compute it. We need a theorem for that. If the surface f(x, y) dips below the xy plane, we ll end up getting net volume - the difference between the volume which lies above the plane, and that which lies below. 185

187 Iterated, double, and triple integrals The double integral and volume (computing) Theorem: The double integral over a region R, R f(x, y) da can be computed using iterated integrals, where the bounds on the iterated integrals describe the region R in the plane. If f(x, y) 0 for all (x, y) R, this will give the volume under the surface z = f(x, y), above the xy plane. (If f(x, y) dips below the xy plane, we get the difference in volume above and volume below.) For rectangular regions a x b, c y d, V = R f(x, y) da = b d a c f(x, y) dy dx = For vertically simple regions a x b, g 1 (x) y g 2 (x), V = R f(x, y) da = b g2 (x) a g 1 (x) For horizontally simple regions h 1 (y) x h 2 (y), c y d, V = R f(x, y) da = d h2 (y) c h 1 (y) d b c a f(x, y) dy dx f(x, y) dx dy f(x, y) dx dy Setting up the region is no different than setting up the region for an area problem - the fact that we re now looking at R f(x, y) da in general instead of R 1 da in particular doesn t affect how you set up the bounds on the iterated integral. That comes from the region R. You ve already practiced the techniques of computing iterated integrals in the form b g2 (x) a g 1 (x) f(x, y) dy dx and so on. Therefore, you already know how to do this (the only thing missing by the time we got to this point was the assertion that you could, in fact, use iterated integrals to compute volume). 186

188 Example: Let R be the region bounded by x = 2 1y, x = y, y =0,y =4. Letf(x, y) =x 2 y 2. Find the volume under the surface f(x, y) over the region R. Start by sketching R and identifying whether vertically or horizontally simple: Set up the iterated integral: And integrate: 187

189 That s pretty much all there is to it, although there are lots of variations in setting up the region (see posted examples). The only thing missing is that we haven t proven the theorem... and you may be relieved to learn that we won t prove it rigorously with the Riemann sums - there are some technical details involving double summations and double limits that are beyond the scope of this class. I will offer a geometric proof, however... Consider the iterated integral b g2 (x) a g 1 (x) f(x, y) dy dx (so we ll look at a vertically simple region as an example - you could repeat this with a dx dy setup as well).when we hold x constant and compute the inner integral g2 (x) g 1 (x) f(x, y) dy we are getting the area of a cross section under the surface of f(x, y), in the plane of x =whateverx we re stuck in (imagine the integration of f over dy as sweeping out that plane). The cross sectional area will vary depending on where x is. So, we can say that A(x) = g2 (x) g 1 (x) is a function that gives cross sectional areas. f(x, y) dy However, we have learned in in single variable Calculus when studying volumes by slicing that b A(x) dx a gives the volume of a solid with cross section area A(x) (you can think of integration with respect to x as sweeping out the volume of the solid). So, b g2 (x) a g 1 (x) f(x, y) dy dx gives the volume of the solid under the surface z = f(x, y) andabover in the xy plane. V = 2 2 x 0 0 (4 x 2 + y 2 ) dy dx 188

190 Iterated, double, and triple integrals Changing the order of integration We have seen that many regions are both vertically and horizontally simple, for example, this one: We could compute the area of that region as either or R R da = da = 1 x 0 x 1 y 0 y 2 dy dx dx dy and find the volume of a solid under a surface f(x, y) and over that region by or R R da = da = 1 x 0 x 1 y 0 f(x, y) dy dx y 2 f(x, y) dx dy The question is, why would we want to? Consider the problem of computing R f(x, y) da = x sin(y 2 ) dy dx This would give the volume under f(x, y) =sin(y 2 ) over the region R: 0 x 1, x y 1. Perfectly nice region, the problem is already set up, so we proceed to the inner integral 1 x sin(y 2 ) dy and start by antidifferentiating sin(y 2 ) with respect to y. What s the problem? 189

191 It is possible (although not guaranteed) that switching the order of integration from to x d h2 (y) c h 1 (y) sin(y 2 ) dy dx sin(y 2 ) dx dy will improve things. Notice that I can t just swap dx and dy in the integral - the bounds are tied to the variables of integration. And we know that 1 1 x 0 sin(y 2 ) dx dy is not allowed - the outer integral must always have constant bounds. We need to take steps to determine exactly what c, d, h 1 (y) andh 2 (y) are. Thewaytodothisisto * Sketch the region described by the original bounds. * And reimagine it. If the bounds were orginally given as y as a function of x, the region was vertically simple. Switch it to a horizontally simple setup, solving the expressions for x as a function of y. * If the bounds were orginally given as x as a function of y, the region was horizontally simple. Switch it to a vertically simple setup, solving the expressions for y as a function of x. * Write the new iterated integral reflecting the new setup. * And try again. 190

192 Example: What is the region described by the bounds on x sin(y2 ) dy dx? Sketch it: Rearrange the expressions so the region reads horizontally, instead of vertically - the left and right bounds should become functions of y, while the top and bottom bounds should become constants: Set up the new iterated integral: And try again with the integration: 191

193 A typical mistake is to do this without sketching, and simply rearrange the expressions and replace them in the integral. Don t - as you switch from a bottom to top setup to a left to right setup, you may find that the top function (at the top of the integral) becomes the left function (at the bottom of the integral). Switching the bounds on regions to change the order of integration is typically done * when the original problem cannot be integrated. * when the original problem is annoying to integrate, and might work better in the other order. * simply because I said to. 192

194 Iterated, double, and triple integrals Double integrals in polar coordinates We ve discussed integration over rectangular regions, and integration over general regions where the bounds for the regions can be expressed as functions of x or functions of y. For regions like the one below, it s difficult to express the bounds as functions of x or y, but simple to express in polar coordinates as functions of a radius r and an angle θ. Furthermore, there are integrals out there that can t be integrated as functions of x and y, but can easily be integrated if we could rewrite in terms of r and θ. To do this, we have to * Recall the conversion between polar and rectangular coordinate systems * Figure out what happens to the expression being integrated. * Figure out how to rewrite the bounds. Converting: A point in the plane may be specified in terms of x and y (horizontal and vertical displacement from the origin), or r and θ (distance from the origin and angle with the positive x axis). This should look familiar - it s identical to what we do with vectors, and the points they point to. You can specify magnitude and direction (r, θ), or x and y components (x, y). x = r cos θ y = r sin θ r 2 = x 2 + y 2 tan θ = y x 193

195 Plotting: Plotting points in polar is a matter of looking in the direction of the angle, and moving out the distance specified by the radius. Polar graph paper with radius rings and marked angles is sometimes used to plot. For example, to plot (r, θ) =(2, π 3 ), move out a distance of 2 along an angle of π 3. The only odd thing to get used to is having negative radius - interpret (r, θ) =( 3, π 4 ) as facing along a line at the angle π 4... andthenwalkingbackwards along that line. Curves in polar coordinates Polar coordinates are well suited for describing circles centered at the origin and lines through the origin. A circle of radius a with equation x 2 + y 2 = a 2 becomes r 2 = a 2, and the curve is described as r = a 0 θ 2π 194

196 You should recognize semicircles from the x 2 + y 2 = a 2 equation solved for either x or y: y = a 2 x 2 radius a. draws the top half of a circle with y = a 2 x 2 with radius a. draws the bottom half of a circle x = a 2 y 2 draws the right half of a circle with radius a. x = a 2 y 2 draws the left half of a circle with radius a. All of these equations are expressed in the form r = a, α θ β, where the range of θ s draws out the correct part of the semicircle. Lines through the origin are expressed in terms of their angle. A line with slope m has tan θ = m, orθ =tan 1 m. r is allowed to vary (by not specfiying anything about r at all, we re implying it runs from to, drawing out the line). The example shown is y = x, withm =1. Soθ = tan 1 1and θ = π 4 is the polar equation of this line. 195

197 Horizontal and vertical lines have more complicated expressions in polar than they do in rectangular (but we may need this for rectangular regions). Vertical: Horizontal: x = a r cos θ = a a r = cos θ r = a sec θ y = b r sin θ = b r = b sin θ r = b csc θ Cartesian equations in general are converted to polar by making the substitutions x = r cos θ y = r sin θ For example, if we can say f(x, y) =x 2 + xy f(r cos θ, r sin θ) = (r cos θ) 2 +(rcos θ)(r sin θ) = r 2 cos 2 θ + r 2 cos θ sin θ = r 2 cos θ(cos θ +sinθ) Example: Express the paraboloid f(x, y) =9 x 2 y 2 as a function of r and θ. 196

198 Regions in polar coordinates Regions in polar coordinates are expressed as inequalities in r and θ. Foraθ - simple region, we have h 1 (θ) r h 2 (θ) α θ β Examples : 0 r 3 π 4 θ 3π 4 1 r 3 0 θ 2π Rectangular regions require a little work to express in polar, since you re slicing radially. The region 0 x 2 0 y 2 is broken into two regions: 0 r 2secθ 0 θ π 4 0 r 2cscθ π 4 θ π 2 197

199 Rewriting integrals using polar coordinates: If f is continuous on a polar region of the form R = {(r, θ) α θ β,h 1 (θ) r h 2 (θ)} then β h2 (θ) f(x, y) da = f(r cos θ, r sin θ)rdrdθ R α h 1 (θ) Note that da becomes rdrdθ. We won t do a full derivation of how this change of variables works for the integrand, but we will at least justify this geometrically. Proceed on to the posted examples of integration. The key to these will be to * Express the bounds of the region in polar coordinates, and put these new bounds on the integrals. * Express the function being integrated as f(x, y) =f(r cos θ, r sin θ). * Integrate β α h2 (θ) h 1 (θ) f(r cos θ, r sin θ)rdrdθ 198

200 Triple integrals The triple integral and volume The development of the triple integral is analogous to that of the double integral, simply moving into more one more dimension, so I m going to go light on the theory here, and simply present several examples. Area Volume: The double integral da over a region in the xy plane gives the area of that region and can be computed with iterated integrals: b g2 (x) R : a x b, g 1 (x) y g 2 (x) A = da = dy dx (vertically simple) R : h 1 (y) x h 2 (y), c y d A = R : h 1 (θ) r h 2 (θ), α θ β A = b g2 (x) a g 1 (x) dy dx = R R R b a da = da = a g 1 (x) d h2 (y) c h 1 (y) β h2 (θ) α h 1 (θ) (g 2 (x) g 1 (x))dx dx dy r dr dθ (horizontally simple) (polar, θ- simple) When we introduce a function f(x, y) as the integrand, with f(x, y) > 0 on R, we can interpret ( or d h2 (y) c h 1 (y) R f(x, y)da = f(x, y)dxdy or b g2 (x) a g 1 (x) β h2 (θ) α h 1 (θ) f(x, y)dy dx as the volume between the surface f(x, y) and the xy plane over R. f(r cos θ, r sin θ)r dr dθ We could express the same quantity using three iterated integrals; performing the inner integration on would immediately give the expression b b g2 (x) f(x,y) a a g 1 (x) g2 (x) g 1 (x) 0 dz dy dx f(x, y)dy dx And I m going to quit with the or at this point; assume that this could be done just as well with a dz dxdy setup or an r dz dr dθ setup. If we use the r dz dr dθ setup, we re using the cylindrical coordinate system, which is simply the polar coordinate system extended into 3D. ) 199

201 Example: Write an expression that would give the volume below the surface f(x, y) = x + y over the region R = {(x, y) 0 x 3, 0 y x} using (a) a double iterated integral, and (b) a triple iterated integral. At this level, it s a matter of perspective - are you integrating a function f(x, y) over a 2D planar region, or are you integrating 1 over a 3D solid region? Same difference. But if we introduce the triple integral, we have a more flexible notation, and can extend things a bit further. Note in the following I m going to switch the notation (and if I d had any sense I d have done this right at the start instead of following the textbook). We re about to have a lot of letters kicking around, so instead of using a s and b s and g 1 (x) s and h 1 (y) s, I m going to switch to subscripting (x 1 x x 2 instead of a x b, y 1 (x) y y 2 (x) instead of g 1 (x) y g 2 (x), and so on). This will make it easier to keep track of everybody, so you aren t looking around going Now, do the h s go with the y s or with the z s? 200

202 Volume The triple integral dv over a rectangular prism D D : x 1 x x 2, y 1 y y 2, z 1 z z 2 in space gives us the volume of that region and can be computed with iterated integrals: x2 y2 z2 V = dv = dz dy dx V = V = (or any permutation that you like). D D D dv = dv = Over a general solid region in space, for example x 1 y 1 z 1 x2 z2 y2 x 1 z 1 y 1 z2 y2 x2 z 1 y 1 x 1 dy dz dx dxdy dz D : x 1 x x 1, y 1 (x) y y 2 (x), 0 z z 2 (x, y), the triple integral gives the volume between z = z 2 (x, y) and the xy plane: and more generally, V = D D dv = dv = x2 y2 (x) z2 (x,y) x 1 y 1 (x) x2 y2 (x) z2 (x,y) x 1 y 1 (x) 0 z 1 (x,y) dz dy dx dz dy dx would give the volume (or net volume if the curves crossed each other) between z 1 and z 2. Setting up and solving these is mostly done live - I ve put a few example pictures in the slides, though, so you can see what we re talking about

203 Example: Let D be the region below the plane z = 4x + 2y + 1 and above the region in the xy plane bounded by x = 0, y = 0, and y = 3 x. D = {(x, y, z) 0 x 3, 0 y 3 x, 0 z 4x + 2y + 1} The integral that would be used to compute the volume of this region would be V = dv = 3 3 x 4x+2y+1 D dz dy dx 202

204 Example: Let D be the region bounded by the paraboloid y = x 2 + z 2 and the plane y = 4. Write the integral that would give the volume of this region. Notice that the way this one is oriented, its trace in the plane y = 4 (which is parallel to the xz plane) is the circle x 2 + z 2 = 4. You can think of this as the way the graph is facing - the best way to set up the region is with y = 4 as the floor. So the best way to describe D is D = {(x, y, z) 2 x 2, x 2 + z 2 y 4, 4 x 2 z 4 x 2 } The iterated integral that would be used to compute the volume of that region is: V = D dv = 2 4 x x 2 x 2 +z 2 dy dz dx 203

205 Basically, we can order our integrals any way we need to describe the region, as long as we follow the rules for two variables on the inner bound, one in the middle, constant on the outer. These are all permissible variations (and there are others): Volume???: V = V = V = V = So, we have this progression... D D D D dv = dv = dv = dv = x2 y2 (x) z2 (x,y) x 1 y 1 (x) z 1 (x,y) z2 x2 (z) y2 (x,z) z 1 x 1 (z) y 1 (x,z) z2 y2 (z) x2 (y,z) z 1 y 1 (z) x 1 (y,z) y2 z2 (y) x2 (y,z) y 1 z 1 (y) x 1 (y,z) dz dy dx dy dxdz dxdy dz dxdz dy x2 x 1 [y 2 (x) y 1 (x)] dx is used in one variable Calculus to give the area between the curves y 1 (x) and y 2 (x) with x 1 x x 2. But we can also write this as R 1dA = x2 y2 (x) x 1 y 1 (x) 1dy dx Drop a function (or difference of functions) in the 1 spot and we ve bumped things up to volume: R [z 2 (x, y) z 1 (x, y)] da = x2 y2 (x) x 1 y 1 (x) [z 2 (x, y) z 1 (x, y)] dy dx gives the volume between z 1 (x, y) and z 2 (x, y) over the region {x 1 x x 2, y 1 (x) y y 2 (x)}, and we can turn that into the triple integral: D 1dV = x2 y2 (x) z2 (x,y) x 1 y 1 (x) z 1 (x,y) 1dz dy dx So, by analogy, we could drop a function in the 1 spot there and move up another dimension (and we ll go back to using f now...i suppose w would be the other alternative after we ve used x, y and z). D f(x, y, z)dv = x2 y2 (x) z2 (x,y) x 1 y 1 (x) z 1 (x,y) f(x, y, z)dz dy dx What does that give? If we continue to extend the geometry by analogy, w = f(x, y, z) is now a four dimensional surface, and we re getting the hypervolume of that surface over the 204

206 region D. A bit tough to visualize, and no, we can t exactly sketch it! Depending on what f represents, however, we may have other interpretations. A common application is that if a region has a variable density, ρ(x, y, z), integrating will give the mass of the solid region. We ll look at a couple applications of that after we finish with the techniques of integrating over 3D regions. The critical thing with these is being able to set up the 3D region that forms the bounds that you re integrating over. Whether the integrand is a 1dV or an f(x, y, z)dv plays no role in setting up that region - that function is simply the function that is being integrated. 205

207 Triple integrals Integrals using cylindrical coordinates Cylindrical coordinates are the extension of polar coordinates to 3D; we know that a point in 2D can be specified by either (x, y), or (r, θ), related by x = r cos θ y = r sinθ x 2 + y 2 = r 2 tan θ = y x In 3D, we may specify a location in similar fashion, by either giving the (x, y, z) coordinates of the point, or the (r, θ, z) coordinates - r and θ specify a distance and angle with the x axis in the xy plane, and the third z coordinate tells you how far up or down from the z axis to proceed. Example: Where would the point (r, θ, z) = (2, π 6, 4) be located? Sketch. What are the (x, y, z) coordinates of the point (r, θ, z) = (2, π 6, 4)? Surfaces in cylindrical coordinates Note that you need to pay attention to the context in order to recognize what geometric object you re looking at. In the context of 2D space, what does x 2 + y 2 = 4 describe? What is the corresponding polar equation? In the context of 3D space, what does the equation x 2 + y 2 = 4 describe? What is the corresponding cylindrical equation? 206

208 Equation forms you should recognize r = C (radius is constant, θ and z are allowed to vary arbitrarily). Cylinder of radius C. Trace in the xy plane is circle r = C; i.e. x 2 + y 2 = C 2. θ = C (angle with x axis is constant, radius and z are allowed to vary). Plane through the origin parallel to z. Trace in xy is the line θ = C; in rectangular, that s the line y = (tan θ)x z = C (z constant, radius and angle vary). Plane through the origin parallel to xy at a height of z. The polar part is causing it to drawn in a disk shape, but that disk extends indefinitely, creating the plane z = C. 207

209 z = Cr (height is proportional to radius, angle varies). Double cone. Traces in planes z = k parallel to the xy plane are circles in the polar form r = k C, rectangular form x 2 + y 2 = ( kc ) 2. The sides of the cone have slope ±C. Integration in cylindrical coordinates The integrating factor for triple integrals in cylindrical coordinates looks like that for double integrals in polar coordinates, because the conversion equations are identical (z remains z). In cylindrical coordinates D f(x, y, z)dv = β r2 (θ) z2 (r,θ) α r 1 (θ) z 1 (r,θ) f(r cos θ, r sinθ, z)r dz dr dθ Note this one s order specific; we don t get into how to switch the order of integration on these things (although it is possible). The inner bounds imply that z can be a function of r and θ, in the middle, r can be a function of θ, and on the outer bounds, θ must be constant. Examples: Examples for this one are all done live; go there now. 208

210 Iterated, double, and triple integrals Spherical coordinates Spherical coordinates give an alternate way of specifying the location of a point based on radius and angle. We use an ordered triplet (ρ, θ, φ) where ρ radius, distance from origin. ρ 0. θ angle with positive x axis, in xy plane. 0 θ < 2π. φ angle with positive z axis. 0 φ π. The best way to visualize (ρ, θ, φ) is as (distance, spin, tilt). How this works Take a segment (or vector) of length ρ. Tilt it down into the xz plane at an angle of φ. Drop a line down to the x axis. Think of this as the base of a triangular wedge. Then, grab the wedge and spin it in the xy plane. That s your θ. 209

211 Converting from spherical (ρ, θ, φ) to rectangular (x, y, z) To derive the conversion factors, first focus on the triangle formed by dropping a line to the xy plane. We see that the angle φ belongs in the top of that triangle. The length of the leg parallel to the z axis is also the distance from the point to the xy plane - it must be the z coordinate of the point. Also, one leg lies in the xy plane - call its length r. z = ρcos φ r = ρsin φ Now, focus just on the xy plane. If you drop a leg parallel to the y axis, and perpendicular to x, you form another right triangle with hypotenuse r and angle θ. x = r cosθ y = r sinθ The conversion factors that will take us back and forth between spherical and rectangular coordinate systems are: Example: x = ρsin φcos θ y = ρsin φsinθ z = ρcos φ Convert the point (ρ, θ, φ) = (5, 3π 4, 5π 6 ) to rectangular coordinates. 210

212 Converting from rectangular to spherical The same equations allow us to convert from rectangular to spherical; to go the other way, say from (x, y, z) = ( 1, 2, 3) into spherical, we could solve simultaneous equations 1 = ρsin φcos θ 2 = ρsin φsinθ 3 = ρcos φ However, it s probably better to go and derive some formulas in the abstract. First, note from the triangles used to set up the conversions, we have x 2 + y 2 = r 2, and then r 2 + z 2 = ρ 2, so ρ 2 = x 2 + y 2 + z 2 Then, to get φ, we have ρ = x 2 + y 2 + z 2 cosφ = z ρ Since the range of the inverse cosine function lines up neatly with the allowable values for φ (0 φ π), we can say φ = cos 1 z ρ Finally, note that if we divide we get y = ρsin φsinθ x = ρsin φcos θ tan θ = y x as with cylindrical (which we should, it s the same θ). Since 0 θ < 2π, be sure when solving that to locate θ in the correct quadrant depending on the signs of x and y. Summary: Rectangular to spherical x = ρsin φcos θ y = ρsin φsinθ z = ρcos φ Spherical to rectangular ρ = x 2 + y 2 + z 2 φ = cos 1 z ρ tan θ = y x 211

213 Example: Convert the point (x, y, z) = ( 1, 2, 3) to spherical coordinates. 212

214 Triple integrals Spherical regions and triple integrals in spherical coordinates Integration using spherical coordinates I m going to get a little ahead of things on this one and start by just showing you what the integral will look like in spherical coordinates: D f(x, y, z)dv = φ2 θ2 ρ2 φ 1 θ 1 ρ 1 f(ρsin φcos θ, ρsin φsinθ, ρcos φ)ρ 2 sinφdρdθ dφ This is a fairly limited form (note all the variables of integration have constant bounds), so we can only use it on shapes that can be expressed as spherical wedges. That makes the region part of the program pretty simple - the main thing that we re interested in expressing in spherical coordinates are spheres and cones. Note: different texts use different orderings and notation conventions (and since all the bounds are constant, you can do these in any order). Don t be surprised if different forms/ordering/notation appear in examples and problems. Spherical regions We ve seen how to plot points in spherical coordinates. The previous integral suggests we are interested in regions in the form {(ρ, θ, φ) ρ 1 ρ ρ 2, θ 1 θ θ 2, φ 2 φ φ 2 } The first thing is to look at a quick gallery of shapes that are easily drawn in spherical coordinates. (Images are in lecture.) The basic shape is a sphere, of course: ρ = C translates as ρ 2 = C 2 and so x 2 + y 2 + z 2 = C 2 ; a sphere with radius C. With θ and φ unspecified, we get the full sphere. The role of φ: Recall that in spherical coordinates, (ρ, θ, φ) is (radius, spin, tilt). φ = C with 0 < φ < π 2 describes a cone (with θ and ρ unspecified): Side note on converting cones back to rectangular: The key one that you should just recognize is that the cone φ = π 4 is the cone z = x 2 + y 2 but it s instructive to look at how that got there so you can generalize it to cones with the sides at an angle other than

215 Using the conversion factors x = ρsin φcos θ y = ρsin φsinθ z = ρcos φ and plugging in φ = π 4, we get x = ρsin π 4 cos θ = 2 2 ρcos θ y = ρsin π 4 sinθ = 2 2 ρsin θ z = ρcos π 4 = 2 2 ρ Then ρ = 2 2 z, and we can plug that back in to the equations for x and y: x = z cos θ y = z sin θ So x z = cos θ, y z = sinθ and ( x ) 2 ( y ) 2 + = 1 z z z 2 = x 2 + y 2 z = x 2 + y 2 since initially we had the top cone. If you want the lower cone, z = x 2 + y 2, specify φ = 3π 4. Cones with other degrees of tilt can be converted in the same way, although it won t work out quite as neatly when cos φ and sinφ have different values. ρ and φ together: Specifying ρ = C and φ 1 φ φ 2 will draw out part of the surface of the sphere, from tilt to tilt. Note that this still isn t giving us a spherical solid, since I have a fixed value for ρ. To draw the ice cream cone solid, we need to fill it in by specifying 0 ρ C, φ 1 φ φ 2. Other solid wedges of spheres can be described by varying the bounds for radius and tilt, and you ll see some variations in the suggested problems. 214

216 The role of θ: The equation θ = C describes a plane perpendicular to the xy plane in exactly the same way it would in cylindrical. Allowing ρ to vary makes it look like it s drawing out a disk, but that disk extends infinitely - so really, a plane. Combined with the other parameters, specifying a range for θ will carve out wedges of the solid; for example 0 ρ 4, 0 φ π 2, 0 θ π 2 will give the eighth of a sphere that lies in the first octant. Integrals in spherical coordinates D f(x, y, z)dv = φ2 θ2 ρ2 φ 1 θ 1 ρ 1 f(ρsin φcos θ, ρsin φsinθ, ρcos φ)ρ 2 sinφdρdθ dφ Coming back to this thing, notice that (as with cylindrical coordinates), we pick up an extra factor in the integral: dv becomes ρ 2 sinφdρdθ dφ. Where that comes from is shown in the picture below, and I m simply going to link to a textbook explanation rather than rewriting the whole thing - it s a bit involved. So, to integrate over spherical regions Describe the region in terms of bounds on ρ, θ and φ. If the integral is given with x, y, z or r, θ, z bounds, sketch and reinterpret in spherical. The integral f(x, y, z)dv = φ 2 D the region. θ2 ρ2 φ 1 θ 1 We can also integrate a function over that region. In the case of D f(x, y, z)dv = φ2 θ2 ρ2 φ 1 be sure to convert the function over to spherical as well. Turn to the live examples to see it in action... θ 1 ρ 1 ρ 1 1ρ 2 sinφdρdθ dφ would give the volume of f(ρsin φcos θ, ρsin φsinθ, ρcos φ)ρ 2 sinφdρdθ dφ 215

217 Applications of double and triple integrals Density, mass, and volume We ve established that the volume of solid a solid region D can be computed from 1dV D where the triple integral itself can be expressed as iterated integrals in rectangular, cylindrical, or spherical - whichever is appropriate for the region. We ve also seen that we can integrate a function over a region in space: f(x, y, z)dv [or f(r, θ, z)dv or f(ρ, φ, θ)dv ] D and the natural question is well, what is that? D The answer so far has been to reason up by analogy - since you can also get volume using a double integral and integrating under a function, as in you can interpret b d z=f(x,y) a c 0 1dz dy dx = D b d a f(x, y, z)dv c D f(x, y)dy dx as the hypervolume under the 4D surface w = f(x, y, z). While a bit difficult to picture, that might be the interpretation you need in some contexts. However, that s not the only interpretation, and it really depends on what f(x, y, z) is representing. Instead of a fourth spatial dimension, it could be representing, say, the temperature of an object as a function of (x, y, z). The interpretation we re going to look at is to consider what you get if f(x, y, z) is a function giving the density of a 3D object at each point in space. 216

218 Constant density, volume, and mass Recall the basic relationship: density (ρ) equals mass (m) over volume (V ), or and therefore ρ = m V m = ρv If a solid region is of uniform (constant) density ρ, we can compute the mass immediately by multiplying ρ by the volume obtained through integration: m = ρ 1dV Notation note: the ρ here has nothing to do with the radius ρ used in spherical coordinates; it s just conventionally used for density. And will probably confuse things a bit if we also happen to be working in spherical. Variable density Now, suppose density varies as a function of (x, y, z): ρ = ρ(x, y, z). Now we have to consider a little arbitrary chunk of mass: D m = ρ(x i, y i, z i ) V Sum up all the little chunks in the usual Riemann sum, smooth it out with the limit, and we get m = ρ(x, y, z)dv D And that pretty much covers it for the concept, and the implementation; you ve already been setting up triple integrals and integrating functions. Only thing new here is a possible interpretation for what that function could be. And while I m using the (x, y, z) form in the explanation, these problems could also be occuring in cylindrical and spherical coordinate systems as well. Keep going for a couple of worked examples, and see the suggested problems for some more. 217

219 Examples: Example 1: Suppose the solid bounded by x = y 2, z = x, z = 0, y = 0, x = 1 has uniform density ρ = 3. Find the mass of the solid. Additional notation note: yes, we lack units. Perk of a math class. If it makes you feel better, imagine the (x, y, z) coordinate system is position in meters, and the density is given in kg m 3. That ll make the mass come out in kg. The hardest thing about this problem is figuring out what that region looks like. If I had to hand sketch, I d note that just x = y 2, y = 0, x = 1 is pretty easy to deal with - it describes a parabola in the xy plane, and so a parabolic cylinder in x, y, z: The plane z = x runs parallel to the y axis and slices through that cylinder, forming the top (the bottom is just z = 0). And if all else fails, there s always Maple (code for plot attached at end). 218