CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES These lecture notes are intended as an outline for both student and instructor; for much more detailed exposition on the topics contained herein, the student should consult Chapters 12-17 of our standard textbook Calculus: Early Transcendentals by Jon Rogawski. Remark 0.1. The reader of these notes should remember that they are intended as an aid to the author s verbal lectures rather than as a comprehensive source of information, and thus most definitions and theorems are stated for a particular dimension of Euclidean space (e.g. for R 2, or R 3, etc.) rather than in their fullest generality. For instance, when we state the chain rule for paths (Theorem 15.1) we work in R 2 as this is the setting we encounter most commonly; conversely when we talk about vector fields (Definition 25.1) we will most commonly work in R 3. As a general (but not absolute) rule, the reader may infer that what theorems and definitions make sense for R 2 or R 3, mostly likely make sense for any R n where n 2 with a few minor adaptations. (One major exception to this is cross products which are strictly a 3-dimensional notion.) 1 Preliminaries Definition 1.1. The symbol R denotes the set of all real numbers. Geometrically we think of R as the set of all points on the number line, or as 1-dimensional space. 2 Vectors in the Plane Definition 2.1. The plane, or Cartesian plane, or xy-plane, is the full set of all ordered pairs of the form (x, y) where x and y are both real numbers. We denote the plane by R 2. We think of R 2 as 2-dimensional space. The following definition is obviously not intended to be mathematically rigorous, but it captures our intuitive notion of what a vector should be, and will be helpful when we consider applications of calculus to physical problems. We will give a more clear, i.e. purely mathematical, definition shortly. (Students who have taken a previous course in linear algebra may be familiar with a certain very general definition of a vector; the definitions we will end up working with here are just a special case of the definition they have already learned.) Definition 2.2 (Naive Definition of a Vector). A vector is an object which consists of both a (strictly positive) length (or magnitude) and a direction. We typically denote vectors as lower-case letters with an arrow hat, e.g. u, v, w, a, b, etc. We denote the length of a vector v by v. We consider two vectors u and v to be equal if they have both the same length and the same direction, and we write u = v. (For equality, note that we do not require u and v to have the same location!) We also allow the existence of a unique zero vector, denoted 0, which we consider to have 0 length and no direction. We may represent a vector v pictorially by drawing an arrow. We refer to pointy part at the end of the arrow as the head, and the base of the arrow as the tail. If P and Q are two points in (two-dimensional or three-dimensional) space, then we denote by P Q the vector which has its tail at P and its head at Q. (Note that unless P = Q, we always have P Q QP.) A scalar is just a magnitude, with no direction. In other words, a scalar is just a real number (a member of the set R). Example 2.3. Using the intuitive definition above, classify the following physical quantities as either a vector or a scalar. (a) A wind blowing southwest at 22 miles per hour. (b) The mass of an apple. (c) The force exerted by gravity on the apple. (d) The temperature in Fargo, ND at 2pm today. 1

2 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES (e) The velocity of a parked car at rest. Definition 2.4 (Naive Definition of Scalar Multiplication). Let v be a vector and let c be a scalar. The scalar product of c and v, denoted c v, is defined as follows: (1) if c > 0 then c v is the vector which points in the same direction as v, and whose length is c times the length of v. (2) if c < 0 then c v is the vector which points in the opposite direction from v, and whose length is c times the length of v. (3) if c = 0 then c v = 0. We call the vector c v a scalar multiple of v. Example 2.5. Draw any non-zero vector v. Then draw the vectors 3 v, 1 2 v, v = ( 1) v, 5 2 v, 0 v, and π v. Remark 2.6. Observe that another way of writing part (b) in the definition above is that for any vector v and any scalar c, c v = c v. This is a helpful identity we will use many times. Definition 2.7. Two vectors are called parallel if one is a non-zero scalar multiple of the other. Definition 2.8 (Naive Definition of Vector Addition and Subtraction). Let u and v be vectors. If the tail of v is placed at the head of u, then the vector sum of u and v, denoted u + v, is the vector whose tail coincides with the tail of u and whose head coincides with the head of v. (Picture helps here.) The vector difference of u and v, denoted u v, is defined to be the vector sum u + ( 1) v. Example 2.9. (a) In general is u + v = u + v? (b) Is the inequality u + v > u + v possible? Fact 2.10 (Triangle Inequality). Let u and v be any vectors. Then u + v u + v. Next we will give a more rigorous definition of vectors and vector operations by coordinatizing them in R 2. Definition 2.11. A vector in the plane is an ordered pair v = (v 1, v 2 ) in R 2. (We think of the tail of v as being the origin (0, 0) and the head as the point (v 1, v 2 ).) Two vectors u = (u 1, u 2 ) and v = (v 1, v 2 ) are considered equal if both u 1 = v 1 and u 2 = v 2, in which case we write v = u. The number v 1 is called the x-component of v and v 2 is called the y-component of v. The magnitude v of a vector v = (v 1, v 2 ) is the number v = v 2 1 + v2 2. Of course by the Pythagorean theorem, the magnitude of v is exactly its genuine geometric length. If c is a scalar, we define the scalar product of c and v to be the vector c v = (cv 1, cv 2 ). We define the vector sum of u and v to be the vector u + v = (u 1 + v 1, u 2 + v 2 ), and the vector difference to be the vector u v = (u 1 v 1, u 2 v 2 ). Remark 2.12. The student should note that our definitions given above (and below) differ slightly from those of Rogawski. There is more than one way to rigorously define a vector in the plane; we have chosen the above since it is consistent with the definitions the student might eventually see in linear algebra or other future math courses. The slight change in definition should present no great difficulty on the homework. Also note that according to our definiton, the collection R 2 is both the set of all points in the plane as well as the set of all vectors in the plane. This duality is intentional. Given a pair (x, y) in R 2, its geometric role as either a point or a vector must be distinguished from the context in which its given.

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 3 Definition 2.13. Let P = (x 1, y 1 ) and Q = (x 2, y 2 ) be two points in R 2. By the notation P Q, we mean the vector P Q = (x 2 x 1, y 2 y 1 ). Example 2.14. Let P = (x 1, y 1 ) and Q = (x 2, y 2 ) be two points in R 2. (a) Sketch a picture of P Q. (b) Compute the magnitude P Q. Definition 2.15. Let u = (u 1, u 2 ) and v = (v 1, v 2 ) be vectors in R 2 and let c be a scalar. Define the vector sum u + v, the vector difference u v, and the scalar product c v as follows: u + v = (u 1 + v 1, u 2 + v 2 ); u v = (u 1 v 1, u 2 v 2 ); c v = (cv 1, cv 2 ). Example 2.16. Let u = ( 1, 2) and v = (2, 3). (a) Evaluate u + v. (b) Write 2 u 3 v as a single vector. (c) Find two distinct vectors half as long as u and parallel to u. Definition 2.17. A unit vector is a vector of length 1. In particular we single out the coordinate unit vectors, which we permanently denote by i = (1, 0) and j = (0, 1). Note that if v = (v 1, v 2 ) is any vector, then we may write v = (v 1, v 2 ) = (v 1, 0) + (0, v 2 ) = v 1 (1, 0) + v 2 (0, 1) = v 1 i + v 2 j. So in general v = (v 1, v 2 ) = v 1 i + v 2 j; this gives us another notation for writing vectors. v Fact 2.18. If v is a non-zero vector, then the vector v = 1 v v has the same direction as v, and v magnitude 1. In other words v is the unique unit vector which points in the same direction as v. Example 2.19. Let P = (1, 2) and Q = (6, 10) be two points in the plane. (a) Find P Q and two distinct unit vectors parallel to P Q. (b) Find two distinct vectors of length 3 parallel to P Q. Example 2.20. Assume the water in a river moves southwest at 4 mi/hr. If a motorboat is traveling due east at 15 mi/hr relative to the shore, determine the speed of the boat and its heading relative to the moving water. Example 2.21. A child pulls a wagon with a force of F = 20 lb at an angle of θ = 40 to the horizontal. Find the force vector F. Example 2.22. A 400-lb engine is suspended from two chains that form 60 angles with a horizontal ceiling. How much weight must each chain withstand? 3 Vectors in Three Dimensions Definition 3.1. The set of all ordered triples (x, y, z) where x, y, and z are all real numbers is called xyz-space or three-dimensional space, and denoted permanently by R 3. Example 3.2. (1) Plot the points P = (3, 4, 5) and Q = ( 2, 3, 4) in xyz-space. (2) Compute the distance between P and Q. Definition 3.3. A vector in three dimensions is an ordered triple in R 3. The magnitude of a vector v = (v 1, v 2, v 3 ) in three dimensions is the quantity v = v 2 1 + v2 2 + v2 3. The real numbers v 1, v 2, and v 3 are called the x-component, y-component, and z-component of v respectively. The notions of being parallel and of vector addition, vector subtraction, and scalar multiplication are defined analogously to the two-dimensional case.

4 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES Definition 3.4. When working in R 3, the unit coordinate vectors are the following three distinguished vectors: i = (1, 0, 0); j = (0, 1, 0); k = (0, 0, 1). Note that for any vector v = (v 1, v 2, v 3 ) we have v = v 1 i + v 2 j + v 3 k. Example 3.5. Let u = (2, 4, 1) and v = (3, 0, 1). (a) Find 4 u + 2 v. (b) Find u v. (c) Find the unique unit vector with the same direction as u v. (d) Write the unit vector from part (c) as a sum of scalar multiples of i, j, and k. 4 Dot Products Definition 4.1. Given two vectors u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ), we define the dot product, denoted u v, to be the scalar given by u v = u 1 v 1 + u 2 v 2 + u 3 v 3. Example 4.2. (a) Let u = ( 3, 1, 0) and v = (1, 3, 0). Compute u v. (b) Let u = (2, 1, 2 3) and v = (2, 2, 3). Compute u v. Fact 4.3 (Properties of the Dot Product). For any vectors u, v, and w, and any scalar c, (i) 0 v = v 0 = 0; (ii) u v = v u (commutative property); (iii) (c u) v = u (c v) = c( u v); (iv) u ( v + w) = u v + u w (distributive property of the dot product over vector addition); and (v) v v = v 2. The usefulness of the dot product emerges in the upcoming Theorem 4.5; in order to prove Theorem 4.5, we need to recall the following fact about triangles. Fact 4.4 (Law of Cosines). If a triangle has angle measures A, B, and C and corresponding opposite side lengths a, b, and c, then the following equalities hold: c 2 = a 2 + b 2 2ab cos C; b 2 = a 2 + c 2 2ac cos B; a 2 = b 2 + c 2 2bc cos A. The following theorem says that we can use dot products to compute the angles between given vectors. Theorem 4.5. Let u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) be non-zero vectors, and let θ be the angle between u and v with 0 θ π. Then u v = u v cos θ. Proof. Consider the triangle which has u and v for two of its sides. The third side is equal as a vector to u v, and hence the Law of Cosines (previous fact) implies that u v 2 = u 2 + v 2 2 u v cos θ. Now solving for u v cos θ in the above, we get: u v cos θ = 1 2 ( u v 2 u 2 v 2 ). Now by the definition of magnitude in R 3, we have u 2 = ( u 2 1 + u2 2 + u2 3 )2 = u 2 1+u 2 2+u 2 3. Similarly we have v 2 = v 2 1 + v 2 2 + v 2 3 and u v 2 = (u 1 v 1 ) 2 + (u 2 v 2 ) 2 + (u 3 v 3 ) 2.

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 5 In that case, expanding out our equality from earlier and simplifying, we get: u v cos θ = 1 2 ( u v 2 u 2 v 2 ) = 1 2 ((u 1 v 1 ) 2 + (u 2 v 2 ) 2 + (u 3 v 2 ) 2 u 2 1 u 2 2 u 2 3 v 2 1 v 2 2 v 2 3) = 1 2 (2u 1v 1 + 2u 2 v 2 + 2u 3 v 3 ) = u 1 v 1 + u 2 v 2 + u 3 v 3 = u v. This proves the equality in the theorem. Corollary 4.6. Let u and v be non-zero vectors, and let θ be the angle between u and v with 0 θ π. Then u v cos θ = u v. Example 4.7. (a) Let u = ( 3, 1, 0) and v = (1, 3, 0). Compute the angle θ between u and v. (b) Let u = (2, 1, 2 3) and v = (2, 2, 3). Compute the angle θ between u and v. Corollary 4.8. If θ is the angle between two vectors u and v, then θ = π 2 u v = 0. (90 degrees) if and only if Definition 4.9. Two non-zero vectors u and v are called orthogonal if u v = 0. (Orthogonal and perpendicular are synonyms in two dimensions.) Definition 4.10. Given two vectors u and v, define the othogonal projection of u onto v, denoted proj v u (Note: Rogawski uses u here), to be the vector component of u which lies in the direction of v. v We have already observed that v is the unit vector which points in the same direction as v. Elementary geometric considerations show that the magnitude of proj v u is u cos θ, where θ is the angle formed by u and v. So we immediately have ( ) proj v u = u cos θ. Theorem 4.11. For any two vectors u and v, proj v u = v v ( ) u v v. v v Example 4.12. Find the projection of u = (5, 1, 3) onto v = (4, 4, 2). 5 Cross Products Definition 5.1. Let u and v be vectors in R 3. Define the cross product u v to be the vector with magnitude given by u v = u v sin θ, where θ is the angle between u and v (0 θ π), and with direction given by the right-hand rule: when you put the vectors tail to tail and let the fingers of your right hand curl from u to v, the direction of u v is the direction of your thumb, orthogonal to both u and v. (Note that the right-hand rule only makes sense if u and v are not parallel; if they are parallel and θ = 0 or θ = π, check that the given magnitude is 0 and hence u v = 0.) Example 5.2. Find the direction and magnitude of u v, where u = (1, 1, 0) and v = (1, 1, 2). Example 5.3. (a) In general is u v = v u? (b) In general is u v = v u?

6 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES Fact 5.4 (Properties of the Cross Product). Let u, v, and w be any vectors in R 3 and let c be any scalar. The following properties all hold. (a) u v = ( v u) (anticommutative property); (b) (c u) v = u (c v) = c( u v); (c) u ( v + w) = ( u v) + ( u w) (left distributive property of cross product over vector addition); and (d) ( u + v) w = ( u w) + ( v w) (right distributive property of cross product over vector addition). Example 5.5. Evaluate all possible cross products of the coordinate vectors i, j, and k. Example 5.6. Let u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ) be any two vectors in R 3. Find a closed formula for the coordinates of u v. Theorem 5.7 (The Cross Product as a Determinant). Let u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ). Then i j k [ ] [ ] [ ] u v = det u 1 u 2 u 3 u2 u = det 3 u1 u i det 3 u1 u j + det 2 k. v 2 v 3 v 1 v 3 v 1 v 2 v 1 v 2 v 3 Remark 5.8. Note that the matrix written in the formula above is not really a matrix in the sense that students will have seen previously, as some its entries are vectors rather than real numbers. Hence the formula above should be read as a formal determinant; that is, equality holds if one ignores the unusual context and just performs the usual determinant computation (using the Laplace cofactor expansion), pretending that i, j, and k are real numbers, and treating each scalar multiplication as though it were just multiplication of real numbers. Example 5.9. Find a vector orthogonal to u = i + 6 k and v = 2 i 5 j 3 k. 6 Vector-Valued Functions Definition 6.1. A vector-valued function is a function which takes a real number for input, and outputs a vector (typically in R 2 or R 3 ). We will typically denote vector-valued functions with the arrow hat notation, e.g. r(t), f(t), etc., where t is regarded as the input variable. If r(t) outputs vectors in R 3, we may write r(t) = (x(t), y(t), z(t)), where x, y, and z are all real-valued functions. In this way we regard r(t) as an ordered triple of real-valued parametric functions. The graph of a vector-valued function r is the set of all possible outputs r(t). (Note that this definition differs from our usual definition of a graph in previous math courses... the input variable is not pictorially represented in the graph!) The graph of a R 3 -valued function is often called a space curve. Example 6.2. Sketch the graphs of the following vector-valued functions. (a) r(t) = (2t + 1, t 2, 3t) (b) f(t) = (cos t, sin t, 5) (c) g(t) = ( sin t, cos t, t) Fact 6.3. A vector-valued function r(t) whose graph is the line in three-dimensional space passing through the point (x 0, y 0, z 0 ) in the direction of the vector v = (a, b, c) is r(t) = r 0 + t v, is given by r(t) = (x 0, y 0, z 0 ) + t(a, b, c), for < t <. Equivalently, the parametric equations of the line for r(t) = (x(t), y(t), z(t)) are x(t) = x 0 + at; y(t) = y 0 + bt; z(t) = z 0 + ct. 7 Calculus of Vector-Valued Functions Definition 7.1. Let a be a real number. A vector-valued function r(t) approaches the limit L as t approaches a, written

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 7 lim = L, t a provided lim r(t) L = 0. t a The function r(t) is continuous at a provided r(a) exists, lim r(t) exists, and lim r(t) = r(a). The t a t a function r(t) is simply called continuous if it is continuous at every point in its domain. Fact 7.2. Let r(t) = (x(t), y(t), z(t)) be a vector-valued function. If lim x(t) = L 1, lim y(t) = L 2, and t a t a lim z(t) = L 3, then t a lim r(t) = (L 1, L 2, L 3 ). t a Also, r(t) is continuous at a point a provided x(t), y(t), and z(t) are all continuous at a. Example 7.3. Consider the function (a) Evaluate lim r(t). t 2 (b) Evaluate lim r(t). t (c) At what points is r continuous? r(t) = cos πt i + sin πt j + e t k for t 0. Definition 7.4. Let r(t) = (x(t), y(t), z(t)) be a vector-valued function. We define the derivative of r(t), denoted r (t), to be r r(t + h) r(t) (t) = lim, h 0 h if the limit exists. If the derivative exists at a point t then we say r is differentiable at t. We also denote the derivative by d dt r(t). Note that the derivative r is a vector-valued function just like r. For any given t, the vector r (t) points in the same direction as the curve given by r(t); for this reason r (t) is called a tangent vector at r(t). We regard r (t) as the rate of change of the function r(t); for example, if r(t) is a position function in three-dimensional space, then r (t) is the associated velocity function. Fact 7.5. Let r(t) = x(t) i + y(t) j + z(t) k, where x, y, and z are all differentiable functions of t. Then r (t) = x (t) i + y (t) j + z (t) k. Example 7.6. Compute the derivative of the following functions. (a) r(t) = (t 3, 3t 2, t3 6 ) (b) r(t) = e t i + 10 t j + 2 cos(3t) k Example 7.7. Observe the behavior of r = (0, t 2, t 3 ) and r at t = 0. Definition 7.8. A vector-valued function r(t) is called smooth on an interval if it is differentiable on that interval, and also r (t) 0 on that interval. Let r(t) be a smooth vector-valued function on some interval [a, b]. The unit tangent vector for r(t) on [a, b] is T (t) = r (t) r (t). Example 7.9. Find the unit tangent vectors for the following functions. (a) r(t) = (t 2, 4t, ln t) for t > 0. (b) r(t) = (10, 3 cos t, 3 sin t) for 0 t 2π. Fact 7.10 (Derivative Rules). Let u(t) and v(t) be differentiable vector-valued functions and let f(t) be a differentiable scalar-valued function. Let a be a constant vector.

8 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES (1) d dt a = 0 (2) d dt ( u(t) + v(t)) = u (t) + v (t) (3) d dt (f(t) u(t)) = f (t) u(t) + f(t) u (t) (Product Rule) (4) d dt u(f(t)) = f (t) u (f(t)) (Chain Rule) (5) d dt ( u(t) v(t)) = u (t) v(t) + u(t) v (t) (Dot Product Rule) (6) d dt ( u(t) v(t)) = u (t) v(t) + u(t) v (t) (Cross Product Rule) Example 7.11. Compute the following derivatives, where u(t) = t i + t 2 j t 3 k and v(t) = sin t i + 2 cos t j + cos t k. (a) d dt v(t2 ) (b) d dt [t2 v(t)] (c) d dt [ u(t) v(t)] Example 7.12. Compute the first, second, and third derivatives of r(t) = (t 2, 8 ln t, 3e 2t ). Definition 7.13. An antiderivative of a vector-valued function r is a function R for which R = r. If r(t) = x(t) i + y(t) j + z(t) k, then R(t) = X(t) i + Y (t) j + Z(t) k, where X, Y, and Z are antiderivatives of x, y, and z respectively. The collection of all antiderivatives R of r is called the indefinite integral of r, and denoted r(t)dt. As is the case with real-valued functions, any two antiderivatives of r differ only by some constant vector C. So if R is any antiderivative of r, we may write r(t)dt = R(t) + C, where C is an arbitrary constant. Example 7.14. Evaluate [ e 3t i + (sin 4t + 1) j + t k ]. t2 + 2 Example 7.15. Find r(t) such that r (t) = (4t + 1, sin 3t, t 4 ) and r(0) = j. Definition 7.16. Let r(t) = f(t) i + g(t) j + h(t) k be a vector-valued function, where f, g, and h are integrable on the interval [a, b]. We define the definite integral of r(t) across [a, b] to be the vector b a r(t)dt = [ b a f(t)dt] i + [ b a g(t)dt] j + [ b a h(t)dt] k. Example 7.17. Compute π 0 [ i + 3 cos( t 2 ) j 4t k]dt. Definition 7.18. We say that a function r(t) or moves on a sphere, if r(t) = r for some fixed radius r 0, for every t in the domain of r. (In other words the graph of r(t) is entirely contained in the sphere of radius r.) Example 7.19. An object moves on a trajectory described by r(t) = (3 cos t, 5 sin t, 4 cos t), for 0 t 2π. (a) Show that the object moves on a sphere, and find the radius of the sphere. (b) Find functions which model the velocity and speed of the object. Theorem 7.20. Suppose r(t) is differentiable and moves on a sphere, i.e. r(t) = r for some fixed r, for all t in the domain of r. Then the position vector r(t) and the velocity vector r (t) are orthogonal at every t in the domain of r.

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 9 Proof. First notice that since r(t) 2 = r(t) r(t) = r 2 is a constant function, we have d dt r(t) r(t) = 0. So, applying the dot product rule for derivatives, we have: 0 = d r(t) r(t) dt = r (t) r(t) + r(t) r (t) = 2 r (t) r(t). So we must have r (t) r(t) = 0, i.e. r and r are orthogonal. 8 Length of Curves Fact 8.1. Let r(t) = (x(t), y(t), z(t)) be a vector-valued function, where x, y, and z are continuous. The graph of r(t) is a space curve in R 3. Assume that the curve is traversed only once for a t b. The arc length s of the curve between the points r(a) and r(b) is s = b a r (t) dt = b a x (t) 2 + y (t) 2 + z (t) 2 dt. We will omit the construction of the above formula here, but please see the intro to Section 8.1 in Rogawski for a plausibility argument (in two dimensions). Example 8.2. Compute the circumference of a circle of radius a. Example 8.3. An eagle rises at a rate of 100 vertical ft/min on a helical path given by r(t) = (250 cos t, 250 sin t, 100t), where r is measured in feet and t is measured in minutes. How far does it travel in 10 minutes? ( 250 2 + 100 2 269.) Definition 8.4. Let r(t) be a smooth vector-valued function for all t a. function (based at a) by s(t) = t a r (u) du. We say that r is parametrized by arc length if s(t) = t a for all t a. Keep in mind for the following fact, we usually define r in such a way that a = 0. Define the arc length The next theorem says that a function is parametrized by arc length if and only if it moves at a constant speed of 1. Theorem 8.5. Let r(t) be a smooth vector-valued function for all t a, and let s(t) be the associated arc length function. Define the velocity function to be v(t) = r (t), and the speed function to be v(t). Then r is parametrized by arc length if and only if v(t) = 1 for all t a. Proof. Suppose r(t) is parametrized by arc length, so for each t a we have s(t) = t v(u) du = t a. a Taking derivatives on both sides above and applying the fundamental theorem of calculus, we get v(t) = 1, as claimed. Conversely, suppose that v(u) = 1 for all u a. Then we have for each t a: s(t) = t a v(u) du = t a 1du = [u]t a = t a. This proves the theorem. Example 8.6. Consider the helix parametrized by r(t) = (cos 4t, sin 4t, 3t) for t 0. (1) Explicitly find the arc length function s(t) for t 0. Is r parametrized by arc length?

10 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES (2) Find a new function r 1 (t) which has the same graph as r(t), but which is parametrized by arc length. Theorem 8.7. Let r(t) be a vector valued function for a t b, and let s(t) be the associated arc length function. Suppose that s is an increasing function of t, so there exists an inverse function s 1 (t) for which s(s 1 (t)) = s 1 (s(t)) = t. Define a new function r 1 by r 1 (t) = r(s 1 (t)). Then r 1 has the same graph as r for s(a) t s(b), and is parametrized by arc length. Proof. It is clear that r has the same outputs for a t b as r 1 does for s(a) t s(b). So the two functions have the same graph on these intervals. So to prove the theorem, it remains only to show that r 1 is parametrized by arc length. To that end, observe that by the chain rule and the formula for derivatives of inverse functions, we have the following: r 1(t) = d dt r(s 1 (t)) = r (s 1 (t)) d dt s 1 (t) = r (s 1 1 (t)) s (s 1 (t)). Note that s in the above expression is just r by the fundamental theorem of calculus. Combining the lines from above, we get: r 1(t) = r (s 1 1 (t)) = 1. r (s 1 (t)) This shows r 1 has constant speed 1, and hence is parametrized by arc length by Theorem 8.5. 9 Curvature Definition 9.1. Recall from Definition 7.8 that the unit tangent vector associated to a smooth vectorvalued function r(t) is the function T (t) = r (t) r (t). The curvature of a smooth parametrized curve, denoted by κ, is the magnitude of the rate of change of T with respect to arc length s. (A picture helps here.) In other words, if s denotes arc length, κ(s) = d T ds. We wish to have a method of computing curvature which works for any parameter t, and not just for the arc length parameter s as above. The following theorem gives us such a method. Theorem 9.2. Let r(t) be a smooth parametrized curve, where t is any parameter. If v = r is the velocity and T is the unit tangent vector, then the curvature is κ(t) = 1 v d T dt = T (t) r (t). Proof. Let s denote arc length of r. Note that the relationship between the derivatives of T with respect to s and to t is given by the chain rule: d T dt = d T ds ds dt. Now by the Fundamental Theorem of Calculus, we have ds dt = v, and hence, solving for d T ds, we get d T ds = 1 v d T dt.

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 11 So by our original definition of curvature, we have as claimed. κ(t) = d T ds = 1 v d T dt = T (t) r Example 9.3. Which has greater curvature: A circle with a large radius or a circle with a small radius? Example 9.4. Compute the curvature function κ(t) for r(t) = (1, e t, t). Theorem 9.5. Let r be a smooth vector-valued function. Let v = r be the velocity function and let a = v = r be the acceleration function. Then the curvature κ of r is Proof. Since T = κ = a v v 3. v v, write v = v T. Differentiate both sides above using the product rule. Next compute the cross product a v: [ ( ) d a v = dt v = a = ( d dt v ) T + v d T dt. ( d dt v T = 0 + v 2 ( d T dt T = v 2 ( d T dt T ] T + v d T [ v T dt ] ) v T + v d T ). ) dt v T Now notice that T and d T dt angle between T and d T dt Putting everything together, we have are orthogonal since T moves in a sphere! (See Theorem 7.20.) So the is π 2, and hence the cross product d T dt T has magnitude d T dt T = d T dt T sin π 2 = d T dt 1 1 = d T dt. a v = v 2 dt dt T = v 2 dt dt = v 3 1 d T v dt = v 3 κ.

12 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES So κ = a v v 3 as claimed. Example 9.6. Find the curvature of the parabola r(t) = (t, at 2 ). Example 9.7. Find the curvature of the helix r(t) = (a cos t, a sin t, bt), where a, b > 0 are real numbers. 10 Motion in Space and Decomposition of Acceleration Example 10.1. Consider the two-dimensional motion given by the position vector r(t) = (3 cos t, 3 sin t) for 0 t 2π. (a) Sketch the trajectory of the object. (b) Find the velocity of the object. (c) Find the speed of the object. (d) Find the acceleration of the object. (e) Sketch the position, velocity, and acceleration vectors for t = 0, t = π 2, t = π, and t = 3π 2. Example 10.2. Find r(t) if a(t) = 2 i + 12t j, v(0) = 7 i, and r(0) = 2 i + 9 k. Fact 10.3. (1) Force (vector) equals mass (scalar) times acceleration (vector), i.e. F = m a. (2) Gravity accelerates objects downward at a rate of approximately 32 ft/s 2, or 9.8 m/s 2. Example 10.4. A bullet is fired from the ground at an angle of 60 above the horizontal. What initial speed v 0 must the bullet have in order to hit a point 150 m high on a tower located 250 m away? Definition 10.5. Let r(t) be a smooth vector-valued function. The unit normal vector N is N(t) = T (t) T (t). Intuitively, the unit normal vector points in the direction the graph of r is curving toward. Example 10.6. Find the unit normal vector at t = π 4 to the helix r(t) = (a cos t, a sin t, bt). Fact 10.7. Let r be a smooth vector-valued function with unit tangent vector T and principal unit normal vector N. Then T and N are orthogonal. Proof. Since T moves in a sphere, it is orthogonal to its own derivative T by Theorem 7.20. Since N points in the same direction as T by definition, T and N are orthogonal. Theorem 10.8. Let r be a smooth vector-valued function describing the motion of an object through space. Then the acceleration of the object a = r has a unique representation as the sum of its tangential component a T and its normal component a N : a = a N N + a T T, where a N = κ v 2 = a v v and a T = s. Proof. We begin with the fact that T = v v and hence v = v T = s T. Differentiating both sides and using the product rule and chain rule respectively, we get a = v = d dt (s T ) = s T + s T = v T + s T.

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 13 Now since N = T T, we have T = T N. Also T = κ v, so T = κ v N. Substituting this equality into that above, we end up with a = κ v 2 N + s T and we are done. Example 10.9. The driver of a car follows the parabolic trajectory r(t) = (t, t 2 ) for 2 t 2 through a sharp bend in the road. Find the tangential and normal components of the acceleration of the car. Theorem 10.10. Let a T and a N be the tangential component and normal component of a for some vector-valued function r, as in the previous theorem. Then a T T = proj v a = ( a v v v ) v and Proof. The first equality is because a N N = a a T T = a ( a v v v ) v. ( ) and hence a T T = ( a T ) T = a v v a T = (a T T + a N N) T = a T ( T T ) + a N ( N T ) = a T T 2 + a N (0) = a T, ( v v = a v 11 Functions of Several Variables v 2 ) v = proj v a. Then the second equality is immediate. Definition 11.1. A function in two variables f(x, y) is a function which assigns a unique (real) output to each input pair (x, y) from a particular set D in R 2. The set D is called the domain of f and is often implicitly (rather than explicitly) described. The range of f is the set of all real numbers z = (x, y) which are assumed as (x, y) ranges over the domain D. Definition 11.2 (Set-Builder Notation). To facilitate the next example and the homework problems, we recall for the student the use of set-builder notation. For convenience we will describe informally using examples, rather than give a formal definition. For an example, suppose we wish to formally describe the set D of all ordered pairs (x, y) for which x is twice y. Then we may write A = {(x, y) : x = 2y}. The above notation should be read as The set of all (x, y) in R 2 such that x = 2y, which clearly and precisely defines our set A. For another example, we could write Y = {x : 5 x < 10}, which reads The set of all x in R such that 5 is less than or equal to x and x is strictly less than 10. The student should easily verify that, using the interval notation, Y = [5, 10). In general, given a set A and a precise mathematical sentence P (x) about a variable x, the set-builder notation should be read as follows. { x : P (x)} The set of all elements x such that sentence P (x) is true for the element x.

14 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES Note that for our present purposes we implicitly assume that variables x range over real numbers in R, pairs (x, y) range over R 2, triples (x, y, z) range over R 3, etc. Example 11.3. Let g(x, y) = 4 x 2 y 2. Find the domain and range of g. Remark 11.4. Notice that for an R 3 -valued function r(t), our definition of the graph of r is {(x, y, z) : there exists a t such that r(t) = (x, y, z)}, i.e. the graph of r equals the range of r. Conversely, for a real-valued function f(x) like those the student will have worked with in previous courses, our definition of the graph of f is {(x, y) : f(x) = y}. Definition 11.5. Let f(x, y) be a function in two variables. Define the graph of f to be the set {(x, y, z) : f(x, y) = z}. Example 11.6. Sketch the graph of g(x, y) = 4 x 2 y 2. Definition 11.7. Given a fixed point P 0 = (x 0, y 0, z 0 ) in R 3 and a non-zero vector n = (a, b, c) in R 2, the set of all points P = (x, y, z) in R 3 for which P P 0 is orthogonal to n is called a plane. (The vector n is called the normal vector to the plane.) Example 11.8. Given a point P 0 and a normal vector n as in the above definition, find an equation in three variables whose graph is the plane determined by P 0 and n. Fact 11.9. The plane passing through P 0 = (x 0, y 0, z 0 ) with normal vector n = (a, b, c) is the set of all points (x, y, z) which satisfy the equation a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. The plane is the graph of a function if and only if c 0, which happens if and only if n is not orthogonal to k = (0, 0, 1). Definition 11.10. Working in R 3 : the xy-plane is the plane passing through (0, 0, 0) with normal vector k = (0, 0, 1); the xz-plane is the plane passing through (0, 0, 0) with normal vector j = (0, 1, 0); and the yz-plane is the plane passing through (0, 0, 0) with normal vector i = (1, 0, 0). These three are called the coordinate planes in R 3. Example 11.11. Find an equation of the plane that passes through the points (2, 1, 3), (1, 4, 0), and (0, 1, 5). Definition 11.12. Two distinct planes are called parallel if their respective normal vectors are parallel. Two planes are orthogonal if their respective normal vectors are orthogonal. Example 11.13. Which of the following distinct planes are parallel and which are orthogonal? Q: 2x 3y + 6z = 12 R: x + 3 2y 3z = 14 S: 6x + 8y + 2z = 1 T : 9x 12y 3z = 7 Example 11.14. Sketch the graph of f(x, y) = 2x 2 + 5y 2. Definition 11.15. If f(x, y) is a function in two variables, then a trace of the graph of f is the intersection of the graph with any plane parallel to a coordinate plane. The traces which lie in the coordinate planes are called the xy-trace, the xz-trace, and the yz-trace respectively. Example 11.16. Use traces to sketch the graphs of the following functions. (a) f(x, y) = x y 2 (b) f(x, y) = x sin y Definition 11.17. Let f(x, y) be a function. Given any fixed real number z 0, the level curve of f at z 0 is the set of all (x, y) in R 2 for which f(x, y) = z 0.

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 15 Example 11.18. Sketch some of the level curves of the following functions. (a) f(x, y) = y x 2 1 (b) f(x, y) = e x2 y 2 12 Limits and Continuity for Functions of Several Variables Definition 12.1. A disk in R 2 is the set {(x, y) : (x, y) (a, b) < r} for some fixed point (a, b) and some fixed radius r. Let R be a subset of R 2. A point P in R is called an interior point if there is some disk about P which is contained entirely in R. A point Q is called a boundary point if every disk containing Q contains both a point in R and a point not in R. A region R is open if it consists entirely of interior points. A region is closed if it contains all its boundary points. Example 12.2. Are the following sets open? Closed? (Both? Neither?) (a) Any disk in R 2. (b) The set R = {(x, y) : 0 < x < 1, 0 < y < 1}. (c) The set R = {(x, y) : 0 x 1, 0 y 1}. (d) The set R = {(x, y) : 0 x < 1, 0 x < 1}. Definition 12.3. A punctured disk in R 2 is a disk with its center point removed. (Note that any punctured disk is still open.) Definition 12.4. Let f(x, y) be a function in two variables. The function f has the limit L as (x, y) approaches (a, b), denoted lim f(x, y) = L, (x,y) (a,b) if for every ɛ > 0, there exists a punctured disk P centered at (a, b) such that f(x, y) L < ɛ whenever (x, y) is in P. and A function f(x, y) is continuous at a point (a, b) provided f(a, b) exists, lim f(x, y) exists, (x,y) (a,b) f(x, y) = f(a, b). We say f is continuous if f is continuous at every point in its domain. lim (x,y) (a,b) Example 12.5. Evaluate Example 12.6. Evaluate Example 12.7. Evaluate Example 12.8. Evaluate 13 Partial Derivatives lim (x,y) (2,8) (3x2 y + xy), if it exists. lim (x,y) (0,0) 3x 2 y 2, if it exists. lim (x,y) (4,1) lim (x,y) (0,0) xy 4y 2 x 2 y, if it exists. (x + y) 2 x 2, if it exists. + y2 Definition 13.1. Let f(x, y) be a function in two variables. The partial derivative of f with respect to x is f x (x, y) = δ δxf(x, y) = lim h 0 f(x + h, y) f(x, y), h provided the limit exists. The partial derivative of f with respect to y is f y (x, y) = δ δy f(x, y) = lim h 0 f(x, y + h) f(x, y), h

16 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES provided the limit exists. Example 13.2. Let f(x, y) = x 2 y 2 + 4. (a) Compute δf δf δx and δy. (b) Evaluate each derivative at (2, 4). Example 13.3. Compute the partial derivatives of the following functions. (a) f(x, y) = sin xy (b) g(x, y) = x 2 e xy Definition 13.4. Let f(x, y) be a function in two variables. The second-order partial derivatives of f are the following four functions (written with both available notations): ) = δ2 f δx of (f 2 x ) x = f xx ; ( δ δf δx δx ( δ δf δy δy ( δ δf δx δy ( δ δf δx δy ) ) ) = δ2 f δy 2 of (f y ) y = f yy ; = δ2 f δxδy of (f y) x = f yx ; = δ2 f δyδx of (f x) y = f xy. The latter two derivatives above are called mixed partial derivatives. Example 13.5. Compute the second-order partial derivatives of f(x, y) = 3x 4 y 2xy + 5xy 3. Example 13.6 (Special Example!). Let xy(x 2 y 2 ) f(x, y) = x 2 + y 2, if (x, y) (0, 0); 0, if (x, y) = (0, 0). (a) Is f continuous? (b) Compute f xy (0, 0) and f yx (0, 0). Are the mixed partial derivatives equal to one another? To answer the above questions we need the following quick lemma, which follows from L Hospital s rule. Lemma 13.7. Let f(x) be a function of one variable, differentiable everywhere except possibly at a single point x = a. Suppose that lim f (x) exists. Then f is differentiable at x = a and f (a) = lim f (x). x a x a Proof. By the definition of the derivative, and an application of L Hospital s rule (applied to the variable h) we have: f f(a + h) f(a) (a) = lim h 0 h f (a + h) 0 = lim h 0 1 = lim f (a + h) h 0 = lim f (x). x a Since the latter limit exists by our hypothesis, this proves the lemma. Solution to Example 13.6. (a) First we prove that f is indeed continuous. Since f is a rational function with positive denominator at every point (x, y) (0, 0), it is clear that f is continuous at every point except possibly the origin. So we just need to check that f is continuous at (0, 0). By our definition of continuity, this amounts to verifying that agrees with the function s limit at (0, 0). lim (x,y) (0,0) xy(x 2 y 2 ) x 2 + y 2 = 0 = f(0), i.e. that the function value

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 17 To see this, we consider approaching (0, 0) along any straight-line path through the origin. The line through the origin at an arbitrary angle θ through (0, 0) is parametrized by r(t) = (t cos θ, t sin θ). Now compute the values of f along the graph of r: f( r(t)) = f(t cos θ, t sin θ) = t cos θt sin θ(cos2 θ sin 2 θ) cos 2 θ + sin 2 θ = t2 ( 1 2 sin 2θ)(cos 2θ) 1 = 1 2 t2 sin 2θ cos 2θ. This means that f( r(t)) = 1 2 t2 sin 2θ cos 2θ 1 2 t2 sin 2θ cos 2θ 1 2 t2 (1)(1) = 1 2 t2 (regardless of the choice of θ!). So let ɛ > 0. Choose P to be a punctured disk of radius R so small that 1 2 R2 < ɛ. Then for every (x, y) in P we have (x, y) = r(t) for some θ and some t < R, and hence This shows that f(x, y) 0 = f(x, y) = f( r(t)) 1 2 t2 1 2 R2 < ɛ. lim f(x, y) = 0 = f(0, 0) as we intended. So f is continuous. (x,y) (0,0) (b) Our next step is to compute the partial derivatives. We start by finding f x. For all (x, y) (0, 0), we just use the quotient rule: f x (x, y) = δ δx ( x 3 y xy 3 ) x 2 + y 2 = (x2 + y 2 )(3x 2 y y 3 ) (x 3 y xy 3 )(2x) (x 2 + y 2 ) 2 = 3x4 y x 2 y 3 + 3x 2 y 3 y 5 2x 4 y + 2x 2 y 3 (x 2 + y 2 ) 2 = x4 y + 4x 2 y 3 y 5 (x 2 + y 2 ) 2. To find the value of f x (0, 0), we first compute lim f x(x, y). As in part (a), consider the approach (x,y) (0,0) along the parametrized path r(t) = (t cos θ, t sin θ) (where θ is arbitrary). We have f x ( r(θ)) = t5 cos 4 θ sin θ + 4t 5 cos 2 θ sin 3 θ t 5 sin 5 θ (cos 2 θ + sin 2 θ) 2 = t 5 (cos 4 θ sin θ + 4 cos 2 θ sin 3 θ sin 5 θ). It follows from the line above that f x ( r(θ)) 6 t 5. So by a similar argument to the one above in part (a), we can check that lim f x(x, y) = lim 6 t 5 = 0. Now Lemma 13.7 implies that f x (0, 0) = 0. So (x,y) (0,0) t 0 we have computed: x 4 y + 4x 2 y 3 y 5 f x (x, y) = (x 2 + y 2 ) 2, if (x, y) (0, 0); 0, if (x, y) = (0, 0).

18 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES An extremely similar computation shows that: x 5 4x 3 y 2 xy 4 f y (x, y) = (x 2 + y 2 ) 2, if (x, y) (0, 0); 0, if (x, y) = (0, 0). Now we are equipped to finish the problem. Check via plug and chug that but f x (0, 0 + h) f x (0, 0) f xy (0, 0) = lim h 0 h h 5 /h 4 = lim = 1 h 0 h f y (0 + h, 0) f y (0, 0) f yx (0, 0) = lim h 0 h h 5 /h 4 = lim = 1. h 0 h This shows that the two mixed partial derivatives f xy and f yx do not agree at the origin (0, 0). The above example shows that in general it is possible for f xy to not be equal to f yx. However, the example given above may seem somewhat pathological, and we hope there is a chance that most natural functions we run across on a day-to-day basis have the nice property that f xy = f yx. The following famous theorem says that this is indeed the case; if the second derivatives are merely continuous, then we have equality of mixed partials. The proof is not terribly difficult, but is beyond the scope of our immediate goals and is omitted. Theorem 13.8 (Clairaut s Theorem). Assume that f is defined on an open set D of R 2, and f xy and f yx are continuous at every point of D. Then f xy = f yx at every point of D. 14 Differentiability and Tangent Planes In this section we wish to give a meaningful definition of a function in multiple variables being differentiable. This notion is a bit more demanding in the multivariable (Cal III) setting than it was in the single-variable (Cal I/II) setting. As a warmup, let s consider an alternative definition of differentiability for single-variable functions, different from what the student would have learned in Cal I. Definition 14.1. A single-variable function f(x) is differentiable at the point a if there exists a tangent line L to f at a. To be precise, L is a function of the form L(x) = m(x a) + f(a), so the graph of L is a line of slope m passing through (a, f(a)), which is tangent to f in the sense that the secant lines connecting (a, f(a)) and (a + h, f(a + h)) get closer to the slope m as h goes to 0, i.e. In this case we set f (x) = m. ( ) f(a + h) f(a) f(a + h) f(a) mh lim m = lim = 0. h 0 h h 0 h

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 19 It is not too hard to check that the previous definition is equivalent to the one the student has already seen, i.e. the tangent line exists at a if and only if f f(a + h) f(a) (x) = lim exists. h 0 h Now consider a multivariable function f(x, y). We want f to be differentiable at a point (a, b) only if there exists a tangent plane L to f at (a, b). The tangent plane at (a, b) should give a good approximation to the secant lines connecting (a, b, f(a, b)) and (a + h 1, b + h 2, f(a + h 1, b + h 2 )) for all choices of (small) h = (h 1, h 2 ). (A picture helps here.) Notice that the slope of this secant line is given by f(a + h 1, b + h 2 ) f(a, b). h So we want our tangent plane to be a planar function passing through (a, b, f(a, b)), i.e. a function of the form L(x, y) = m 1 (x a) + m 2 (x b) + f(a, b) and with the property that the secant lines connecting (a, b, f(a, b)) and (a + h 1, b + h 2, f(a + h 1, b + h 2 )) get closer and closer to their corresponding ones on L, as h = (h 1, h 2 ) goes to (0, 0) (for all choices of h = (h1, h 2 )). Now given any particular h, check that the slope of the associated secant line on L is L(a + h 1, b + h 2 ) L(a, b) h = m 1h 1 + m 2 h 2. h Check that if such an approximation happens, it must be the case that f x (a, b) and f y (a, b) both exist, and that m 1 = f x (a, b) and m 2 = f y (a, b). So we arrive at the following definition, which will be our permanent one. Definition 14.2. Let f(x, y) be a function in two variables. We call f differentiable at a point (a, b) if f x (a, b) and f y (a, b) exist, and in addition f(a + h 1, b + h 2 ) f(a, b) (f x (a, b)h 1 + f y (a, b)h 2 ) lim = 0. (h 1,h 2) 0 (h 1, h 2 ) If this is the case the tangent plane to f at (a, b) is the graph of the function L(x, y) = f x (a, b)(x a) + f y (a, b)(y b) + f(a, b). As was the case in Clairaut s theorem, we get nice results if the partial derivatives are continuous on an open set. Theorem 14.3. Let f(x, y) be a function in two variables. If f x and f y exist and are continuous at every point of some open set D, then f(x, y) is differentiable at every point of D. Example 14.4. Show that f(x, y) = 5x + 4y 2 is differentiable, and find the equation of the tangent plane at (2, 1). 15 The Chain Rule for Paths Theorem 15.1 (Chain Rule for Paths). Let f be a differentiable function in two variables and let r(t) = (x(t), y(t)) be a differentiable vector valued function. Then the composition f( r(t)) is differentiable, and has derivative d dt f( r(t)) = f x( r(t))x (t) + f y ( r(t))y (t). Written in shorthand with the Leibniz notation, we have

20 CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES d δf dx f( r(t)) = dt δx dt + δf dy δy dt. Example 15.2. Let f(x, y) = x 2 3y 2 + 20 and let r(t) = (2 cos t, 2 sin t). Find d dtf( r(t)) and evaluate it at t = π 4. Example 15.3. Compute d dt f( c(t)), where f(x, y, z) = xy + z2 ) and c(t) = (cos t, sin t, t). 16 Directional Derivatives and the Gradient Definition 16.1. Let f be differentiable at (a, b) and let u = (cos θ, sin θ) be a unit vector in R 2. The directional derivative of f at (a, b) in the direction of u is f(a + h cos θ, b + h sin θ) f(a, b) D u f(a, b) = lim, h 0 h provided the limit exists. Theorem 16.2. Let f be differentiable at (a, b) and let u = (u 1, u 2 ) be a unit vector in R 2. Then D u f(a, b) = (f x (a, b), f y (a, b)) (u 1, u 2 ). Proof. Define a new function g(s) (real inputs and real outputs) by the rule g(s) = f(a + su 1, b + su 2 ). Geometrically, the graph of g is a cross-section of the graph of f passing through the point (a, b) parallel to vector u. It is clear from the definition of the directional derivative that D u f(a, b) = g (0). Now setting x(s) = a + su 1 and y(s) = b + su 2 (so g(s) = (x(s), y(s)) and applying the chain rule for paths, we get D u f(a, b) = g (0) = f x (x(0), y(0))x (0) + f y (x(0), y(0))y (0) = f x (a, b)u 1 + f y (a, b)u 2 ) = (f x (a, b), f y (a, b)) (u 1, u 2 ). Example 16.3. Consider the paraboloid z = f(x, y) = 1 4 (x2 +2y 2 )+2 and the unit vectors u = ( 1 2, 1 2 ) and v = ( 1 2, 3 2 ). (a) Find the directional derivative of f at (3, 2) in the directions of u and v. (b) Graph the surface and interpret the directional derivatives. Definition 16.4. Let f(x, y) be differentiable. The gradient of f at (x, y) is the function f(x, y) = (f x (x, y), f y (x, y)). Remark 16.5. Observe that f is our first example of a vector-valued function in two variables, i.e. it takes vectors in R 2 for input and returns vectors in R 2 as output. We will study functions of this type more closely when we look at vector fields later in the course. Fact 16.6 (Rephrasing of the Chain Rule for Paths). If f is a differentiable function of two variables and r is a differentiable R 2 -valued function, then f( r(t)) is differentiable and d dt f( r(t)) = f( r(t)) r (t). Example 16.7. Find f and f(3, 2) for f(x, y) = x 2 + 2xy y 3.

CALCULUS III MATH 265 FALL 2014 (COHEN) LECTURE NOTES 21 Example 16.8. Let f(x, y) = 3 x2 10 + xy2 10. (a) Compute f(3, 1). (b) Compute D u f(3, 1) where u = ( 1 2, 1 2 ). (c) Compute the directional derivative of f at (3, 1) in the direction of the vector (3, 4). Theorem 16.9. Let f be differentiable at (a, b). (1) f has its maximum rate of increase at (a, b) in the direction of the gradient f(a, b). The rate of increase in this direction is f(a, b). (2) f has its maximum rate of decrease at (a, b) in the direction of f(a, b). The rate of decrease in this direction is f(a, b). (3) If u is orthogonal to f(a, b), then D u f(a, b) = 0. Proof. For any unit vector u, we have D u = f(a, b) u = f(a, b) u cos θ = f(a, b) cos θ, where θ is the angle between f(a, b) and u. Then cos θ is maximized when θ = 0 and minimized when θ = π, which proves statements (1) and (2) above. If f(a, b) and u are orthogonal then θ = π 2 and hence cos θ = 0; this shows statement (3). Example 16.10. Consider the bowl-shaped paraboloid z = f(x, y) = 4 + x 2 + 3y 2. (a) If you are located at the point (2, 1 2, 35 4 ) on the paraboloid, in which direction should you move in order to ascend the surface at the maximum rate? How quickly will you ascend? (b) If you are at the point (3, 1, 16), in which directions may you walk in order to neither gain nor lose height? 17 Optimization in Several Variables Definition 17.1. Let f be a function in two variables. We say f has a local maximum at (a, b) is there is some disk D containing (a, b) such that f(a, b) f(x, y) for all (x, y) in D. We say that f has a local minimum at (a, b) if there is some disk D containing (a, b) such that f(a, b) f(x, y) for all (x, y) in D. In either case, we say that f has a local extremum at (a, b). A point (a, b) is a critical point of f if either (1) f x (a, b) = f y (a, b) = 0 or (2) one (or both) of f x and f y does not exist at (a, b). Fact 17.2. If f has a local maximum or minimum at (a, b), then (a, b) is a critical point of f. Example 17.3. Find the critical points of f(x, y) = xy(x 2)(y + 3). Definition 17.4. Let f be a function in two variables. We say that f has a saddle point at (a, b) if (a, b) is a critical point, but for every disk D containing (a, b) there are points (x, y) in D for which f(x, y) > f(a, b) and points (x, y) in D for which f(x, y) < f(a, b) (in other words f has neither a min nor a max at (a, b)). Theorem 17.5 (Second Derivative Test). Suppose that the second partial derivatives of f(x, y) are continuous in a disk containing (a, b), where f x (a, b) = f y (a, b) = 0. Set D(x, y) = [f xx f yy f 2 xy](x, y). (1) If D(a, b) > 0 and f xx (a, b) < 0, then f has a local maximum value at (a, b). (2) If D(a, b) > 0 and f xx (a, b) > 0, then f has a local minimum value at (a, b).