Vector Calculus Lecture Notes

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1 Vector Calculus Lecture Notes by Thomas Baird December 5, 213 Contents 1 Geometry of R Coordinate Systems Distance Surfaces Vectors Geometric approach Algebraic Approach Dot Product Cross Product Equations for a Line Parametric equations Symmetric equations Equations for a Plane Parametric equations Symmetric equation Vector-valued Functions of a Single Variable and Space Curves Derivatives Arc Length Curvature Scalar-valued Functions of Several Real Variables Visualizing multi-variable functions Limits and Continuity [Review] Partial Derivatives [Review] Higher derivatives [Review] Partial Differential Equations [Review] Harmonic Functions and the heat equation [Review] Wave equation [Review] Tangent Planes and Linear Approximation Chain Rule [Review] Directional Derivatives and Gradients

2 3.8 Maxima and Minima [Review] Absolute maxima and minima [Review] Lagrange Multipliers Multiple constraints Integrating Multi-variable Scalar Functions Integrating Two Variable Scalar Functions Integrating over regions in the plane [Review] Polar coordinates General change of coordinates in dimension two Integrating Three Variable Scalar Functions Change of variables in dimension three Cylindrical coordinates Spherical coordinates Calculus of Vector Fields Line Integrals and the Fundamental Theorem Green s Theorem Parametrized Surfaces Surface Area and Surface Integrals Integrating Vector Fields over Surfaces: Flux Curl and Stokes Theorem Curl Stokes Theorem Div and the Divergence Theorem The divergence of a vector field Brief digression on differential forms Back to the Divergence Theorem Applications to Electromagnetism The Lorentz Force Maxwell s Equations Electrostatics Electro-Magnetic Waves Potentials Geometry of R Coordinate Systems Recall that the Cartesian plane is the set of ordered pairs of real numbers. R 2 R R {(a, b) a, b R} 2

3 For a point p (a, b) R 2 the numbers a, b are called the x- and y-coordinates of p respectively. Geometrically, R 2 is represented as a plane with a horizontal x-axis and vertical y-axis Proceeding by analogy, Cartesian 3-space is the set of ordered triples of real numbers: R 3 R R R {(a, b, c) R]} For a point p : (a, b, c) R 3 the numbers a, b, c are the x-, y- and z-coordinates of p respectively. R 3 is represented geometrically with three coordinate axes. 3

4 The point O (,, ) is called the origin. A plane spanned by any two axes is called a coordinate plane; there are three: the xy-plane, the xz-plane and the yz-plane. More generally, one may consider Cartesian n-space, R n, for any positive integer n. In this course we will mainly be concerned with doing calculus in R 3, but many of the ideas extend directly to any n Distance For and Q be two points in R 3, denote by P Q the straight line segment joining them The (Euclidean) distance between P : (x 1, y 1, z 1 ) and Q : (x 2, y 2, z 2 ) is defined by the formula P Q (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2 This quantity corresponds to the usual geometric notion of length of the line segment P Q by the following argument. Consider the diagram 4

5 The rectangle is drawn with sides parallel to the coordinate planes. The side lengths satisfy P A x 1 x 2, AB y 1 y 2 and BQ z 1 z 2. The points P, A, B form the vertices of a right triangle in a plane parallel to xy-plane, so using the usual Pythagorean theorem P B P A 2 + AB 2 (x 1 x 2 ) 2 + (y 1 y 2 ) 2 The points P, B, Q also form the vertices of a right triangle so Surfaces P Q P B 2 + BQ 2 (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2 Recall that the solution set of an equation in variables x and y (normally) determines a curve in R 2. An equation in the variables x, y, z (normally) determines surface in R 3. Problem 1. Graph the solution set of y 7 5

6 The sphere in R 3 centered at C with radius R is by definition the set {P R 3 P C R}. Using the distance formula, this sphere is the solution set of the equation R (x a) 2 + (y b) 2 + (z c) 2 or alternatively R 2 (x a) 2 + (y b) 2 + (z c) 2 Problem 2. Graph the solution set of 4 (x 1) 2 + (y) 2 + (z + 2) Vectors Geometric approach Given two points P, Q in R 3 (or R n ), let P Q denote the arrow pointing from P to Q. 6

7 This arrow represents a vector in R 3. The data defining a vector consists of its length or norm P Q P Q and its direction. In particular, two arrows that are related by a translation represent the same vector. There are two basic operations that can be performed on vectors. Vector addition: Given two vectors u and v in R 3 we form a new vector, u + v, by the triangle rule: In words, translate v so that its base is at the tip of u and make u + v the arrow spanning from the base of u to the tip of v. Notice that by the following diagram 7

8 u + v v + u. (this is sometimes called the parallelogram law). The unique vector of length zero is denoted and satisfies + v v. Scalar multiplication: If c R and u a vector, then we may form a new vector c u called the scalar product of u with c. The magnitude of c u satisfies c u c u. If c then c u is the vector with the same direction as u and if c < then c v points in the opposite direction to u. Observe that u + ( 1) u. The set of vectors in R 3 equipped with these operations is called the space of vectors in R Algebraic Approach Observe now that any arrow in R 3 can be translated to a unique arrow based at the origin. 8

9 Consequently, there is a one-to-one correspondence { vectors in R 3 } R 3 It is customary to identify the set of vectors in R 3 with R 3 itself, even though R 3 as a vector space is different from R 3 as a coordinate space. We will often blur the distinction between points and vectors, but when I want to distinguish the two I will write a 3-tuple < a, b, c > when it is a vector and (a, b, c) when it is a point. In terms of coordinates, the basic operations are: Vector addition: < a 1, a 2, a 3 > + < b 1, b 2, b 3 >< a 1 + b 1, a 2 + b 2, a 3 + b 3 > Scalar multiplication: c < a 1, a 2, a 3 >< ca 1, ca 2, ca 3 > Norm: < a 1, a 2, a 3 > a a a 2 3 The above formulas generalize naturally to R n for any positive integer n. Here are some properties of R n as a vector space. Proposition 1.1. Let a, b, c be vectors in R n and c, d scalars. Then i) a + b b + a v) (c + d) a c a + d a ii) ( a + b) + c a + ( b + c) iii) a + a iv) a + ( 1) b vi) (cd) a c(d a) vii) c( a + b) c a + c b viii) 1 a a i), iii) and iv) we ve already seen geometrically. We do v) algebraically as an illustration and leave the rest as an exercise. (c + d) a < (c + d)a 1,..., (c + d)a n > < ca 1 + da 1,..., ca n + da n > < ca 1,..., ca n > + < da 1,..., da n > c a + d a When working in R n it is useful to distinguish a set of standard basis vectors. In R 3 these are the vectors i :< 1,, > j :<, 1, > k :<,, 1 >. Any other vector in R 3 can be expressed as a linear combination of i, j, k as follows: 9

10 < a 1, a 2, a 3 > a 1 < 1,, > +a 2 <, 1, > +a 3 <,, 1 > a 1 i + a 2 j + a 3 k A unit vector is a vector of length one. If a is a non-zero vector, the vector 1 a is the a unique unit vector pointing in the same direction as a. Problem 3. Find the unit vector u pointing in the same direction as a 4 i 2 j + k. a ( 2) , so u 1 21 a 4 21 i j k 1.3 Dot Product The dot product is a function that inputs a pair of vectors and outputs a real number. For vectors a :< a 1, a 2, a 3 > and b < b 1, b 2, b 3 > the dot product is defined a b a 1 b 1 + a 2 b 2 + a 3 b 3 The dot product makes sense in any R n by the general formula a b n a i b i i1 The geometric meaning of the dot product is captured by the following theorem Theorem 1.2. Let a and b be two vectors in R 3 ( more generally R n ), and let θ be the angle between them. Then Proof. See the textbook. a b a b cos(θ) Corollary 1.3. The angle θ between vectors a and b is given by the formula cos(θ) a b a b We say that two vectors are perpendicular or orthogonal if the angle between them is 9 degrees. Corollary 1.4. Vectors a and b are orthogonal if and only if a b. Problem 4. Show that 7 i 2 j + k is orthogonal to i + 2 j 3 k. (7 i 2 j + k) ( i + 2 j 3 k)

11 Definition 1. Let a and b be vectors in R n. The projection of a onto b a vector defined by the formula ( a b ) proj b ( a) b b 2 Proposition 1.5. Geometrically, the projection of a vector can be understood by the following picture: Proof. ( a b ) b proj b ( a) b b a cos(θ) u where θ is the angle between a and b and u is the unit vector in the direction u. The assertion follows from the diagram. Exercise 1. Show that ( a proj b ( a)) b. Consequently the sum a proj b ( a) + ( a proj b ( a)), decomposes a into sum of a vector orthogonal to b and one parallel to b. 11

12 1.4 Cross Product The cross product is a function that inputs two vectors in R 3 and outputs a vector in R 3. Unlike the dot product, the cross product is special to R 3. If a, b R 3, then the cross product is a b (a 2 b 3 a 3 b 2, a 3 b 1 a 1 b 3, a 1 b 2 a 2 b 1 ) (1) The formula for the cross product is most easily remembered and understood as the determinant of a matrix. Consider the matrix, i j k a 1 a 2 a 3 (2) b 1 b 2 b 3 with the standard basis vectors occuring as entries in the first row, and a and b occuring as rows two and three respectively (it may seem strange that vectors are occuring both as rows and as matrix entries, but bear with me). Because scalars and vectors can be multiplied, it makes sense to take the determinant of (2), i j k i a 2 a 3 b 2 b 3 j a 1 a 3 b 1 b 3 + k a 1 a 2 b 1 b 2 a 1 a 2 a 3 b 1 b 2 b 3 giving a new formula for the cross product. (a 2 b 3 a 3 b 2 ) i (a 1 b 3 a 3 b 1 ) j + (a 1 b 2 a 2 b 1 ) k a b Corollary 1.6. Suppose a and b are linearly dependent, i.e. for some scalar c, c a b or a c b. Then a b. Proof. In the first case, a c b then by row deduction a i j k b a 1 a 2 a 3 b 1 b 2 b 3 i j k cb 1 cb 2 cb 3 b 1 b 2 b 3 i j k b 1 b 2 b 3 The case b c a is similar.. (3) Proposition 1.7. For any three vectors a, b, c R 3 we have ( a c 1 c 2 c 3 b) c a 1 a 2 a 3 b 1 b 2 b 3 (4) Equation 4 is sometimes called the scalar triple product of a, b and c. 12

13 Proof. ( a b) c (a 2 b 3 a 3 b 2 )c 1 (a 1 b 3 a 3 b 1 )c 2 + (a 1 b 3 a 3 b 1 )c 3 a c 2 a 3 1 b 2 b 3 c 2 a 1 a 3 b 1 b 3 + c 3 a 1 a 2 b 1 b 2 c 1 c 2 c 3 a 1 a 2 a 3 b 1 b 2 b 3 In some sense formula (4) is more fundamental than (1). That is, a b should be defined to be the vector satisfying (4), and formula (1) worked out as a calculation. Corollary 1.8. The crossproduct a b is orthogonal to both a and b. Proof. By Corollary (1.3) this equivalent to showing ( a b) a ( a b) b. Using formula 4 ( a a 1 a 2 a 3 b) a a 1 a 2 a 3 b 1 b 2 b 3 because two of the rows are equal. Similarly, for ( a b) b. Corollary 1.8 determines the direction of a b up to a sign. The sign is determined as follows. If a and b are linearly independent, then they span a plane in R 2. From one side of the plane, the angle sweeping counter-clockwise from a to b is less than 18 degrees. This is the side a b points into. This can remembered using the right-hand rule. Exercise 2. Illustrate the right hand rule with the example i j k. It remains to understand the norm a b geometrically. Theorem 1.9. For a, b R 3, let θ [, π] be the angle between a and b. Then a b a b sin(θ) (5) 13

14 Proof. a b 2 (a 2 b 3 a 3 b 2 ) 2 + (a 3 b 1 a 1 b 3 ) 2 + (a 1 b 2 a 2 b 1 ) 2 (a a a 2 3)(b b b 2 3) (a 1 b 1 + a 2 b 2 + a 3 b 3 ) 2 a 2 b 2 a b a 2 b 2 (1 cos 2 (θ)) Taking (positive) square roots completes the argument. a 2 b 2 sin 2 (θ) Corollary 1.1. Vectors a and b in R 3 are parallel if and only if a b. Even more geometrically, (5) can be interpreted as the area of the parallelogram spanned by a and b as shown. Proposition Given a, b, c R 3, the quantity ( a b) c is equal to the volume of the parallel-piped spanned by a, b and c. Proof. Drawing the parallel-piped the volume V, satisfies V Ah where A is the area of the base pararellogram spanned by a and b and h is the height measure orthogonal to the base. By earlier results we have A a b and h proj a b ( c). Thus V Ah a b ( a b) c a b a b ( a b) c. 2 14

15 One application of the cross product in physics is the concept of torque. Consider a force f acting on a rigid body in R 3 at a point r. The torque of this action (relative to the origin) is the cross product τ r f. If there are multiple forces f 1,..., f n acting at positions r 1,..., r n then the total torque is a sum of vectors τ n r i f i i1 The analogue of Newton s second law of motion for says that the angular velocity of a rigid body will accelerate around the axis spanned by τ, at a rate proportional to the norm of τ. In particular, if the rigid body is in equilibrium, then τ. Problem 5. Consider see-saw, with one arm of length 3m and the other of length 5m. Suppose a person weighing 12 pounds sits on the long end of the see-saw. How much weight must be placed at the other end if the see-saw is to remain balanced. Does this weight depend on the angle of inclination of the see-saw? We list some properties satisfied by the cross product. Theorem Let a, b, c be vectors in R 3 and let λ R be a scalar. Then i) (λ a) b a (λ b) λ( a b) ii) a ( b + c) a b + a c iii) ( a + b) c a c + b c iv) a b b a v) a ( b c) ( a b) c The first three properties can be summarized: the cross product is linear in both arguments. Proof. Of v) only. By earlier formulas ( a b) c 15 c 1 c 2 c 3 a 1 a 2 a 3 b 1 b 2 b 3

16 and a ( b c) ( b c) a a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 The first matrix can be transformed into the other using two row transpositions. It follows that determinants differ by a factor of ( 1) 2 1 and thus are equal. It is worthwhile noting that the following expressions are not equal in general: Consider the counterexample a ( b c) ( a b) c. ( i i) j j which is not equal to i ( i j) i k j. We say that the cross product is not associative. 1.5 Equations for a Line A line L in R 3 is completely determined by a point in L, and the direction of L. Using vectors, this geometric fact can be used to produce equations for L Parametric equations In the following diagram, let r denote a vector based at the origin and with tip on L, and let v be a vector pointing parallel to L. Using the triangle rule for vector addition, it is clear that for any other vector r L, the difference r r must be parallel to L thus also to v. So for some scalar c R we have r r t v. In particular the equation r r + t v (6) 16

17 has a unique solution t R for any r L. We call (6) a vector parametric equation for L and t is called the parameter. If v < a, b, c > and r < x, y, z >, then L is equal to the set of points (x, y, z) satisfying x x(t) x + ta, y y(t) y + tb, z z(t) z + tc for some t R. These are called scalar parametric equations for L. It is sometimes useful to think of t as a time variable and the parametric equations as describing the trajectory of a particle moving along L (without acceleration). Example 1. Produce parametric equations for the line L in R 3 passing through the points r : (1, 2, ) and r 1 : (, 3, 1). What values of t describe points on the line segment between r and r 1? The difference v r 1 r < 1, 5, 1 > is parallel to L, so we obtain the vector parametric equation r r + t v < 1, 2, > +t < 1, 5, 1 >< 1 t, 2 5t, + t > (7) In coordinates, L is the set of 3-tuples (x, y, z) satisfying x 1 t, y 2 5t, z t for some t R. Thinking in terms of a trajectory, notice that (7) equals r when t and r 1 when t 1, so the line segment between and 1 correspond to those values t 1. This equation for the line can be rewritten as follows r r + t v r + t( r 1 r ) (1 t) r + t r 1 (8) putting r and r 1 on a more even footing. Vectors r satisfying (8) for for t 1 (i.e. those lying on the line segment) are called convex linear combinations of r and r Symmetric equations It is possible to define a line L using equations in x, y, z without introducing an extra parametric variable like t. Consider again the parametric equations x x + ta, y y + tb, z z + tc If none of a, b, c are zero, then manipulating to isolate t, we get 17

18 x x y y z z a b c the solutions of which define L. These are called symmetric equations for L. If a and b, c do not, then we instead get equations y y x x, z z b c which means that the line lies in the plane x x. If a, b then get equations describing a line parallel to the z-axis. x x, y y. Example 2. i) Find parametric and symmetric equations for the line L passing through points (, 1, 1) and (1, 2, ). ii) At what point does L intersect the x z plane i) so x t, y 1 3t, z 1 t, and r <, 1, 1 > +t(< 1, 2, > <, 1, 1 >) <, 1, 1 > +t < 1, 3, 1 > < + t, 1 3t, 1 t > x y 1 3 z 1 1 ii) L interesects the x z-plane where y 1 3t. Thus t 1/3 and the intersection point is (1/3,, 2/3). 1.6 Equations for a Plane Parametric equations A plane P is determined by a point r in P and by two linearly independent vectors u and v which are parallel to P. This gives rise to a parametric vector equation for P : Every point r on P can be expressed 18

19 r r + s u + t v for a unique choice s, t R. Expressing this equation in coordinates give scalar parametric equations for P. If r < x, y, z >, u < u 1, u 2, u 3 > and v < v 1, v 2, v 3 > then P consists of points (x, y, z) satisfying: x x(s, t) x +su 1 +tv 1, y y(s, t) y +su 2 +tv 2, z z(s, t) z +su 3 +tv 3. for some s, t R. We think of the parameters s and t as defining a new set of coordinates for P. Example 3. Find parametric equations for the plane P containing the points (1,, ), (, 1, ), (, 1, 2). Any difference of distinct points in the plane determines a vector parallel to the plane. Thus u < 1,, > <, 1, >< 1, 1, > and v < 1,, > <, 1, 2 >< 1, 1, 2 > are parallel to P and are linearly independent. This gives parametric equations for P : and r < 1,, > +s < 1, 1, > +t <, 1, 2 > Symmetric equation x 1 + s, y s + t, z 2t Like in the case of lines, it is possible to define a plane using equations in x, y, z without introducing auxiliary variables like s, t. In fact, a single equation suffices to define a plane. A normal vector n to a plane P is a vector that is perpendicular to P (i.e. perpendicular to all vectors that are parallel to P ). For instance, if P is given by a parametric equation r r + s u + t v 19

20 then the cross product n : u v is perpendicular to both u and v and thus to the entire plane P. Given a vector r < x, y, z > lying on P and a normal vector n < a, b, c >, it is clear that any other vector r < x, y, z > in P must satisfy or in coordinates n ( r r ) so P is equal to the solutions of this equation. a(x x ) + b(y y ) + c(z z ) (9) Example 4. Find a parametric equation for the plane determined by 5x 2y + z 4 Solution: First manipulate the expression to put it into the form of (9). For example equation (1) is equivalent to 5x 2y + (z 4) (1) So the plane is normal to n < 5, 2, 1 > and passes through the point (,, 4). To get two vectors parallel to the plane, we choose vectors whose dot product with n is zero, say u :< 1,, 5 > and v :<, 1, 2 >. Then a parametric equation for P with parameters s, t R r :<,, 4 > +s <, 1, 2 > +t < 1,, 5 > 2 Vector-valued Functions of a Single Variable and Space Curves A vector valued function of a real variable is a function whose input is a real number and whose output is a vector (we will focus on this case that the output is a vector in R 3 ). That is, a vector valued function r : I R 3 associates to any real number in the domain interval t I R, a vector r(t) R 3. If f(t), g(t) and h(t) are the components of the vector r(t) then we write 2

21 r(t) < f(t), g(t), h(t) > f(t) i + g(t) j + h(t) k where f(t), g(t), h(t) are real valued functions. We define the limit of a vector-valued function to be the limit of its components: lim < f(t), g(t), h(t) >< lim f(t), lim g(t), lim h(t) >. t t t t t t t t A vector valued function is continuous if lim t t ( r(t)) r(t ) for all t R. This is equivalent to the components f(t), g(t), h(t) being continuous. Given a continuous vector valued function of one variable r :< f(t), g(t), h(t) > the set C of points (x, y, z) R 3 satisfying x f(t), y g(t), z h(t) for some t I is called a space curve. The vector-valued function r(t) is a called a parametrization of C and t is the parameter. Example 5. The parametric equation for a line r(t) : r +t v introduced in section determines a parametrized line. Problem 6. Sketch the curve parametrized by r(t) :< sin(t), cos(t), t >. Solution: Using the identity sin 2 (t) + cos 2 (t) 1, we deduce that the curve lies on the cylinder x 2 +y 2 1 winding around at a regular pace. The z coordinate also increases at a constant rate. The curve spirals upwards around the cylinder, forming a helix. Problem 7. Find a parametrization for the curve C defined by the equations x 2 + y 2 1 and x + z 3. Solution:The equation x 2 + y 2 1 determines a cylinder. The equation x + z 3 defines a plane orthogonal to < 1,, 1 > containing the point (,, 3). Plotting: 21

22 We have seen that the x sin(t), y cos(t) will define a curve winding around the cylinder. To control the z coordinate, solve sin(t) + z 3 to z 3 sin(t). Thus a parametrization is r(t) < sin(t), cos(t), 3 sin(t) >. If we want to parametrize the curve so that each point is counted only once, we can restrict the domain t < 2π. 2.1 Derivatives A vector-valued function r(t) is called differentiable at t if the limit ( r) (t) d r(t) dt : lim h r(t + h) r(t) h exists and we call ( r) (t) the derivative of r(t). In terms of coordinates, r(t) < x(t), y(t), z(t) > is differentiable if and only all of x(t), y(t), and z(t) are all differentiable and we have the equality ( r) (t) < x (t), y (t), z (t) >. We call ( r) (t) the derivative of r(t). If we interpret r(t) as the trajectory of a moving particle, ( r) (t) is is the velocity vector of the particle and the norm ( r) (t) is the speed of the particle. In the geometric terms, d r/dt is a limit of scalar multiples of secant vectors to the path. This means in particular that d r/dt is tangent to the curve. The tangent line to the curve parametrized by r(t) at the point r(t ) for some constant t, is the line with parametrization l and parameter s: l(s) : r(t ) + s d r(t ) dt This line provides the best linear approximation to the curve near the point r(t ). 22

23 Problem 8. For the curve with parametrization r(t) :< sin(t), cos(t), t > calculate: i) The derivative d r(t)/dt. ii) The linear approximation of r(t) at (1,, π/2). Solution: i) Simply differentiate coordinate-wise. d r(t)/dt < dsin(t)/dt, dcos(t)/dt, dt/dt >< cos(t), sin(t), 1 >. ii) r(π/2) (1,, π/2) to t π/2. Then the tangent line is parametrized by l(s) : (1,, π/2) + s < cos(π/2), sin(π/2), 1 > (1,, π/2) + s <, 1, 1 >. Theorem 2.1. Let u(t) and v(t) be vector valued functions, c R a scalar and f(t) a real-valued function. The following differentiation rules hold. (i) d [ u(t) + v(t)] d u(t)/dt + d v(t)/dt dt (ii) d [c u(t)] c(d u/dt) dt (iii) d dt [f(t) u(t)] f (t) u(t) + f(t)( u) (t) (iv) d dt [ u(t) v(t)] ( u) (t) v(t) + u(t) ( v) (t) (v) d dt [ u(t) v(t)] ( u) (t) v(t) + u(t) ( v) (t) ( addition rule) (scalar multiplication rule) (more general scalar product rule) (dot product rule) (cross product rule) (vi) d dt [ u(f(t))] ( u) (f(t))f (t) f (t)( u) (t) (chain rule) Proof. All are proven by passing to coordinates and applying the standard differentiation rules for real-valued functions. For example, to prove iv) let u < u 1, u 2, u 3 > and v < v 1, v 2, v 3 >, then d dt [ u(t) v(t)] d dt [u 1v 1 + u 2 v 2 + u 3 v 3 ] u 1v 1 + u 1 v 1 + u 2v 2 + u 2 v 2 + u 3v 3 + u 3 v 3 u 1v 1 + u 2v 2 + u 3v 3 + u 1 v 1 + u 2 v 2 + u 3 v 3 ( u) v + u ( v) 23

24 Problem 9. Show that if the norm of a vector valued function is constant, r(t) c, then the derivative ( r) (t) is perpendicular to r(t) for all t. Solution: Abbreviate r r(t). If r(t) c then r r c 2 is constant and Thus r ( r) and r is perpendicular to ( r). 2.2 Arc Length d dt ( r r) ( r) r + r ( r) 2 r ( r). Let C be a curve in R 3 with differentiable parametrization r(t). The arc length of C is defined to be the limit of lengths of inscribed polygonal paths, as illustrated below: This definition can made more precise using Riemann sums, but we will avoid this language. Instead, we will define length using a version of the fundamental theorem of calculus. Because C admits a differentiable parametrization r(t), it is possible to express the arc length using an integral equation. Recall that the derivative ( r) (t) may be interpreted as the velocity of a particle moving along C and ( r) (t) as the speed. Consequently, it makes sense that the length L of the curve between points r(a) and r(b) is equal to the integral of the speed function, L b a ( r) (t) dt (11) Observe that ( r) (t) is a real number, so the expression (11) is simply the integral of a real valued function. If r(t) < x(t), y(t), z(t) > then (11) can be rewritten L b a x (t) 2 + y (t) 2 + z (t) 2 dt We can also define the arc-length function by making one limit of integration a parameter s(t) : t a ( r) (h) dh. 24

25 Example 6. Determine the length of the curve C parametrized by r(t) :< sin(t), cos(t), t > between the points (, 1, ) and (1,, π/2). Solution: r(t) passes (, 1, ) at t a and crosses (1,, π/2) at t b π/2. Thus the length of the arc equals π/2 ( r) (t) dt π/2 < cos(t), sin(t), 1 > dt π/2 cos 2 (t) + sin 2 (t) + 1 dt π/2 2dt π/ 2 A parametrization u(t) of C is called an arc-length parametrization if the norm of the velocity vector equals 1 for all t: ( u) (t) 1 that is, the particle moves with unit speed. Let r(t) be a differentiable parametrization of C and suppose that ( r) (t) > for all t (such a parametrization is called regular or smooth). Then the arc-length function s(t) is a strictly increasing function of t, and it is possible to find an inverse function so t t(s). Define the arc-length parametrization u(s) by the formula u(s) r(t(s)) To see that u(s) has unit speed, differentiate: so d ds [ u(s)] d ds [ r(t(s))] ( r) (t) dt ds d ds [ u(s)] ( r) (t) dt/ds ds dt dt ds 1 Example 7. Consider the case r(t) < sin(t), cos(t), t >. We showed before that is a constant. The arc-length function is ( r) cos 2 (t) + sin 2 (t) s(t) t 2dt 2t. 25

26 The inverse function is t(s) 1 2 s so the arc-length parametrization of the curve is u(s) r(t(s)) < sin( 1 s), cos( 1 1 s), s > Curvature Suppose that r(t) is a regular vector-valued function. Geometrically, this implies that the curve parametrized by r(t) has no corners or cusps For such a curve, it is possible to define the arc-length parametrization u(s). The curvature of C at the point u(s) is defined to be the scalar κ : d2 u(s) ds 2. This measures how quickly the unit tangent vector is rotating along the curve as the point varies with unit speed, or in other words how much C curves. The formula above is conceptually useful, but it can be complicated to calculate directly because computing the arc-length parametrization explicitly can be difficult. Thus it is useful to have a formula that works for any (regular) parametrization r(t). Proposition 2.2. The curvature κ of a curve with regular parametrization r(t) satisfies κ ( r) ( r) ( r) 3 Proof. First, notice that d u/ds is simply the unit vector pointing in the same direction as ( r) (t). Thus, T : d u/ds ( r) ( r). We call T the unit tangent vector to the curve. Now by the chain rule, κ d 2 u/ds 2 d T ds d T dt Now we simplify even more by using some tricks. orthogonal to its derivative T. It follows that dt ds T ds/dt T ( r) T T T T sin(π/2) T 26 (12) Since T is a unit vector, it is

27 Substituting into (12) we get κ T T. (13) ( r) This isn t much of a simplication so far, but bear with me. Observe that ( r) ( r) T ds dt T. (14) Differentiating both sides with respect to t, and employing the (scalar) product rule ( r) d2 s dt 2 T + ds dt T. (15) Take the cross product of (14) and (15) and use the fact the T T : thus ( r) ( r) ds d 2 s T dt dt T + 2 ( ds ) 2 T T dt T T ( r) ( r) ( r) 2 which plugged into (13) completes the proof. ( ds ) 2 T T ( r) 2 T T dt Example 8. Compute the curvature of the curve with parametrization r(t) < sin(t), cos(t), t >. Solution: We have ( r) (t) < cos(t), sin(t), 1 > and ( r) (t) < sin(t), cos(t), > Thus r r < + cos(t), sin(t), cos 2 (t) sin 2 (t) > < cos(t), sin(t), 1 > and r r cos 2 (t) + sin 2 (t) We saw before that r cos 2 (t) + sin 2 (t) , so κ r r r 3 2 ( 2)

28 To further explore the geometric meaning of κ we need more terminology. The principal unit normal vector N N(t) to the curve r is the unit vector pointing in the same direction as T : N : T Of course, this is only defined when T. The binormal vector B B(t) is defined by T B T N Since T and N are orthogonal unit vectors, B is a unit vector orthogonal to both T and N. Thus { T (t), N(t), B(t)} form an orthonormal basis for all t (when they are defined). A nice video illustrating how this basis varies with t can be found here. The plane spanned by N(t) and B(t) is called the normal plane to the curve at r(t). The plane spanned by T (t) and N(t) is called the osculating plane from the Latin osculum, meaning to kiss. Now we can explain the geometric meaning of the curvature. In the osculating plane at r(t), draw a circle with radius 1/κ(t) which is passes through r(t) and whose centre is in the direction of N(t). This circle is not only tangent to the curve, but has the same curvature, normal and binormal at r(t). This is called the osculating circle, or kissing circle. Here are some videos of osculating circles in two dimensions and in three dimensions. A parametric equation c(t) for the osculating circle to the curve r(t) at time t can be found as follows. The centre of the circle is r(t ) + 1 κ(t ) N(t ) and the circle lies in the plane spanned by orthonormal vectors T (t ) and N(t ), so 28

29 c(t) : r(t ) + 1 κ(t ) N(t ) + 1 κ(t ) (cos(t) T (t ) + sin(t) N(t )) r(t ) + 1 κ(t ) (cos(t) T (t ) + (1 + sin(t)) N(t )) is a constant speed parametrization of the osculating circle with period 2π. Example 9. Find the principal unit normal and binormal vectors to the curve r(t) (sin(t), cos(t), t), and find a parametrization of the osculating circle at the point (, 1, ). Solution: ( r) < cos(t), sin(t), 1 > has constant norm 2, so T (t) 1 2 < cos(t), sin(t), 1 >. Thus T (t) 1 2 < sin(t), cos(t), > and Taking cross products N(t) < sin(t), cos(t), >. B(t) T N 1 2 < cos(t), sin(t), 1 >. The point (, 1, ) r(), so there the osculating plane is spanned by T () 1 2 < 1,, 1 > and N() <, 1, >. The curvature is κ 1/2 (from Example 8), so the osculating circle has radius 2 and is centred at and we get the equation 2 N() + r() 2 <, 1, > + <, 1, ><, 1, >. c(t) <, 1, > (cos(t) < 1,, 1 > +(1 + sin(t)) <, 1, >) 2 cos(t) 2 2 < cos(t) 2 2 < 1,, 1 > +3 + sin(t) 2 3 sin(t),, cos(t) > <, 1, > 29

30 3 Scalar-valued Functions of Several Real Variables Definition 2. A scalar-valued function of several real variables f is a function whose input is a point (x 1,..., x n ) R n and whose output is a scalar f(x 1,..., x n ) R. The set of points in R n where the function is defined is called the domain of f. The set of values in R attained by f is called the range of f. Example 1. The function f(x, y) 2 + x xy + x 3 is a function of two variables. 3.1 Visualizing multi-variable functions A two variable function can be visually represented in a number of ways. Definition 3. The graph of a two variable function f(x, y) is the set of points in (x, y, z) R 3 satisfying the equation z f(x, y). This determines a surface in R 3 (if f is continuous). Example 11. Suppose that (x, y) are latitude and longitude coordiates on a map, and the h(x, y) equals the height above sea level of the land at the point (x, y). Then graph z h(x, y) recreates the mountains and valleys of the landscape. Example 12. Draw the graph of the function f(x, y) 2 + x y. Solution: This means plot the solutions to the equation z 2 + x y. This equation is equivalent to x y z + 2 < x, y, z 2 > < 1, 1, 1 > (< x, y, z > <,, 2 >) < 1, 1, 1 > which is the equation of the plane through (,, 2) with normal vector < 1, 1, 1 >. 3

31 Example 13. Draw the graph of the function f(x, y) 1 x 2 + y 2, with domain {(x, y) x 2 + y 2 1}. Solution: Solve the equation z 1 x 2 + y 2. This is equivalent to z 2 1 x 2 y 2 and z. Equivalently x 2 + y 2 + z 2 1 and z. These equation describes the upper hemisphere of the sphere with radius one, centered at the origin Another common way to represent a function of two variables is using level curves. Definition 4. A level curve of a function f of two variables is the set of solutions in R 2 of the equation f(x, y) k, where k is constant scalar in the range of f. Level curves are also called contour lines. Example 14. Recall the example h(x, y) where (x, y) are longitude and latitute on a map and h(x, y) is the height function. If we draw enough level curves, we get a topographical map. The contour lines describe paths of constant height. Example 15. Level curves are also frequently used on weather maps. In these examples (x, y) are again longitude and latitute, f(x, y) might be the temperature, or the air pressure. Often, colour or shading is put in between contour lines to indicate the value of the function in between. Example 16. Draw level curves for the function f(x, y) 4x 2 + y 2. Solution: The level curves are those of the form 4x 2 + y 2 k for k a constant scalar. Such a curve is only non-empty when k. At k get a single solution, the origin. When k > get a one dimensional curve - an ellipse centred at the origin. 31

32 We will also be interested in scalar valued functions of three variables: f(x, y, z). This is a function whose input is an element of R 3 and whose output is a scalar in R. In this case it is not possible to plot the graph w f(x, y, z) because this would live in R 4. We can however consider level surfaces which are solutions to the equation k f(x, y, z) in R 3, where k is a constant. Example 17. Draw level surfaces of the function f(x, y, z) x 2 + y 2 + z 2. Solution: The equation k x 2 + y 2 + z 2 has solutions only when k. When k the origin is the only solution. When k > the level surface is the sphere centered at the origin of radius k. 3.2 Limits and Continuity [Review] Definition 5. Let A (a 1,...a n ) R n and let f be a scalar valued function of n-variables whose domain contains points arbitrarily close to A. We say that f has limit L R at A, denoted lim X A f(x) L if for every ɛ > there exists a δ > depending on ɛ, such that if X (x 1,..., x n ) R n satisfies X A (x 1 a 1 ) (x n a n ) 2 < δ then f(x) L < ɛ. This can be understood using the diagram 32

33 Example 18. Find the limit lim (x,y) (,) xy 2 x 2 + y 2 if it exists. Solution: Observe that y 2 x 2 + y 2 for any ordered pair (x, y). Thus, if (x, y) (, ) we have Thus for any ɛ > choose δ ɛ. Then if then So the limit exists and equals. xy2 x 2 + y 2 x y 2 x 2 + y 2 x x 2 + y 2 < (x, y) (, ) x 2 + y 2 < δ ɛ xy2 x 2 + y 2 x 2 + y 2 < ɛ Limits of multivariable functions can be related to limits of single variable functions in the following way. Suppose for concreteness that F (x, y, z) is a three variable function and lim F (x, y, z) L (x,y,z) (a,b,c) and let (f(t), g(t), h(t)) be a continuous parametrized curve or path in R 3 which crosses (a, b, c). I.e., for some constant value t R, we have (f(t ), g(t ), h(t )) (a, b, c). Then the limit lim t t F (f(t), g(t), h(t)) L. Example 19. Determine whether the function f(x, y) xy2 has a limit at (x, y) x 2 +y 4 (, ). Solution: First we try approaching (, ) along straight lines. We parametrize a line of slope m in the x y plane by (t, mt) and this passes through (, ) at time t. Then m 2 t 3 lim f(t, mt) lim t t t 2 + m 4 t lim m 2 t 4 t 1 + m 4 t 2 33

34 which is consistent with the limit being. Consider now a path approaching along the parabola y 2 x, say (t 2, t). Then lim f(t 2 t 2 t 2, t) lim t t t 4 + t lim 1 4 t 2 1/2 So we get different limits by approaching (, ) along different paths, so lim (x,y) (,) f(x, y) does not exist. Now that we know what limits are, we can speak of continuous functions. Definition 6. Let f be a scalar valued function of n-variables and let A (a 1,...a n ) R n be a point in the domain of f. We say that f is continuous at A, if lim f(x) f(a). X A The function f is called continuous if it is continuous at all points in its domain. Theorem 3.1. Let f(x) and g(x) be continuous real valued functions of several variables X R n and let h(x) be a continuous function of one-variable. Then, (i) Constant functions are continuous. (ii) Functions of the form F (x 1,..., x n ) h(x i ) are continuous. (iii) The sum f(x) + g(x) is continuous. (iv) The product f(x)g(x) is continuous. (v) The quotiennt f(x)/g(x) is continuous at all points where g(x). Proof. Identical to one-variable case. Example 2. Show that the function f(x, y) x2 + sin(y) x 4 y is continuous. Solution:The numerator x 2 + sin(y) is a sum of one-variable continuous functions, hence continuous. The function x 4 y 2 is a product of one-variable continuous functions, hence continuous and x 4 y 2 +1 is a sum of continuous, hence continuous. Note x 4 y 2 +1 > so the quotient x2 +sin(y) is continuous and everywhere well-defined. x 4 y Partial Derivatives [Review] Let f(x, y) be a function of two variables. We want to develop a notion of derivative for f(x, y). The derivative should be the rate of change of the function f(x, y), but the rate of change with respect to what? Consider the following diagram 34

35 We may try taking the rate of change of f(x, y) along a parametrized curve (x(t), y(t)). As yesterday, we may compose these functions to get f(x(t), y(t)), a scalar valued function of a single variable - and this we know how to differentiate. Partial derivatives are obtained by applying this construction to arc-length parametrizations parallel to the coordinate axes. Definition 7. Let f(x, y) be a scalar-valued function of two variables. The partial derivative with respect to x of f at the point (a, b) R 2 is the limit f x (a, b) f f(a + h, b) f(a, b) (a, b) : lim x h h if this exists. Similarly, the partial derivative with respect to y of f at (a, b) is the limit f y (a, b) f f(a, b + h) f(a, b) (a, b) : lim y h h Notice that both f x and f y are both derivatives with respect to h of functions of the form f(x(h), y(h)) where (x(h), y(h)) (a + h, b) in the first case and (a, b + h) in the second. By varying (a, b), partial derivatives are made into scalar-valued functions themselves: and f x (x, y) f f(x + h, y) f(x, y) (x, y) : lim x h h f y (x, y) f f(x, y + h) f(x, y) (x, y) : lim y h h In practice, when calculating partial derivatives, we don t need to introduce the dummy variable h. To differentiate f(x, y) with respect to x, simply treat y like a constant and differentiate f(x, y) as though it were simply a function of x. Likewise to differentiate with respect to y. Example 21. Find the partial derivatives of the function f(x, y) x 2 + 2xy 3 y. Solution: To calculate f x, we treat y like a constant and differentiate with respect to x. f x (x, y) 2x + 2y 3 + 2x + 2y 3 To calculate f y, treat x like a constant and differentiate with respect to y. f y (x, y) + 6xy 2 1 6xy

36 Example 22. Find the partial derivatives of the function f(x, y) sin( x 1+y ). Solution: This time we must use the chain rule. and f x ( x (sin x 1 + y ) ( x ) ( x ) ( x ) 1 ) cos cos 1 + y x 1 + y 1 + y 1 + y f y ( x ) ( x ) ( x ) ( x ) ( x) y (sin ) cos cos 1 + y 1 + y y 1 + y 1 + y (1 + y) 2 It is useful to picture geometrically what the partial derivatives are measuring. The partial derivatives of f x and f y are measuring the slopes of the tangent lines to the graph z f(x, y) lying parallel to the x z-plane and y z-plane respectively. These two lines span a plane, called the tangent plane that we will study in greater depth later. All this makes extends without effort to scalar-valued functions of three or more variables: Definition 8. The partial derivatives of a scalar-valued function f(x, y, z) of three variables are defined: f x f f(x + h, y, z) x lim h h f y f f(x, y + h, z) y lim h h f z f f(x, y, z + h) z lim h h whenever these limits are defined Higher derivatives [Review] Since the output of a partial derivative is a scalar-valued function, it makes sense to iterate the process. For instance, the second order partial derivatives of f(x, y) are: (f x ) x f xx ( f ) 2 f x x ( x) 2 36

37 (f x ) y f xy ( f ) 2 f y x y x (f y ) x f yx 2 f x y (f x ) x f xx 2 f ( x) 2 An example of a third order partial derivative is: ((f x ) x ) y f xxy 3 f y( x) 2 Example 23. Calculate the second order partial derivatives of f(x, y) x 2 + xy 3 y. Solution: f x 2x + y 3 f xx 2, f xy 3y 2 f y 3xy 2 1 f yx 3y 2, f yy 6xy Notice that in this example f xy f yx. This is an example of Theorem 3.2 (Clairaut s Theorem). Let f be defined on a disk D containing the point (a, b) and suppose both f xy and f yx are continuous on D. Then Proof. Skipped f xy (a, b) f yx (b, a) Lets consider now the geometric meaning of the second order partial derivatives. Consider a graph z f(x, y) The partial derivative f xx and f yy describe the convexity of the surface along the x and y directions respectively. The meaning of f xy f yx is more subtle, and will be explained later when we study max-min problems. 37

38 3.4 Partial Differential Equations [Review] This subsection is to motivate some of these concepts and will not be tested Harmonic Functions and the heat equation [Review] Suppose that u(x, y, z) describes the heat density in a uniform medium ( say a block of metal). Heat tends to flow from hot to cold, so u will vary with time. Thus we can think of u u(x, y, z, t) as depending on four variables, with t time. The flow of heat satisfies the heat equation: c u t 2 u x + 2 u 2 y + 2 u 2 z 2 where c is some constant depending on the medium. Under reasonable conditions, this equation has only one solution given initial conditions u(x, y, z, ). If the domain of u is not all of R 3, then we must also impose boundary conditions. Example 24. Suppose take a (spherical) roast at room temperature (7 degrees F), and put it in an oven at 35 degrees F. This can be modeled as a region in R 3 for which initial conditions u(x, y, z, ) 7 for x 2 + y 2 + z 2 < 1 and boundary conditions u(x, y, z, t) 35 for x 2 + y 2 + z 2 1. The heat equation describes how the roast heats up in time. In the long run when t gets large, the heat function of the roast will approach the constant function u(x, y, z) 35. This is called the equilibrium state. In general, the equilibrium state is the solution to the Laplace equation 2 u x + 2 u 2 y + 2 u 2 z 2 which exists and is unique for a given set of boundary conditions. A function u(x, y, z) satisfying the Laplace equations is called harmonic. The Laplace equation comes up in many other contexts and is of great importance in physics and engineering. 38

39 3.4.2 Wave equation [Review] Now suppose that u(x, t) describes the motion of a vibrating string Then u(x, t) satisfies the wave equation: 2 u t 2 u 2 a2 x 2 where a is a positive constant. An solution to the wave equation is given by u(x, t) cos(x at). If the argument is set to zero, x at we see that x/t a so a is the speed of propagation of the wave. The wave equation also makes sense with more variables: 2 u ( 2 t u 2 a2 x + 2 u ) 2 x 2 This equation can describe the vibrations of a drum, or waves in deep water. 3.5 Tangent Planes and Linear Approximation The graph of a continuous two variable function, z f(x, y) determines a surface in R 3. It is often useful to try to approximate a surface near a point (x, y, z ) by a tangent plane. If the partial derivative f x and f y are continuous at (x, y ) then the tangent plane to the graph at (x, y, z ) exists and satisfies the equation: z z f x (x, y )(x x ) + f y (x, y )(y y ). To see why this plane works, observe that the plane is the graph of the function 39

40 p(x, y) : f x (x, y )(x x ) + f y (x, y )(y y ) + z Plugging in p(x, y ) z so the graph passes through (x, y, z ). Furthermore, taking partial derivative: p x (x, y) f x (x, y ), p y (x, y) f y (x, y ) so it has the same partial derivatives as f at (x, y ). The function p(x, y) is called the linear approximation of f(x, y) near (x, y ). The tangent plane may also be expressed using parametric equations. The tangent vectors (1,, f x (x, y )) and (, 1, f y (x, y )) are linearly independent so we get a parametrization: r(s, t) (x, y, f(x, y )) + s(1,, f x (x, y )) + t(, 1, f y (x, y )) Example 25. Let f(x, y) x 2 + 2xy 3 + y. Calculate both symmetric and parametric equations for the tangent plane to the graph of f at the point (1,, 1). Use the linear approximation to estimate the value of f(1.1,.1). Solution: First calculate partial derivatives: f x (x, y) 2x + 2y 3, f x (1, ) 2 f y (x, y) 6xy 2 + 1, f y (1, ) 1 Thus the tangent plane at (1,,1) has equation z 1 f x (1, )(x 1) + f y (1, )(y ) 2(x 1) + y and has parametric equation r(s, t) : (1,, f(1, ))+s(1,, f x (1, ))+t(, 1, f y (1, )) (1,, 1)+s < 1,, 2 > +t <, 1, 1 >. so The linear approximation is p(x, y) z 2(x 1) + y + 1 f(1.1,.1) p(1.1,.1) 2(.1) Informally, we say a scalar-valued multivariable function f(x, y) is differentiable at a point (a, b) if the linear approximation p(x, y) exists at (a, b) and is good near (a, b). A more precise definition is the following. Definition 9. A scalar-valued function of two variable f(x, y) is called differentiable, if the partial derivative f x (a, b) and f y (a, b) exists, so we can define a linear approximation and if the difference p(x, y) f(a, b) + f x (a, b)x + f y (a, b)y f(x, y) p(x, y) ɛ 1 (x, y)(x a) + ɛ 2 (x, y)(y b), where ɛ 1, ɛ 2 are continuous functions satisfying ɛ 1 (a, b) ɛ 2 (a, b). A function f(x, y) is simply called differentiable if it is differentiable at all points in its domain. 4

41 The following criterion is an easy way to verify the a function is differentiable. Theorem 3.3. A function f(x, y) is differentiable at (a, b) if the partial derivatives f x (x, y) and f y (x, y) exist near (a, b) and are continuous at (a, b). Caution: the converse of Theorem 3.3 is not true. I.e., it is possible for f to be differentiable even if the partial derivatives are not continuous. Example 26. Show that the function f(x, y) x 2 y 2y 2 is differentiable everywhere. Solution: Take partial derivatives f x (x, y) 2xy, f y (x, y) x 2 4y both of which are continuous (they are sums and products of one-variable continuous functions), so f is differentiable by Theorem 3.3. Example 27. Consider the function { yx 2 sin( 1 x f(x, y) ) + y if x y if x Show that f(x, y) is differentiable at (, 1). Solution: Taking partial derivative with respect to x at points x : ( 1 ( 1 )( 1 ) ( 1 ( 1 f x (x, y) 2yxsin + yx x) 2 cos 2xysin ycos x x x) x) 2 Observe that f x is not continuous at (, 1) because the term ycos( 1 ) rapidly between x ±1 as (x, 1) goes to (, 1), so we cannot apply Theorem 3.3. On the other hand, computing directly using the limit definition: f x (, 1) h 2 sin( 1 lim ) h (h,1) (,1) h by the squeeze theorem, because hsin( 1 ) h. Also h f y (, 1) lim h f(, 1 + h) lim h 1 + h 1 h Thus the linear approximation p(x, y) y. Sure enough, where lim (h,1) (,1) hsin( 1 h ) f(x, y) p(x, y) f(x, y) y yɛ(x, y) ɛ(x, y) { x 2 sin( 1 x ) if x if x 1. is continuous and equal to zero at (, 1). So f is differentiable at (, 1). All this stuff generalizes easily to three and more dimensions. 41

42 Proposition 3.4. If f(x, y, z) is a scalar-function, then The linear approximation at a point (a, b, c) is p(x, y, z) f(a, b, c) + f x (a, b, c)(x a) + f y (a, b, c)(y b) + f z (a, b, c)(z c). f is called differentiable if f(x, y, z) p(x, y, z) ɛ 1 (x a) + ɛ 2 (y b) + ɛ 3 (z c) where ɛ 1, ɛ 2, ɛ 3 are continuous functions vanishing at (a, b, c). f is differentiable at (a, b, c) if (but not only if) the partial derivatives f x, f y, f z exist and are continuous at (a, b, c). 3.6 Chain Rule [Review] Theorem 3.5. Let f(x, y) be a scalar valued function, which is differentiable at (a, b). Suppose (x(t), y(t)) is a differentiable path such that (x(t ), y(t )) (a, b) for some number t. Then composing we have f(x(t), y(t)) is single-variable function which is differentiable at t t and df dt f dx x dt + f dy y dt with functions evaluated at (a, b) and t as appropriate: df(x(t ), y(t )) dt f(a, b) dx(t ) + x dt f(a, b) dy(t ) y dt Proof. We use the limit definition for derivatives: df dt f(x(t + h), y(t + h)) f(x(t ), y(t )) tt lim. h h Before proceeding, we introduce simplifying notation. Let t t t h which we think of as the change in t. Let x x(t + t) x(t ) be the change in x, and y and f be the change in y and f. In this notation, df dt lim f t t. By assumption, f is differentiable at (a, b) so, f(x, y) f(a, b) + ( f x (a, b) + ɛ 1(x, y) ) x + ( f y (a, b) + ɛ 2(x, y) ) y. where lim (x,y) (a,b) ɛ i for both i 1, 2. Thus 42

43 f lim t t f(x(t), y(t)) f(a, b) t ( ( f ) x ( f ) y ) lim t x + ɛ 1 t + y + ɛ 2 t ( f ) lim t x + ɛ x ( f ) 1 lim t t + lim t y + ɛ 2 f x dx dt + f y dy dt lim t y t Example 28. If z x 2 y + xy 3 and x sin(3t) and y cos( t) determine dz/dt at the time t. Solution: Employing the chain rule: dz dt dz dx dx dt + dz dy dy dt (2xy + y3 )3cos(3t) + (x 2 + 3xy 2 )sin(t) At t we have x sin() and y cos() 1, so dz dt t Example 29. Determine if the function { y 3 if (x, y) (, ) x f(x, y) 2 +y 2 if (x, y) (, ) is differentiable at (x, y) (, ). Solution: Compute partial derivatives using the limit approach: f x (, ) lim h h h 3 /h 2 f y (, ) lim h h lim h 1 1 So the partial derivatives exist and determine a linear approximation p(x, y) y. Now consider the differentiable path (x(t), y(t)) (t, t), which passes through the origin at t. Composing with f and differentiating: df dt d ( t 3 ) d (t/2) 1/2 dt t 2 + t 2 dt However, according to the chain rule if f(x, y) is differentiable at (, ) df dt f dx x dt + f dy y dt /2 which is a contradiction, so f is not differentiable at (, ). 43

44 The chain rule works in more general situations. Suppose that f f(x, y) is a differentiable function of x, y and that x x(s, t) and y y(s, t) are differentiable functions of s, t. By composing we may consider f(x(s, t), y(s, t)) as a two variable function. In this expression, s, t are sometimes called independent variables and x, y are called intermediate variables. The chain rule for partial derivatives in this case is: f s f x x s + f y y s, f t f x x t + f y y t One way of remembering these formulas is using a tree diagram: Example 3. Let f f(x, y) x 2 4xy 4 and x e st y sin(s + t) determine the partial derivative f s as functions of s, t. Solution: f s f x x s + f y y s (2x 4y4 )te st + 16xy 3 cos(s + t) (2e st 4sin 4 (s + t))te st + 16e st sin 3 (s + t)cos(s + t) A multivariable scalar-valued function is differentiable at a point if its linear approximation exists and is a good approximation at that point, in similar fashion to two-variable functions. The chain rule extends to differentiable multi-variable, scalar-valued functions as follow: Theorem 3.6. Let f(x 1,..., x n ) be a differentiable function of n variables, and for each i 1,..., n let x i x i (t 1,..., t m ) be differentiable in m variables. Then the composition is differentiable function in m variables and the partial derivatives satisfy for each k 1, 2,..., m. f t k f x 1 x 1 t k f x n x n t k Proof. Since the partial derivative of f with respect to t i is defined by setting the other variables t j, j i to be constant, this follows immediately from Theorem

45 3.7 Directional Derivatives and Gradients The partial derivatives of a scalar function can be packaged together into a single vectorvalued function, called the gradient. Definition 1. Let f(x 1,..., x n ) be a multivariable scalar-valued function, differentiable at the point A (a 1,..., a n ). The gradient of f at A is a vector in R n defined by f(a) < f x1 (A),..., f xn (A) >. If f is differentiable on its domain, the gradient f is a vector-valued multivariable function f < f x1, f x2,..., f xn > In geometric terms, the gradient vector points in the direction of steepest ascent for the function, and the length of the vector is equal to the rate of change (or slope) in that direction. The gradient of a scalar function is our first example of a vector field. We will explore these in greater depth later. A vector field V is a function that associates to each point A in ( a region of ) R n a vector V (A) V A in R n. This can be represented visually when n 2 as follows: Example 31. Let f(x, y) 2x + y. Calculate the gradient f. Draw this vector field. Solution: The gradient is a constant function. f < f x, f y >< 2, 1 > Example 32. Let f(x, y) 1 2 (x2 + y 2 ). Calculate the gradient and draw the vector field. Solution: The gradient is f < f x, f y > 1 2 < 2x, 2y >< x, y >. 45

46 The gradient vector field consists of arrows radiating outwards and away from the origin, and growing longer the farther away. One of the nice properties of the gradient is that it provides a slick restatement of the Chain rule. Proposition 3.7. Let f(x 1,..., x n ) be a differentiable function, and let r(t) : (x 1 (t),..., x n (t)) be a differentiable path in R n taking values in the domain of f. Then the derivative of f with respect to t equals the dot product: Proof. By the chain rule as previously formulated f t ( f) r (t) (16) f t f x 1 f x n x 1 t x n t. The right hand side is equal to the dot product < f,..., f >< x 1 x 1 x n t,..., x n t > ( f) r (t) Observe that the formula (16) depends only on the velocity vector of the path r(t) and not on the path itself. It follows from Proposition 3.7 that the partial derivatives of f can be recovered from the gradient f. In two dimensions, we have < 1, > f 1f x + f y f x, <, 1 > f f x + 1f y f y Similarly, in three dimensions i f < 1,, > < f x, f y, f z > f x, j f f y, k f fz More generally a similar result holds for directional derivatives. Definition 11. Let u be a unit tangent vector in R n and let f(x 1,..., x n ) be a scalar function which is differentiable at a point A (a 1,..., a n ). The directional derivative of f at A in the direction u, is a real number denoted D u (f)(a) : u ( f). 46

47 By the Proposition 3.7, we may interpret D u (f)(a) as the rate of change of f at A in the direction u. The geometric interpretation of the gradient offered earlier can now be proven. Corollary 3.8. The gradient f points in the direction of steepest ascent for the function f and the magnitude f is the directional derivative in this direction. Proof. Let u be a unit vector and let θ be the angle between f and θ. Then D u (f) u f u f cos(θ) f cos(θ) which is maximized when θ, so when u points in the same direction as f, in which case D u (f) f. Corollary 3.9. The level sets of a differentiable scalar function f(x 1,..., x n ) are orthogonal to f. Proof. If r(t) is a differentiable path lying on a level set f(x 1,..., x n ) k for some constant k, then the composition f(r(t)) k is constant and has zero derivative. Thus f t ( f) r (t) Thus the velocity vector r (t) which is tangent to the level curve is orthogonal to f. Example 33. Recall the example f(x, y) 1 2 (x2 + y 2 + z 2 ). The gradient vector field looks like: The level curves must look be concentric circles centered at the origin. This is true because f(x, y) equals half the distance between (x, y) and the origin, so the level curves should be circles centered at the origin. This actually provides a helpful strategy to draw level curves, because gradients tend to be easy to compute. 47

48 3.8 Maxima and Minima [Review] Definition 12. Let f(x 1,..., x n ) be a scalar valued function. We say that f has a local maximum at A (a 1,..., a n ) if for all points X (x 1,..., x n ) in a neighbourhood of A, we have f(a) f(x). Similarly, we say that f has a local minimum at A if f(a) f(x) for all X in a neighbourhood of A. In both cases, we say that f has a local extremum at A. If the above inequality holds for all points X in the domain of f, we say that f has an absolute maximum, absolute minimum or absolute extremum respectively. Proposition 3.1. If f has a local extremum at A, f is defined in a neighbourhood of A and the partial derivatives are defined at A, then f(a). Proof. Consider the n 2 case for simplicity, so f f(x, y). If f has a local extremum at A (a, b) then the single variable function f(t, b) has a local extremum at t a. Fermat s theorem then implies that Similarly, f y (a, b), so df dt ta f x (a, b). f(a, b) < f x (a, b), f y (a, b) >. Definition 13. A point A in the domain of a differentiable function f is called a critical point if f(a). It follows from Proposition 3.1 that if f is differentiable everywhere, then local maxima and minima must occur at critical points. The converse is not true. Example 34. Find the critical points of f(x, y) x 2 y 2. Find all local maxima and minima. Solution: The gradient f < f x, f y >< 2x, 2y > is defined everywhere and vanishes only at (, ), so (, ) is the only critical point. However f(, ) while f(x, ) x 2 > for x near and f(, y) < for y near, so f(, ) is neither a local maximum, nor a local minimum. 48

49 A critical point like the one in this example is called a saddle point, because its graph is shaped like a saddle. One way to check whether a critical point is local minimum, maximum or saddle point is the second derivative test. Proposition 3.11 (Second Derivative Test). Suppose that f(x, y) is a scalar function of two variables, for which f(a, b) and the partial derivatives f x and f y are defined and continuous in a neighbourhood of (a, b). Define ( ) fxx (a, b) f D det xy (a, b) f f yx (a, b) f yy (a, b) xx (a, b)f yy (a, b) [fxy(a, b)] 2 If f xx (a, b) > and D > then f has a local minimum at (a, b). If f xx (a, b) < and D > then f has a local maximum at (a, b). If D < then f has a saddle point at (a, b) and it is not a local extremum. Otherwise, the test is inconclusive. Proof. Idea of proof: Consider the situation where f xy (a, b) f yx (a, b). Then the determinant simplifies to D f xx (a, b)f yy (a, b) If f xx (a, b) > and D > then f yy (a, b) >, so f is concave up in both the x and y-directions, and we expect f to have a local minimum at (a, b). Similarly, if f xx (a, b) < and D > then f yy (a, b) <, so f is concave down in both the x and y-directions, and we expect f to have a local maximum at (a, b). If D < then the concavity in the x and y directions differ, so we know that f does not have a local extremum at (a, b), and is shaped like a saddle. The general case can be reduced to the situation f xy (a, b) by a change of coordinates argument that we will omit. Example 35. Use the second derivative test to classify the critical points of the function Solution: Calculate the gradient, f(x, y) 2x 3 + xy 2 + 5x 2 + y 2. f < 6x 2 + y 2 + 1x, 2xy + 2y >< 6x 2 + y 2 + 1x, 2y(x + 1) >, which exists and is continuous everywhere. 49

50 Solving for f, we have 2y(x + 1), so y or x 1. If y, then 6x 2 + 1x so x or x 5/3. If x 1 then y 2 4 so y ±2. We obtain four critical points: (x, y) (, ), ( 5/3, ), ( 1, 2), ( 1, 2) Now calculate the second order derivatives, f xx 12x + 1, f xy f yx 2y, f yy 2(x + 1). At each critical point, form the matrix ( fxx f xy f yx f yy At (, ) get ( 1 2 f xx (, ) 1 > and D 2 > so f has a local minimum at (, ). At ( 5/3, ) get ( 1 ) 4/3 f xx 1 < and D 4/3 > so f has a local maximum at (, ). At ( 1, 2) get ( ) f xx 2 < and D 8 < so f has a saddle point at (, ). At ( 1, 2) get ( ) f xx 2 < and D 28 < so f has a saddle point at (, ). ) Absolute maxima and minima [Review] Recall the Extreme Value Theorem for single variable, scalar functions. Theorem Let f(x) be a continuous scalar function defined on a closed, bounded interval [a, b] R. Then f achieves an absolute maximum and absolute minimum value in [a, b]. This means there exists c, d [a, b] such that for every value x [a, b], ) f(c) f(x) f(d). This theorem extends to multi-variable scalar functions as follows. 5

51 Theorem Let f(x 1,..., x n ) be a continuous, scalar function defined on a closed, bounded region R. Then f achieves an absolute maximum and absolute minimum value on. This means there exists C (c 1,..., c n ) and D (d 1,..., d n ) such that for every point X (x 1,..., x n ), f(c) f(x) f(d). Of course, we need to define what closed and bounded means. We concentrate on the case n 2. A region R 2 is bounded if it can be covered with a sufficiently large disk: A region R 2 is closed if it contains all of its boundary points. A boundary point of a region is a point A for which every disk centered at A intersects both and its complement. All other points in are called interior points. Proposition Let R 2 be a closed and bounded region, and let f be a continuous, differentiable scalar function on. The absolute minimum and maximum values of f occur either at critical points of f in the interior of, or on the boundary of. Proof. Since an absolute extremum is also a local extremum, this follows from Proposition 3.1. Example 36. Let f(x, y) 4x+6y x 2 y 2. Find the absolute maximum and minimum values of f on the region : {(x, y) R 2 x 4, y 5}. Solution: Calculate f(x, y) < 4 2x, 6 2y >. Then the only critical point is (x, y) (2, 3). 51

52 The boundary in broken into four line segments: L 1, L 2, L 3, L 4 pictured below. Along L 1, we have x, y 5, so f(, y) 6y y 2. The only critical value of this one dimensional function is y 3. Of course, f might be extremized on L 1 at one of the end points (, ) or (, 5). Continuing in this fashion, we find eight possible locations of absolute extrema along the boundary. The critical points along L 1, L 2, L 3, L 4 : (, 3), (2, 5), (4, 3), (2, ) and the end points of these line segments, which are also the corners of, (, ), (, 5), (4, 5), (4, ) To find the absolute maximum and minimum of f, we check all of these values: f(2, 3) 13, f(, 3) 9, f(2, 5) 9, f(4, 3) 9, f(2, ) 4, f(, ), f(, 5) 5, f(4, 5) 5, f(4, ). So f has absolute maximum 13 achieved at the point (2, 3) and absolute minimum achieved at the point (, ) and (4, ). 3.9 Lagrange Multipliers We now consider the problem of maximizing a scalar function subject to a constraint. Consider the following example: Example 37. Maximize the value of the function f(x, y) xy amongst points (x, y) satisfying x 2 + y 2 1. That is, maximize f along the unit circle. 52

53 The region over which we want to maximize the function is closed and bounded, so the Extreme Value Theorem tells us that the maximum occurs somewhere. However, now every point in the region is a boundary point Proposition 3.14 isn t much help. For problems of this type, it is better to use the Method of Lagrange Multipliers. We begin with some geometric motivation. We plot some level curves f(x, y) xy k for several values of k. The symmetric nature of the curves around the line x y should quickly become apparent, as should the fact that f is maximized on the circle at points ( 1 1 2, 2 ) and ( 1 2, 1 2 ), and is minimized at the points ( 1 2, 1 2 ) and ( 1 1 2, 2 ). The important takeaway observation from this example is that at an extreme value, the level curve and the constraint curve are tangent to one another. Since by Corollary 3.9, we know that the gradient vector is perpendicular to the level set, this motivates: Proposition 3.15 (Method of Lagrange Multipliers). Suppose that f(x 1,..., x n ) is a differentiable scalar function and S is a level set defined by an equation g(x 1,..., x n ) k where k is a constant and g is a differentiable function for which g for all points on S. If f(x 1,..., x n ) achieves a local extremum on S at the point (a 1,..., a n ) S, then for some scalar λ R. f(a 1,..., a n ) λ g(a 1,..., a n ) 53

54 Example 38. Find the extreme values of the function f(x, y) x 2 y, subject to the condition x 2 + 2y 2 6. Solution: First calculate the gradients f < 2xy, x 2 > g < 2x, 4y > Now by equating the gradients and imposing the constraint, we get three equations in three variables. 2xy 2λx x 2 4λy x 2 + 2y 2 6 By the first equation, we know that either x or λ y. Case 1: If we set x then 2y 2 6 so y ± 3 and λ. Case 2: If we set λ y then x 2 4y 2, so 6y 2 6, so y ±1 and x ±2. We obtain six solutions: (x, y) (, ± 3), (±2, ±1) Now we can plug these into f(x, y) to get f(, 3), f(±2, 1) 4, f(±2, 1) 4. Thus f achieves its minimum of 4 at the points (±2, 1) and its maximum at the points (±2, 1). Example 39. Find the extreme values of the function f(x, y, z) xyz subject to the constraint that x 2 + 2y 2 + 3z 2 6. Solution: Let g(x, y, z) x 2 + 2y 2 + 3z 2. Calculate the gradients, f < yz, xz, xy > g < 2x, 4y, 6z > This provides four equations an four unknowns: yz 2λx xz 4λy xy 6λz x 2 + 2y 2 + 3z 2 6 First observe that if any of x, y or z equals zero, then f(x, y, z). Now assume that x, y, z. Then isolating λ in each equation we get 2yz x xz y, 3yz x xy z 54

55 then cross-multiplying gives and canceling gives 2y 2 x 2, 3z 2 x 2 Plugging back into the constraint equations gives 3x 2 6, so x ± 2. Similarly y ±1 and z ± 2/3. Thus we have extreme values 2 3 and 2 3, each achieved at four different points. Example 4. Maximize the function f(x, y) 4x + 3y 2 over the region : {(x, y) R 2 x 2 + y 2 1}. Solution: Calculate the gradient f < 4, 6y >. Observe that f vanishes nowhere, so f has no critical points in the interior of. Thus the extreme values all occur on the boundary g(x, y) x 2 + y 2 1, so we apply the method of Lagrange multipliers to locate them. Then equating f λ g we get equations g < 2x, 2y > 4 2λx 6y 2λy x 2 + y 2 1 Case 1: If y does not equal, then dividing the second equation by y get λ 3, so x 2/3 so (2/3) 2 + y 2 1 and y ± 5/3. Case 2: If y then x 2 1 and get solutions (±1, ). Checking these points, we get f(1, ) 4, f( 1, ) 4, f(2/3, ± 5/3) 8/3 + 5/3 13/ /3 So f is maximized in at the points (2/3, ± 5/3) where f(2/3, ± 5/3) 13/ Multiple constraints The method of Lagrange multipliers also applies to situations in which there is more than one constraint. So suppose that we want maximize a differentiable scalar function f(x 1,..., x n ) subject to the conditions that g(x 1,..., x n ) C and h(x 1,..., x n ) D for constants C and D and that g and f are linearly independent along the constraint set, then the necessary condition for a local extremum is that for some constants g and h. f λ g + µ h 55

56 Example 41. Find all critical values of f(x, y, z) x 2 y 2 + z 2, subject to the conditions that x + y 2 and y + z 4. Solution: The constraint equations are g(x, y, z) x + y 2 and h(x, y, z) y + z 4 so our Lagrange equations are < 2xy 2, 2x 2 y, 2z > f λ g + µ h λ < 1, 1, > +µ <, 1, 1 > combined with the original constraint equations we get five equations in five unknowns: x + y 2 y + z 4 2xy 2 λ 2x 2 y λ + µ 2z µ Equation 1,2 and 5 give x 2 y, z 4 y and µ 2z 2(4 y). Plugging into equations 3 and 4 gives 2(2 y)y 2 λ 2(2 y) 2 y λ + 2(4 y) Subtracting one from the other and dividing by 2 reduces to one variable equation (2 y) 2 y (2 y)y 2 4 y. Pushing all to one side and simplifying gives the equivalent equation 4 y + (2 y)y 2 (2 y) 2 y 4 y + 2y 2 y 3 4y + 4y 2 y 3 4 5y + 6y 2 2y 3 Unfortunately, at this point we are faced with a cubic equation which is tricky to solve exactly. It turns out that the equation has a single root approximately equal to y 2.3. Plugging in the other values, we get (x, y, z) (2 2.3, 2.3, 4 2.3) (.3, 2.3, 1.7). 56

57 4 Integrating Multi-variable Scalar Functions 4.1 Integrating Two Variable Scalar Functions Integrating over regions in the plane [Review] We begin with a review of the two variable case. Let f(x, y) be a continuous, scalar function and let be a region in R 2. The integral of f over is a real number f(x, y)da V 1 V 2 where V 1 is the volume of the solid bounded above by the graph of f and below by lying in the x y coordinate plane, and V 2 is the solid bounded below by the graph of f and bounded above by lying in the x y coordinate plane. Example 42. For some constant R >, let f(x, y) : R 2 x 2 y 2 and let : {(x, y) R 2 x 2 + y 2 R 2 }. Then the graph of f is the upper hemisphere of the sphere of radius R centred at the origin. Consequently, the value of the integral f(x, y)da 1 4πR πR3 3. Similarly, since the graph of f(x, y) is the lower hemisphere of the same sphere, we have f(x, y)da 2πR

58 A rigourous definition of integration makes use of Riemann sums. The idea is to approximate the region by a disjoint union of small rectangles, and then approximate f by a function r which is constant over each rectangle. The solid bounded by the graph of r looks like a union of boxes Since the volume of a box is easy to understand (height x length x depth), the integral of r is easy to determine and this provides an estimate of f(x, y)da. If the rectangles are made smaller, we expect the estimate to improve and we define f(x, y)da to be the limit of these estimates as the rectangles shrink to zero. In fact, the expression f(x, y)da may be decomposed into da is the area of an infinitesimally small rectangle (the base of a box in the Riemann sum) f(x, y) is the height of the box (which might be positive or negative). The (signed) volume of the box is then f(x, y)da means summing up the contributions of these boxes over the region. We say that the integral exists if the limit of Riemann sums converges. Proposition 4.1. The integral f(x, y)da exists if f is continuous and is closed and bounded. Proof. Omitted. Proposition 4.2. Let f(x, y) and g(x, y) be continuous scalar functions defined on a closed and bounded region. cf(x, y)da c f(x, y)da for c a constant scalar. f(x, y) + g(x, y)da f(x, y)da + g(x, y)da If is equal to a disjoint union of bounded functions 1 2, then f(x, y)da 1 f(x, y)da + 2 f(x, y)da Proof. Since the integral is defined in terms of a limit of Riemann sums, it is enough to verify these statements after replacing by a dijoint union of rectangles and f and g by functions that are constant on those rectangles. We leave this as an exercise. 58

59 The definition of integrals in terms of Riemann sums is theoretically quite useful, but rather cumbersome for the purposes of calculation. In practice, multivariable integrals are usually reduced to an interation of single variable integrals using Fubini s Theorem. Theorem 4.3. Let f(x, y) be a continuous scalar function defined on the rectangular region : {(x, y) R 2 a x b, c y d}. Then we have equality between the multivariable integral and the iterated single variable integrals: f(x, y)da d c ( b a ) f(x, y)dx dy b a ( d c ) f(x, y)dy dx The expression ( d b )dy f(x, y)dx means that for each value of y we define a func- c a tion F (y) : b f(x, y)dx, treating y as a constant in this integral, and then integrate a F (y)dy. The brackets are normally dropped so that we write d c d c ( b a ) f(x, y)dx dy d b c a f(x, y)dxdy Idea of Proof. Assume for simplicity that f(x, y), so the integral f(x, y)da is simply the volume the region, For any fixed value y y, the value of the function F (y ) f(x, y a b )dx is simply the area of the two dimensional cross section lying in the plane y y. Thus ( d ) b f(x, y)dx dy d F (y)dy is integrating the cross sectional areas and this should c a c give the volume. Example 43. Evaluate the iterated integrals 3 1 (1+4xy)dxdy and 1 3 (1+4xy)dydx. 1 1 Solution: (1 + 4xy)dxdy [x + 2x 2 y] 1 dy 1 + 2ydy [y + y 2 ] 3 1 ( ) 1 59

60 1 3 1 (1 + 4xy)dydx 1 1 [(y + 2xy 2 ] 3 1dx (2 + 2x(9 1))dx [2x + 8x 2 ] 1 1 Notice that the value of the two iterated integrals is the same, as they must be according to Theorem 4.3 Fubini s Theorem also generalizes to integrals over non-rectangular regions. Theorem 4.4. Let f(x, y) be a continuous scalar function defined on a closed and bounded region R 2. If : {(x, y) R 2 a x b, l(x) y u(x)} for some continuous single variable functions l(x) and u(x), then f(x, y)da b u(x) a l(x) f(x, y)dydx. Similarly, if {(x, y) R 2 c < y < d, l(y) x u(y)} then f(x, y)da d u(y) c l(y) f(x, y)dxdy. Example 44. Evaluate the integral f(x, y)da, when f(x, y) : xcos(y) and is the region bounded by curves, y, y x 2, x 1 Solution: First, sketch the region. 6

61 We see that {(x, y) x 1, y x 2 }. By Fubini we have f(x, y)da 1 x xcos(y)dydx [xsin(y)] x2 dx xsin(x 2 )dx Now we use a substitution u u(x) x 2, du 2xdx. Then u(1) 1 and u(), so 1 xsin(x 2 )dx u(1) u() 1 sin(u) du 2 1 sin(u)du 2 [ cos(u)] 1 1 cos(1) Polar coordinates In single variable calculus, there is a technique of integration called the substitution rule, which works like the chain rule in reverse. Suppose that we have an integral of the form b f(x)dx. The substitution rule introduces a new variable t defined by an equation a x x(t), where x(t) is a differentiable function and for which there exist c, d R such that x(c) a and x(d) b. Then we have f(x) f(x(t)) so dx dx dt dt x (t)dt b a f(x)dx x(d) x(c) f(x)dx d c f(x(t))x (t)dt Let s explore the geometric meaning of the substitution rule. The equation x x(t) is called a change of variables, and may be thought of as defining a function sending the interval [c, d] to the interval [a, b]: 61

62 If f(x) is a real-valued function defined on [a, b], then the composition f(x(t)) is a function on [c, d]. In forming Riemann sums, we find that the heights of rectangles for f(x) agree with the heights of rectangles for f(x(t)), but that the base of the rectangles might not agree: The change of variables may stretch or contract small intervals. To compensate for this, we introduce a correction factor x (t). Example 45. Integrate 2 x2 dx using the substitution x 2u. Solution: This integral is of course easy to do without the substitution: 2 x2 dx [x 3 /3] 2 8/3. Now consider how this integral behaves under the substitution x x(t) 2t. Notice that an interval in the t variable corresponds to an interval in the x-variable of twice the length. That is the geometric meaning of dx x (t)dt 2dt. In effect, we are multiplying the composed function by 2, doubling the heights of the rectangles to compensate for the fact that the intervals are half as long. Applying the substituion rule and integrating : 2 x2 dx x(1) x() x2 dx 1 2 (2t)2 2dt 8 1 t2 dt 8[t 3 /3] 1 8/3 A similar principle applies to multi-variable functions. In this case however, the change of variables is sometimes motivated by the shape of the region of integration. Consider a region of the form : {(x, y) R 2 1 x 2 + y 2 4, y }: 62

63 In order to integrate this in the coordinates x, y, we must subdivide into regions 1, 2, 3 and the determine equation y l(x) and y u(x) for the upper and lower boundary curves. This is going to be messy and a lot of work. For integrals over regions like, it is usually better to use polar coordinates. Introduce coordinates r and θ 2π and functions x rcos(θ), y rsin(θ). Geometrically Figure 1: Polar Coordinates so we call r the radial coordinate and θ the angle coordinate. These coordinates satisfy equations In these new coordinates, r 2 x 2 + y 2, tan(θ) y/x (when x ) {(x, y) 1 x 2 + y 2 4, x } {(r, θ) 1 r 2, θ π} So is a rectangle when expressed in polar coordinates. So we expect that it should be easier to calculate integrals over in terms of the r, θ coordinates. Of course, perform such a change of variables, we must introduce a correction factor the compensate for the fact that the function (r, θ) (x, y) does not preserve the areas of small rectangles. Consider the following geometric argument. Suppose we have a small rectangle in the r θ plane. It s image in the x y plane looks like: 63

64 If the rectangle is very small, this looks like a rectangle of height r and base r θ, so In the infinitesimal limit, we get so the correction factor is just r. A x y r r θ. da dxdy rdrdθ, Proposition 4.5. Let f(x, y) be a continuous function defined on a closed, bounded region R 2. We have an equality of integrals: f(x, y)da f(x(r, θ), y(r, θ))da f(x(r, θ), y(r, θ))rdrdθ Example 46. Evaluate the integral e x2 y 2 da over the region bounded between the semi-circle x 4 y 2 and the y-axis. Solution: Sketch the region, Switch to polar coordinates, 64

65 e (x2 +y 2) da e r2 rdrdθ π π 2 2 π/2 π/2 2 u(2) u() e r2 rdθdr e r2 rdr (u r 2, du 2rdr) e u du π 2 [ e u ] 4 π 2 (1 e 4 ) General change of coordinates in dimension two More generally, suppose we have a pair of differentiable functions x x(s, t) and y y(s, t). We think of this as a function T, assigning to every point in the s t-plane and point in the x y-plane: T (s, t) (x(s, t), y(s, t)) Given a region in x y-plane, we say that x(s, t), y(s, t) are a change of coordinates for if there is a region in s t-plane for T restricts to a one-to-one and onto function from to. The Jacobian for a change of variables (x(s, t), y(s, t)) is ( x ) ( ) x xs x t J(s, t) : det s y s t y t det The absolute value J(s, t) provides the correction factor for integration. Theorem 4.6. Let f(x, y) be a continuous function defined on a closed, bounded region R 2. We have an equality of integrals: f(x, y)da f(x(s, t), y(s, t))da f(x(s, t), y(s, t)) J(s, t) dsdt y s y t 65

66 Proof. The correction factor should equal the ratio between the area of a small rectangle in the s t-plane, and the area of its image under T. Consider arc-length parametrizations of the lines parallel to the coordinate axes in the s t-plane. Composing these with T we obtain parametrized curves in the x y plane, with velocity vectors equal to < x, x s t y > and <, y s t > (by the chain rule). The area of a small square based at (s, t), with side length l is l 2. The image of this square should be approximated by the parallel-piped with sides l < x, x > and s t l < y, y > based at the point (x(s, t), y(s, t)). Thus the area of the image of an l l s t square is approximately: ( ) lxs lx det t l 2 J(s, t) ly s ly t so the correction factor is equal to J(s, t) as desired. Example 47. Find the Jacobian for a polar change of coordinates. Solution: Here x(r, θ) rcos(θ) and y(r, θ) rsin(θ). Calculating partial derivatives, Thus, x r cos(θ), y r sin(θ), x θ rsin(θ) y θ rcos(θ). ( xr x J(r, θ) det θ y r y θ ) ( cos(θ) rsin(θ) det sin(θ) rcos(θ) so the correction factor is J(r, θ) r r as expected. ) rcos 2 (θ) + rsin 2 (θ) r Example 48. Integrate y2 da over the region lying in the first quadrant and bounded by the curves xy 1, xy 2, xy 2 1, xy 2 2. Solution: The region is : {(x, y) R 2 1 xy 2, 1 xy 2 1}. This suggests making an change of coordinates u xy and v xy 2. Solving for x and y we get y vu 1 (xy 2 )/(xy), x u 2 v 1 (xy) 2 /(xy 2 ) 66

67 Observe that these equations are valid if xy and xy 2 are non-zero, which will always be the case for (x, y). The Jacobian for this change of variables is ( xu x J(x, y) : det v y u y v ) x u y v x v y u (2uv 1 )(u 1 ) ( u 2 v 2 )( vu 2 ) v 1 Notice that J(u, v) v 1 J(u, v) because v is bounded between 1 and 2, hence positive on. Thus, I y 2 dxdy vu 2 dudv 2 1 (vu 1 ) 2 (v 1 dudv) [ vu 1 ] 2 1dv v( 1/2 + 1)dv 1 4 [v2 ] Example 49. Evaluate the integral e(x+y)/(x y), where is the trapezoidal region (1,), (2,) (,-2) and (,-1). Solution: Begin by sketching the region, The integral I e(x+y)/(x y) dxdy is awkward to handle directly, so we introduce coordinates u x + y and v x y. Now solving for x and y gives x 1 (u + v), 2 y 1 (u v). 2 The Jacobian for this transformation is ( ) xu x J(u, v) det v x y u y u y v x v y u ( 1 v 2 )(1 2 ) (1 2 )( 1 2 ) 1 2 So we have I 1 2 e u/v dudv which seems simpler. We still have to express in terms of the u, v-coordinates. 67

68 Observe that the bounding curves for are the lines with equations y x 1 x y 1, x y 2, x and y. These same lines in the u, v - coordinates are v 1, v 2, u + v, and u v. Sketching in the u v-plane gives This region is most easily described {1 v 2, v u v}. So we set up an iterated integral, I v v e uv 1 dudv 1 2 (ve 1 ve 1 )dv (e e 1 )[v 2 ] (e e 1 ) 1 [ euv 1 v 1 ]v vdv (e e 1 )vdv 4.2 Integrating Three Variable Scalar Functions We now consider integration of three variable functions. These are integrals of the form f(x, y, z)dv where is a region in R 3, f(x, y, z) is a scalar function defined on and dv is the infinitesimal volume. For example, if f is the constant function 1 then 1dV is simply the volume of. Another example is if represents some physical object and f(x, y, z) is respectively the mass density, the heat density, or the charge density, then f(x, y, z)dv is the total mass, the total heat, or the total charge of. There is no serious conceptual leap in passing from two to three variables, but it does get harder to picture what is going on. We begin with integrals over rectangular solids. Example 5. Integrate the function f(x, y, z) xy 2 z 1 over the region : {(x, y, z) R 3 x 2, y 1, 1 z 3}. Solution: We denote this integral by I xy 2 z 1 dv 68

69 We think of dv is the volume of an infinitesimal cube in R 3. Thinking geometrically in terms of Riemann sums, I is the signed volume of four dimensional solid. It is probably better not to try to picture this though. Instead, we compute I using the three dimensional version of Fubini s theorem. I.e., I is equal to the iterated integral: I xy 2 z 1 dxdydz This iterated integration can be done in any order of the variables. I [(x 2 /2)y 2 z 1 ] 2 dydz 2y 2 z 1 dydz [(2y 3 /3)z 1 ] 1 dz 1 z 1 dz 2 3 [ln(z)] ln(3) (ln(3) ln(1)) 3 3 In this example, it is also possible to use a short cut. Since when we integrate with respect to a particular variable, the other variables are treated like constants, we are allowed to pull them outside the integral signs y 2 z 1 ( 2 ) xdx dydz 3 1 ( 1 ( 2 z 1 y 2 ) ) xdx dy dz Because the intervals of integration are constants (independent of any variables), we may also pull the differentials through the integral signs, to get ( 3 )( 1 )( 2 z 1 dz y 2 dy 1 as before. ) xdx [ln(z)] 3 1[y 3 /3] 1 [x 2 /2] 2 (ln(2))(1/3)(2). Fubini s theorem also works for non-rectangular regions. For instance suppose a 3- dimensional region is defined {(x, y, z) R 3 a x b, l(x) y u(x), b(x, y) z t(x, y)} (17) where l, u, b, t are continuous scalar functions on one and two variables respectively. Then for a continuous scalar function f(x, y, z) defined over, we have equality of integrals: 69

70 f(x, y, z)dv b u(x) t(x,y) a l(x) b(x,y) f(x, y, z)dzdydx Example 51. Compute the volume of the solid region bounded between the coordinate planes and the plane x + y + z 1. Solution: We begin by sketching the region. It is defined by the inequalities {(x, y, z) x, y, z, and x + y + z 1} Finding the volume of a solid is the same as integrating the constant function 1 over it, so we want to compute I 1dV. To use Fubini s Theorem, we need to find a description of in the form of (17). Certainly the lower bounds of all the variables is, so we have x, y, z in. The upperbounds are determined by the inequality x + y + z 1. First of all, the largest value of x achieved in is 1, so we get bounds x 1. For any fixed value of x, we have x + y x + y + z 1 and so y 1 x. For any fixed value of x, y we have x +y +z 1 so we get bounds z 1 x y. Combined, we get, Consequently, I {(x, y, z) x 1, y 1 x, z 1 x y} 1dV 1 1 x x 1 x y [z] 1 x y dydx [y xy y 2 /2] 1 x dx ((1 x) 2 /2)dx [x x2 + x 3 /3] 1 1/6 1dzdydz 1 1 x 1 1 (1 x y)dydx ((1 x)(1 x) (1 x) 2 /2)dx (1 2x + x 2 )dx 7

71 4.2.1 Change of variables in dimension three We also have a change of variables formula for three variable integrals. integral f(x, y, z)dxdydz Consider an over a closed and bounded region R 3. A change of coordinates for is a set of three differentiable functions x x(u, v, w), y y(u, v, w), z z(u, v, w) determining an (almost-everywhere) one-to-one and onto transformation for. Then we have an equality of integrals f(x, y, z)dxdydz f(x(u, v, w), y(u, v, w), z(u, v, w)) J(u, v, w) dudvdw where J(u, v, w) is the Jacobian for the change of variables, and is defined by x u x v x w J(u, v, w) : det y u y v y w z u z v z w Cylindrical coordinates Cylindrical coordinates are appropriate for integrating over regions with rotational symmetry around the z-axis. They are, x rcos(θ), y rsin(θ), z z. where r and θ [ π, π] (we may also choose θ [, 2π] as appropriate). This is essentially just polar coordinates in the x-y-plane, because we leave z as it was. Calculating the Jacobian J(r, θ, z) : det x r x θ x z y r y θ y z z r z θ z z det cos(θ) rsin(θ) sin(θ) rcos(θ) 1 r(cos 2 (θ)+sin 2 (θ)) r. Observe that this is the same correction factor as for polar coordinates. This makes sense because we left z unchanged. Example 52. Find the volume of the solid that lies within both the cylinder x 2 + y 2 1 and the sphere x 2 + y 2 + z 2 4. Solution: Sketch the region, 71

72 In cylindrical coordinates, the equations for the cylinder and spere are r 2 1 and r 2 + z 2 4. Consequently, the region satisfies, Thus the volume of equals {(r, θ) r 1, z (4 r 2 ) 1/2 V 2π 2π 1dV 1 (4 r 2 ) 1/2 2π 1 1 [rz] (4 r2 ) 1/2 dr (4 r 2 ) 1/2 rdr rdθdzdr Use a substitution u 4 r 2, du 2rdr, to get V 2π u(1) u() Spherical coordinates u 1/2( 1) 2 du π[ 2 3 u3/2 ] 3 4 2π 3 (8 (3)3/2 ) Spherical coordinates are useful for integrating over regions that are radially symmetric, such as sphere or cones. Spherical coordinates consist of two angle coordinates θ, φ and one length coordinate ρ as depicted in the following diagram: Geometrically it is clear that we should restrict θ 2π, φ π and ρ. To get the change of variable formulas, it is actually easier to first pass through cylindrical coordinates z ρcos(φ), r ρsin(φ), θ θ so plugging these into the cylindrical change of coordinates formula, we get 72

73 Figure 2: Spherical Coordinates x rcos(θ) ρsin(φ)cos(θ), y rsin(θ) ρsin(φ)sin(θ), z ρcos(φ) Now to compute the Jacobian. Method 1: J(ρ, φ, θ) : det det x ρ x φ x θ y ρ y φ y θ z ρ z φ z θ ρ 2 sin(φ) det sin(φ)cos(θ) ρcos(θ)cos(φ) ρsin(φ)sin(θ) sin(φ)sin(θ) ρcos(φ)sin(θ) ρsin(φ)cos(θ) cos(φ) ρsin(φ) sin(φ)cos(θ) cos(θ)cos(φ) sin(θ) sin(φ)sin(θ) cos(φ)sin(θ) cos(θ) cos(φ) sin(φ) [ ( ) sin(φ)sin(θ) cos(φ)sin(θ) ρ 2 sin(φ) sin(θ) det cos(φ) sin(φ) ( sin(φ) cos(φ) ρ 2 sin(φ)[ (sin 2 (θ) + cos 2 (θ)) det cos(φ) sin(φ) ρ 2 sin(φ)( 1)( sin 2 (φ) cos 2 (φ)) ρ 2 sin(φ) ( sin(φ)cos(θ) cos(θ cos(θ) det cos(φ) s Method 2: We can make use of the fact that we have already calculated the Jacobian for cylindrical coordinates to save ourselves some trouble. Observe that dv dxdydz rdrdθdz r J dρdθdφ 73 ) ]

74 where J is the Jacobian for the change of variables from cyclindrical coordinates to spherical coordinates. I.e. From which we deduce that J : det det r ρ r φ r θ z ρ z φ z θ θ ρ θ φ θ θ sin(φ) ρcos(φ) cos(φ) ρsin(φ) 1 ρ(sin 2 (φ) + cos 2 (φ)) ρ dv rρdρdθdφ ρ 2 sin(φ)dρdθdφ Example 53. Evaluate (9 x2 y 2 )dv where is the solid hemisphere defined by x 2 + y 2 + z 2 9 and z. Solution: Sketch the region: Observe that x 2 + y 2 r 2 (ρsin(φ)) 2 so, I 9 x 2 y 2 dv (9 ρ 2 sin 2 (φ))dv (9 ρ 2 sin 2 (φ))ρ 2 sin(φ)dρdφdθ In the new coordinates {(ρ, θ, φ) ρ 3, φ π/2, θ 2π} I 3 π/2 2π 9ρ 2 sin(φ)dθdφdρ 3 π/2 2π ρ 4 sin 3 (θ)dθdφdρ I 1 I 2 I 1 ( 3 )( π/2 9ρ 2 dρ [3ρ 3 ] 3 [ cos(φ)] π/2 [θ] 2π (3 4 )(1)(2π) 2(3 4 )π 74 )( 2π ) sin(φ)dφ 1dθ

75 I 2 So I I 1 I π ( 3 )( π/2 )( 2π ρ 4 dρ (1 cos 2 (φ))sin(φ)dφ [ρ 5 /5] 3 [ cos(φ) + cos 3 (φ)/3] π/2 [θ] 2π (3 5 /5)(2/3)(2π) (8/5)3 4 π ) 1dθ 5 Calculus of Vector Fields A vector field F is a function that assigns to every point p in a domain D R n a vector F(p) in R n. Concretely, any vector field F in R 2 can be described in coordinates F(x, y) < f(x, y), g(x, y) > where f and g are continuous real valued functions defined on D R 2. Similarly, a vector field on R 3 may be expressed F(x, y, z) < f(x, y, z), g(x, y, z), h(x, y, z) > for some continuous scalar functions f, g, h defined on D R 3. Example 54. The gradient f of a differentiable scalar function f is a vector field. Example 55. Physical examples include: currents in a fluid, electric field, magnetic field, gravitational force field, etc Line Integrals and the Fundamental Theorem Let r(t) (x 1 (t),..., x n (t)) be a differentiable path in R n. If F is a vector field on R n, we define the line integral by r(b) r(a) F dr b a (F r (t))dt Example 56. Find the line integral of the vector field F(x, y) < x 2, xy > with respect to the path r(t) (x(t), y(t)) (cos(t), sin(t)), t π/2. Solution: Begin with a rough sketch of the vector field and the curve, 75

76 Since the angle between F(x, y) and r(t) is always obtuse along the curve, we know the integral should be negative. Calculating, r(t) < sin(t), cos(t) > r(b) r(a) F dr π/2 π/2 π/2 < x(t) 2, x(t)y(t) > < sin(t), cos(t) > dt < cos 2 (t), cos(t)sin(t) > < sin(t), cos(t) > dt 2cos 2 (t)sin(t)dt 2 3 [cos3 (t)] π/2 2 3 ( 1) 2 3 Although, the line integral is defined using a parametrized curve (a.k.a. path) the value of the line integral depends only on the curve (and its orientation), not on the parametrization. Proposition 5.1. The line integral of a vector field along a curve is independent of the parametrization of the curve. Proof. Let r(t) be a differentiable curve defined on the interval [a, b], and let t(u) be a differentiable function such that t(c) a and t(d) b so r(t(u)) is a reparametrization. We get d c F dr du du d c d c t(d) t(c) r(b) r(a) ( dr dt ) F du dt du ( F dr ) dt dt du du ( F dr ) dt dt F dr Using (in order) the chain rule, linearity of the dot product, change of variables formula for single variable integrals. If the vector field is the gradient of a scalar function, the line integral only depends on the end points of the curve. This result is known as the First Fundamental Theorem of Line Integrals. 76

77 Theorem 5.2. If F f is the gradient of a differentiable scalar function f, then f(r(b)) f(r(a)) r(b) r(a) for any continuous and piecewise differentiable path r(t). Proof. Denote r(t) : (x(t), y(t), z(t)). Then F dr r(b) r(a) F dr b a b a b a b a ( f) r dt < f x, f y, f z > < x t, y t, z t > dt ( f x x t + f y y t + f z ) dt z t df dt dt f(b) f(a) df f(b) f(a) Reduces to FTC for one variable via the chain rule. Example 57. To motivate Theorem 5.2, it is helpful to consider the one variable case. Let f f(x) be a differentiable function with gradient F f < f (x) >. Consider a curve r(t) (x(t)) (t) for a t b. Then the theorem says: f(b) f(a) r(b) r(a) F dr b a < f (x(t)) >< 1 > dt b a f (t)dt which we recognize as the first fundamental theorem of (single variable, scalar function) calculus. Example 58. Consider f(x, y) y, so F(x, y) f <, 1 >. Consider the curve r(t) (t, t) from t 1, joining (, ) to (1, 1). Integrating: 1 f r dt 1 <, 1 > < 1, 1 > dt 1 1dt 1 f(1, 1) f(, ). Similarly, if we consider the path r(t) (t, t 2 ) from t 1 also joining (, ) to (1, 1) then, 1 f r dt 1 <, 1 > < 1, 2t > dt 1 2tdt [t 2 ] 1 1 f(1, 1) f(, ). 77

78 Definition 14. A vector field F is called conservative if it satisfies the property that any line integral r(b) F dr depends only on the value of the endpoints r(b) and r(a). r(a) Remark 1. Important examples of conservative vector fields in nature include (static) electric fields, magnetic fields and gravitational fields. It follows from Theorem 5.2 that any gradient vector field f is conservative. The Second Fundamental Theorem of Line Integrals is the converse of that statement. Theorem 5.3. If F is a conservative vector, then there exists a differentiable function f (called the potential of F) such that F f. Proof. Using Theorem 5.2, we know that for any pair of points ā (a 1,.., a n ), b (b 1,..., b n ) in the domain of F, the difference f( b) f(ā) F dr for some path r joining ā to b. Observe that this quantity is well-defined because the line integral is independent of the choice of r. This formula determines f up to an arbitrary constant ( much like antiderivatives in single variable calculus are determined up to a constant). Remark 2. Important physical examples of potentials include the electric potential and the gravitational potential. In these contexts the potential is closely related to the concept of potential energy. In contrast with single variables, in dimension greater than one most vector fields are not gradients of scalar functions, so most vector fields are not conservative. Is it possible to determine whether a vector field is conservative, without checking every line integral and constructing a scalar anti-derivative? The answer is yes (with some conditions). Proposition 5.4. Let F(x, y) :< P (x, y), Q(x, y) > be a conservative and continuously differentiable vector field. Then Q x P y Q x P y. Proof. Because F is conservative, we know by the second Fundamental Theorem, that F f < f x, f y > for some differentiable scalar function f(x, y). Consequently, P f x and Q f y and by Clairaut s Theorem. Q x P y f yx f xy The function Q x P y is evidently interesting, so we give it a name. 78

79 Definition 15. We define the curl of a two-dimensional vector field F < P, Q > to be the scalar function curl(f) Q x P y. We caution the reader that the term curl is sometimes reserved for the three variable version of this concept, defined in... Example 59. The curl of a vector field has a geometric meaning as the amount of circulation. For example, consider the vector field < y, x >. This describes the current of a fluid flow that rotates around the origin in the counterclockwise direction. The curl of the vector field is curl(< y, x >) (x) x ( y) y is constant and positive. The vector field < y, x > describes rotation around the origin in the clockwise direction. It has curl 2 curl(< y, x >) ( x) x (y) y 2 is constant and negative. Proposition 5.4 has a partial converse, Theorem 5.5. Suppose F is a continuously differentiable vector field defined on a rectangular region D R 2 or defined everywhere on R 2. Then F is conservative if and only if curl(f). Proof. This will follow as a consequence of Green s Theorem in the next section. Why is the constraint on the domain D necessary? Example 6. Consider the vector field F : 1 x 2 +y 2 < y, x >< P, Q >. In this case Thus, Q x x(x2 + y 2 ) 1 x P y y(x2 + y 2 ) 1 y (x 2 + y 2 ) 1 2x 2 (x 2 + y 2 ) 2 (x 2 + y 2 ) 1 + 2y 2 (x 2 + y 2 ) 2 curl(f) Q x P y 2(x 2 + y 2 ) 1 (2x 2 + 2y 2 )(x 2 + y 2 ) 2. On the other hand, if we consider the graph of the vector field, 79

80 It is clear that taking the line integral counter-clockwise around a circle centered at the origin with give a positive number (Exercise: Find this number). This does not contradict Theorem 5.6, because F is not defined at the origin. In fact, if we integrate around any loop that does not contain the origin, we will get zero. Definition 16. A simple, closed curve C in R 2 is a curve that doesn t intersect itself anywhere except at its endpoints. That is, C admits a parametization r(t), a t b with r(a) r(b) but r(t 1 ) r(t 2 ) for any a t 1 < t 2 < b. Every simply connected curve in R 2 divides its complement into an bounded interior region and an unbounded exterior region (Aside: this fact is actually surprisingly difficult to prove and is called the Jordan curve theorem). Definition 17. A region R is called simply connected if for every simple closed curve C, the region bounded by C is contained in. The second fundamental theorem of calculus generalizes to Theorem 5.6. Suppose F is a continuously differentiable vector field defined on a simply connected region D R 2. Then F is conservative if and only if curl(f). Proof. This will follow as an easy consequence of Green s Theorem (Theorem 5.7). Example 61. Let us return now to Example 6. In this case we found that the vector field 1 F : x 2 + y < y, x > 2 8

81 was not conservative despite having vanishing curl. However, if we restrict to a simply connected region by (say) removing the positive part of the x-axis, then by the Fundamental Theorem there must be a potential. In this case the potential is simply the angular coordinate function θ. We have θ arctan(y/x) (for x ) and θ < θ x, θ y > < (y/x) 2 ( y x 2 ), < y x 2 + y 2, x x 2 + y 2 > F 1 ( 1 ) > 1 + (y/x) 2 x The reason that θ doesn t work as a potential globally is that it is only defined up to adding 2π. Indeed, the fundamental theorem is telling us that the integral F along a curve wrapping once around the origin is 2π. 5.2 Green s Theorem What if the curl of a vector field is nonzero? Is it still telling us something interesting? Theorem 5.7 (Green s Theorem). Let C be a simple closed piecewise differentiable curve with counter-clockwise orientation and suppose F is a vector field defined and continuously differentiable on C and on the region bounded by C. Then there is an equality of integrals: F dr curl(f)da C Proof. The idea of the proof is to decompose the region into little squares n so that curl(f)da curl(f)da. n n 81

82 Each n is bounded by a square curve C n. Because changing the orientation of a line integral introduces a minus sign, we get, F dr F dr. C n C n Finally, argue that for sufficiently small n, we have F dr curl(f)da. C n n To do this, consider the small rectangle n with corners (x, y ), (x + x, y ), (x + x, y + y), and (x, y + y). On a small rectangle, we can use the linear approximation of the vector field L(x, y) F(x, y) where L(x, y) F(x, y ) + (x x )F x (x, y ) + (y y )F y (x, y ) where F x : (P x, Q x ) and F y (P y, Q y ). Computing C n F drda C n L dr x x x + y x L(x + t, y ) < 1, > dt + y L(x + t, y + y) < 1, > dt L(x + x, y + t) <, 1 > dt y (L(x + t, y ) L(x + t, y + y)) < 1, > dt L(x, y + t) <, 1 > dt (L(x + x, y + t) L(x + t, y + y))) <, 1 > dt ( yf y (x, y )) < 1, > dt + y x y( P y (x, y ) + Q x (x, y )) x y(curl(f)(x, y )) curl(f)da. n xf x (x, y ) <, 1 > dt Remark 3. Observe that if curl(f) on a simply connected region, then Green s Theorem tells us that integrals around simple closed curves must vanish. This is essentially equivalent to the Second Fundamental Theorem of Line Integrals. Example 62. Integrate C F dr where F < x4, xy > where C is the triangle curve oriented clockwise, with vertices (, ), (1, ), (, 1). Solution: First sketch the curve 82

83 Rather than do line integrals for each edge, we use Green s theorem: I : F dr curl(f)da C where the minus sign comes from the fact that curve is oriented clockwise. Calculate so I yda 1 1 x curl(f) (xy) x (x4 ) y ydydx 1 y (1 x) 2 /2dx [ (1 x) 3 /6] 1 1/6 Green s Theorem can also be adapted to work over a region with holes, by dividing into simply connected regions. Example 63. Consider the region depicted below, with boundary consisting of two simple closed curves C 1 and C 2. Divide into two regions 1 and 2 with simple closed curves as boundaries. Then for any continuously differentiable vector field F defined over, we have curl(f)da curl(f)da + curl(f)da 1 2 F dr + F dr 1 2 F dr F dr C 1 C 2 where i denotes the boundary of i and the minus sign in the last line arises because C 2 is oriented clockwise. 83

84 Example 64. Let F 1 < y, x >. Show that the value of the integral F dr is x 2 +y 2 C the same for every simple, closed curve C, oriented counter-clockwise and bounding the origin. Solution: This vector field was shown before to be defined everywhere except the origin, and to satisfy curl(f). Thus, if we take two circles C 1 and C 2 centred at the origin with different radii, then by applying Green s Theorem to the region bounded between the curves (an annulus), we discover that curl(f)da F dr F dr C 1 C 2 from which we conclude that F dr F dr. C 1 C 2 Now given an arbitrary simple closed curve C bounding the origin, we may choose a circle C 1 containing C. By the same argument F dr F dr. C Parametrized Surfaces Recall that parametrized curve or path in R 3 is simply a continuous function C r(t) (x(t), y(t), z(t)) defined on some interval t [a, b] R. Similarly, a parametrized surface in R 3 is defined to be a continuous two variable function r(u, v) (x(u, v), y(u, v), z(u, v)) defined on some region R 2. This concept is illustrated in the following diagram In the same way we can do calculus on parametrized curves (tangent lines, arc length, line integrals), we can do calculus on parametrized surfaces (tangent planes, surface area and surface integrals). We say r(u, v) is differentiable if the coordinate functions x(u, v), y(u, v), z(u, v) are all continuous. We define vector-valued functions r u (u, v) < x u, y u, z u >, r v (u, v) < x u, y u, z u > 84

85 Definition 18. Let r(u, v) (x(u, v), y(u, v), z(u, v)) be a differentiable parametrized surface S in R 3. Given a constant (a, b) in the domain of r, the tangent plane to S at r(a, b) is parametrized in variable s, t by: P (s, t) r(a, b) + r u (a, b)s + r v (a, b)t Example 65. The graph of a two variable scalar function f(x, y) is a surface with parametrization r(u, v) (u, v, f(u, v)). Example 66. Find a parametrization for the surface z 2 x 2 + 2y 2 and z, and find a parametrization for the tangent plane at the point (1,, 1). Solution: Since we are only considering points with z, z 2 x 2 + 2y 2 z x2 + 2y 2. Thus a good parametrization is with the domain of r equal to all of R 2. derivatives, r(u, v) (u, v, (u 2 + 2v 2 ) 1/2 ), To find the tangent plane, we take partial r u < 1,, (u 2 + 2v 2 ) 1/2 u >, r v <, 1, 2(u 2 + 2v 2 ) 1/2 v > In this parametrization r(1, ) (1,, 1). Thus the formula for the tangent plane is, p(s, t) r(1, ) + r u (1, )s + r v (1, )t (1,, 1) + s < 1,, 1 > +t <, 1, > (1 + s, t, 1 + s) Example 67. Identify and sketch the surface with parametrization r(u, v) : (2cos(u), v, 2sin(u)). Find the tangent plane at the point ( 2, 3, 2). Solution: The surface is a cylinder of radius 2 centred around the y-axis. The point ( 2, 3, 2) is realized when (u, v) (π/4, 3). The tangent plane has parametrization p(s, t) r(2, 3) + sr u (2, 3) + tr v (2, 3) < 2, 3, 2 > +s < 2,, 2 > +t <, 1, > < 2(1 s), 3 + t, 2(1 + s) >. 85

86 5.3.1 Surface Area and Surface Integrals Recall that when we performed change of variable for two variable functions, we needed to introduce the Jacobian as a correction factor. Geometrically, the Jacobian measures the area of a small rectangle changes in the new coordinates. To do integrals over parametrized surfaces, we must do something similar. Given a differentiable parametrized surface r(u, v) in R 3, consider a square in the u-v plane, based at the point (a, b), with sides the unit vectors < 1, > and <, 1 >. Under the linear approximation to r(t), this square is sent to the parallel-piped in the tangent plane to r(a, b), with sides r u (a, b) and r v (a, b). The area of the parallel-piped is equal to the norm of the cross-product: Area r u (a, b) r v (a, b) so this is the appropriate correction factor for integrating scalar functions over parametrized surfaces in R 3. In particular, Proposition 5.8. The surface area of a surface S with differentiable parametrization r(u, v) with domain is equal to the two variable, scalar integral Area(S) da r u r v dudv S Remark 4. Observe the similarity of between this formula and the formula for arc-length : Arc-length b a r (t) dt. Example 68. Find the surface area of the surface z x 2 +y 2 bounded between z 1. Solution: Draw a sketch We can parametrize the surface by r(u, v) (u, v, u 2 + v 2 ) with : {(u, v) u 2 + v 2 1}. Taking partial derivatives gives, r u (u, v) < 1,, 2u >, r v (u, v) <, 1, 2v > 86

87 Forming the cross-product i j k 1 2u 1 2v < 2u, 2v, 1 >, so that r u r v (4u 2 + 4v 2 + 1) 1/2. Thus the surface area is equal to the integral A (4u 2 + 4v 2 + 1) 1/2 dudv. Since is a disk of radius 1 we change to polar coordinates, u rcos(θ), v rsin(θ), A 2π 2π 1 2π 1 t(1) t() π 4 [(2/3)t3/2 ] 5 π 6 53/2 (4r 2 + 1) 1/2 rdθdr (4r 2 + 1) 1/2 rdr (t 1/2 /8)dt, (t 4r 2 + 1, dt 8dr) More generally, we can integrate a scalar function f(x, y, z) over a surface with as follows by f(x, y, z)da f(r(u, v)) r u r v dudv. e.g. Evaluate the surface integral S S zda where S is the hemisphere satisfying x 2 + y 2 + z 2 1 and z. Solution: We use cylindrical coordinates r (x, y, z) where x cos(θ)sin(φ), y sin(θ)sin(φ) and z cos(φ). Then and Thus r θ ( sin(θ)sin(φ), cos(θ)sin(φ), ) r φ (cos(θ)cos(φ), sin(θ)cos(φ), sin(φ)). 87

88 r θ r φ ( cos(θ)sin 2 (φ), sin(θ)sin 2 (φ), sin 2 (θ)sin(φ)cos(φ) cos 2 (θ)sin(φ)cos(φ)) ( cos(θ)sin 2 (φ), sin(θ)sin 2 (φ), sin(φ)cos(φ)) sin 4 (φ)(cos 2 (θ) + sin 2 (θ)) + sin 2 (φ)cos 2 (φ) sin 4 (φ) + sin 2 (φ)cos 2 (φ) sin 2 (φ) sin(φ) Thus the integral becomes S zda 2π π/2 2π π/2 cos(φ)sin(φ)dθdφ cos(φ)sin(φ)dφ 2π[sin 2 (φ)/2] π/2 dφ π[sin 2 (π/2) sin 2 ()] π 5.4 Integrating Vector Fields over Surfaces: Flux Now we learn how to integrate vector fields over surfaces. Let F(x, y, z) be a vector field on R 3 and S R 3 a surface. An orientation at a point ā S is a choice of unit normal vector n to the surface at ā. An orientation on S is a continuously varying choice of orientations at all points. Figure 3: Orientation of a Surface in R 3 Example 69. If S is the unit sphere then there are two possible orientations: normal vectors either point out or in. 88

89 Example 7. A mobius strip S does not admit any orientation. Figure 4: Mobius Band Definition 19. Let F be a vector field in R 3 and let S be a surface in R 3 with orientation n. Then we define the flux of F over the S to be the integral F ds : (F n)da. S If we imagine the vector field F as the current of a flowing fluid, then the flux S (F n)da equals the rate at which the fluid passes through the surface in the direction n. Example 71. Suppose S is the radius R disk defined by x 2 + y 2 R 2 and z with (constant) normal vector n <,, 1 >. S If F <,, l > is constant vector field consisting of length l vectors pointing in the z-direction, the flux is S (F n)da S (l)da lπr2, the length of the vector times the area of the surface. If F <, 1, > points in the y-direction, then the flux is zero (no fluid is passing through the surface). 89

90 If the surface S is curved, computing the flux is best accomplished using a parametrization of S. So suppose that r(u, v) is a parametrization of S with domain R 2. Then from the last section we know, F lux (F n)da (F n) r u r v dudv. S This formula can be simplified. Recall that the cross-product r u r v is perpendicular to the surface, so it is a scalar multiple of the unit normal n. Thus r u r v ± r u r v n where the ambiguity of sign is a result of two possible choices of orientation. It follows that F lux (F n) r u r v dudv (F ( r u r v n))dudv ± F (r u r v )dudv Example 72. Let F :< xze y, xze y, z > and let S be the part of the plane x+y+z 1 lying in the first octant and oriented downwards in the z direction. Calculate the flux. Solution: Sketch the surface, The surface is the graph z 1 x y, restricted to the domain x + y 1 and x, y. This suggests a parametrization, r(u, v) (u, v, 1 u v) with domain : {(u, v) u, v u}. Calculating r u < 1,, 1 >, r v <, 1, 1 >, r u r v < 1, 1, 1 > Observe that the cross product r u r v points in the opposite direction from the desired orientation of S. Thus 9

91 F lux F (r u r v )dudv < u(1 u v)e v, u(1 u v)e v, 1 u v > < 1, 1, 1 > dudv (1 u v)dudv 1 1 u 1 1 (1 u v)dvdu [(1 u)v v 2 /2] 1 u du (1 u) 2 /2du [ (1 u) 3 /6] 1 1/6 Proposition 5.9. Flux integrals are linear in the sense that if F 1 and F 2 are vector fields and λ 1 and λ 2 are scalars then ((λ 1 F 1 + λ 2 F 2 ) n)da λ 1 (F 1 n)da + λ 2 (F 2 n)da S Proof. This follows in two steps from the linearity of the dot product and the linearity of integrals: ((λ 1 F 1 + λ 2 F 2 ) n)da (λ 1 (F 1 n) + λ 2 (F 2 n))da S S (λ 1 (F 1 n))da + (λ 2 (F 2 n))da S S λ 1 (F 1 n)da + λ 2 (F 2 n)da S S S S 5.5 Curl and Stokes Theorem In this section we prove a version of the Fundamental Theorem of Calculus called Stokes Theorem that relates the flux of vector field through a surface with the line integral of the vector field around the boundary of the surface. This is basically just a 3-dimensional version of Green s Theorem. First we need the 3-dimensional version of curl Curl Definition 2. Let F : P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k be a differentiable vector field on R 3. The curl of F is the vector field on R 3 defined by 91

92 ( R curl(f) : y Q ) ( P i + z z R ) ( Q j + x x P ) k y This formula is pretty complicated, but there is a trick for remembering it. We introduce the differential operator (pronounced del or nabla ) by i x + j y + k z. The meaning of this expression is that when we apply to a scalar function f(x, y, z) we get the gradient f i f x + j f y + k f z f x i + f y j + f z k. Thinking of as a vector with components / x, / y, / z can form the cross product of with the vector field F to get i j k F det x y z P Q R ( R y Q ) i + z curl(f) So we can Definition 2 can be summarized ( P z R x ) ( Q j + x P ) k y curl(f) : F. (18) Example 73. Suppose that the vector field F is every tangent to the horizontal planes z constant and are independent of z. I.e., F has the form Then curl(f) F ( y Q z ( Q x P y F P (x, y)i + Q(x, y)j + k. ) ( P i + ) k z x ) ( Q j + x P ) k y So we end up with the vertical unit vector field k times the scalar curl of the 2-dimensional vector filed < P (x, y), Q(x, y) >. In this sense the 2 dimenisional scalar-valued curl is a special case of the 3-dimensional curl. 92

93 More generally, we can understand 3-dimensional curl as composed of three twodimensional curls: curl(< P, Q, R >) curl(< Q, R >)i + curl(< R, P >)j + curl(< P, Q >)k. This helps explain the geometric meaning of curl (see Figure 5). Recall that in 2- dimensions, we interpreted the scalar curl as a kind of infinitesimal circulation of the vector field around a small rectangle. In three dimensions we can still interpret curl as infinitesimal circulation, but now direction of the vector curl(f) tells us the axis around which the fluid rotates and the norm curl(f) indicates the rate of circulation. For this reason, a vector field F is called irrotational if curl(f). Figure 5: Geometry of Curl We have an analogue of Proposition 5.4 Theorem 5.1. If f(x, y, z) is a scalar function with continuous second order partial derivatives then curl( f). Equivalently: A conservative vector field is irrotational. Proof. curl( f) f i j k det x f x y f y ( 2 f y z 2 f z y z f z ) i + ( 2 f z x 2 f ) ( 2 f j + x z y x 2 f ) k x y 93

94 where the last equality follows from Clairaut s Theorem 3.2. Example 74. Show that the vector field F(x, y, z) :< xy 2 z 3, x 3 yz 2, x 2 y 3 z > is not conservative. Solution: We compute the curl i j k F x y z xy 2 z 3 x 3 yz 2 x 2 y 3 z ( (x 2 y 3 z) (x3 yz 2 ) ) ( (xy 2 z 3 ) i + (x2 y 3 z) ) ( (x 3 yz 2 ) j + y z z x x (3x 2 y 2 z 2x 3 yz)i + (3xy 2 z 2 2xy 3 z)j + (3x 2 yz 2 2xyz 3 )k < 3x 2 y 2 z 2x 3 yz, 3xy 2 z 2 2xy 3 z, 3x 2 yz 2 2xyz 3 > (xy2 z 3 ) ) k y Since the curl(f) < 3x 2 y 2 z 2x 3 yz, 3xy 2 z 2 2xy 3 z, 3x 2 yz 2 2xyz 3 > is non-zero, it follows from Theorem 5.1 that F is not conservative. Theorem 5.1 has a partial converse (see Corollary 5.13) stating that irrotational vector fields are conservative under some conditions. This is a consequence of Stokes Theorem that we study in the next section Stokes Theorem Given a surface S with orientation n and boundary curve S, the positive orientation of S is determined by the right hand rule. If your open right hand points in the direction of n then your fingers point in the direction of orientation of the boundary. Equivalently, if n is pointing towards you, then the boundary is oriented counter-clockwise. Theorem 5.11 (Stokes Theorem). Let S be a piecewise-smooth surface with orientation n that is bounded by a simple, closed, piecewise smooth boundary curve S with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region of R 3 that contains S. Then F dr (curl(f) n)da. S S In other words, the line integral of F around S is equal to the flux of curl(f) through S. Proof. The basic idea is to use a parametrization to reduce this to Green s Theorem. We will only prove a special case, when S is the graph of a scalar function. Suppose that F(x, y, z) < P, Q, R > is the vector field and that S has parametrization < x, y, z > r(u, v) < u, v, z(u, v) > 94

95 for (u, v) lying in a domain and z(u, v) a differentiable function. Then r u < 1,, z u > r v <, 1, z v > r u r v < z u, z v, 1 > If we assume for simplicity that the orientation n agrees with the orientation of r u r v we get (curl(f) n)da (curl(f) r u r v )dudv S (< R y Q z, P z R x, Q x P y > < z u, z v, 1 >)dudv ( R z y u + Q z z u P z z v + R z x v + Q x P ) dudv y ( chain rule Q u P v + R z x v R z ) dudv y u where in the last step we use the chain rule and the fact that x/ u y/ v 1 and x/ v y/ u to obtain equalities and similarly Q u Q x x u + Q y y u + Q z z u Q x + Q z z u P v P x x v + P y y v + P z z v P x + P z z v On the other hand, if we let (u(t), v(t)) for t [a, b] be a counter-clockwise parametrization of the boundary of then < u(t), v(t), z(u(t), v(t)) > is a positive parametrization of the boundary S we have S F dr b a b a b a < P, Q, R > < du dt, dv dt, z du u dt + z v ( P du dt + Qdv dt + R z du u dt + R z dv v dt ( < P + R z ) (, Q + R z ) u v ( < P + R z ) (, Q + R z ) > dr u v dv dt > dt ) dt >< du dt, dv dt > dt To replace the line integral around S with a two dimensional line integral around. Applying Green s Theorem we get 95 (19) (2)

96 F dr S ( curl(< P + R z ) (, Q + R z ) >)dudv u v ( Q + R z ) ( P + R z ) dudv u v v u ( Q u + R z u v + R 2 z u v P v R z v u R 2 z ) dudv v u ( Q u P v + R z x v R z ) dudv y u ( Q u P v + R z u v R z ) dudv v u Using the chain rule we have equations similar to (19) and (2) R u R x + R z and R v R x + R z z v Plugging these in we get ( Q F dr S u P v + R z x v + R z z z u v R z x u R z z ) dudv z v u ( Q u P v + R z x v R z ) dudv x u z u Remark 5. The special case proof above can be turned into a general proof without too much difficulty, using something called the Implicit Function Theorem. This says that a smooth surface in R 3 can be broken into pieces where each piece looks like the graph of a function of the kind z f(x, y), y g(x, z) or z h(x, y). We will not prove the Implicit Function Theorem in this course though. A surface S R 3 is called closed if it is bounded, oriented, and has no boundary. Some examples of closed surfaces include a sphere, a torus, and the surface of a cube. Corollary Let F curl(a). The flux of F through an oriented surface S depends only on the boundary S, not on S itself. In particular, the flux of F through a closed surface is always zero. Proof. By Stokes theorem (curl(f) n)da F dr, depends only on the boundary. S S If S is closed, then the boundary S is empty so the integral is zero. 96

97 Figure 6: Closed Surfaces Corollary 5.12 should be compared to the fact (Theorem 5.2) that the line integral of a gradient vector field along a path depends only on the endpoints of the path. By analogy, we call we call A a vector potential for F curl(a). Speaking of gradients, we are now able to prove a conditional converse to Theorem 5.1. A region R 3 is called simply connected if every simple closed curve in C is the boundary of a disk in D. For example R 3 itself is simply connected, and R 3 \ {} is simply connected, but R 3 with a coordinate axis removed is not simply connected because a loop that winds around the axis is not the boundary of a disk. Corollary Let F be a vector field with continuous partial derivatives defined on a simply connected open region R 3. If curl(f), then F is conservative and there exists a potential function f such that f F. In other words: irrotational vector fields on simply connected regions are always conservative. Proof. The vector field F is conservative if and only if the line integral around any simple closed curve C is zero. In a simply connected region, every simple closed curve C is the boundary of some orientable surface S. Thus, if curl(f), then F dr (curl(f) n)da da. C S S Example 75. Evaluate the line integral C F dr, where F(x, y, z) < y2, x, z 2 > and C is the curve formed by intersecting the plane y + z 2 with the cylinder x 2 + y 2 1 oriented clockwise when viewed from above. 97

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