Vector Calculus lecture notes

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1 Vector Calculus lecture notes Thomas Baird December 13, 21 Contents 1 Geometry of R Coordinate Systems Distance Surfaces Vectors Geometric approach Algebraic Approach Dot Product Cross Product Equations for a Line Equations for a Plane Vector-valued Functions of a Single Variable and Space Curves Derivatives Arc Length Curvature Scalar-valued Functions of Several Real Variables Limits and Continuity Partial Derivatives Higher derivatives Partial Differential Equations Harmonic Functions and the heat equation Wave equation Tangent Planes and Linear Approximation Chain Rule Directional Derivatives and Gradients Maxima and Minima Absolute maxima and minima Lagrange Multipliers

2 4 Integrating Multi-variable Scalar Functions Integrating Two Variable Scalar Functions 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 Integrating Vector Fields over Surfaces: Flux 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} 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 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: 2

3 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. The point O = (,, ) is called the origin. The 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 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 transparently to all n Distance For p and q be two points in R 3, denote by pq the straight line segment joining them 3

4 The (Euclidean) distance between p := (x 1, y 1, z 1 ) and q := (x 2, y 2, z 2 ) is defined by the formula pq = (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 pq by the following argument. Consider the diagram The rectangle is drawn with sides parallel to the coordinate planes. The side lengths satisfy pa = x 1 x 2, ab = y 1 y 2 and q = 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 P Q = P B 2 + BQ 2 = (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2 4

5 1.1.2 Surfaces Recall that solution sets of equations in x and y (normally) determine curves in R 2. Equations in the variables x, y, z determine surfaces in R 3. Example 1. Graph the solution set of y = 7 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 Example 2. Graph the solution set of 4 = (x 1) 2 + (y) 2 + (z + 2) 2 5

6 1.2 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. 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: 6

7 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 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. 7

8 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. 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. To distinguish the two, I will write a 3-tuple < a, b, c > when I wish to think of it 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. 8

9 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,..., 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: < 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. Example 3. Find the unit vector u pointing in the same direction as a = 4 i 2 j + k. a = ( 2) = 21, 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 9

10 The dot product makes sense in any R n by the general formula a b = n a i b i i=1 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. Got this far last time. 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 =. Example 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) = = 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: 1

11 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. 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 11

12 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 3 a 3 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 = The case b = c a is similar. cb 1 cb 2 cb 3 b 1 b 2 b 3 i j k b 1 b 2 b 3 =. (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. 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. 12

13 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 9 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 Proof. a b = a b sin(θ) (5) 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. 13

14 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 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 i=1 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 =. Example 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? 14

15 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 = and c 1 c 2 c 3 a 1 a 2 a 3 b 1 b 2 b 3 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. 15

16 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 an equation for L. 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) has a unique solution t R for any r L. We call (6) a vector parametric equation for L. 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 + ta, y = y + tb, z = 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 tranjectory of a particle moving along L (without acceleration). Example 6. 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) 16

17 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 segment can be rewritten as follows r = r + t v = r + t( r 1 r ) = (1 t) r + t r 1 (8) for t 1, putting r and r 1 on a more even footing. Vectors r satisfying (8) are called convex linear combinations of r and r 1. 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 x x = y y = z z a b c the solutions of which define L. These are called the symmetric equations of 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 7. 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) r =< + t, 1 3t, 1 t > so and x = t, y = 1 3t, z = 1 t, x = y 1 3 = z ii) L interesects the x z-plane where y = = 1 3t. Thus t = 1/3 and the intersection point is (1/3,, 2/3). 17

18 1.6 Equations for a Plane 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 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 + su 1 + tv 1, y = y + su 2 + tv 2, z = z + su 3 + tv 3. for some s, t R. We think of s and t as defining a new set of coordinates, or parameters for P. Example 8. 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 > x = 1 + s, y = s + t, z = 2t 18

19 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 be a parametric equation r = r + s u + t v 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 9. 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 > 19

20 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 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 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 t t < f(t), g(t), h(t) >=< lim t t f(t), lim t t g(t), lim t t h(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 R is called a space curve. The vector-valued function r(t) is a called a parametrization of C and t is the parameter. Example 1. Sketch the curve parametrized by r(t) :=< sin(t), cos(t), t >. Solution: 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. 2

21 Example 11. 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: 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 We call ( r) (t) the derivative of r(t). In terms of our interpretation of r(t) describing the trajectory of a moving particle, ( r) (t) is is the velocity vector of the particle. The norm ( r) (t) is the speed. 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: 21

22 l(s) := r(t ) + s d r(t ) dt This line provides the best linear approximation to the curve near the point r(t ). Example 12. For the curve with parametrization r(t) :=< sin(t), cos(t), t > calculate: i) The derivative d r(t)/dt. ii) A parametrization of the tangent line 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) 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 22

23 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) Example 13. 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 the language of Riemann sums, but we will avoid this language. 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) =< f(t), g(t), h(t) > then (11) can be rewritten L = b a f (t) 2 + g (t) 2 + h (t) 2 dt 23

24 Example 14. 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 For a chosen constant a R, define the arc-length function s(t) := t a ( r) (t) dt 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 is one for all t: ( u) (t) = 1 i.e. 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(t) 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 24

25 Example 15. 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) + 1 = 2 s(t) = t 2dt = 2t. 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). 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 25 Since T is a unit vector, it is

26 Substituting in we get κ = T T. ( r) This isn t much of a simplication so far, but bear with me. Observe that ( r) = ( r) T = ds dt T. (12) Differentiating both sides with respect to t, and employing the (scalar) product rule ( r) = d2 s dt 2 T + ds dt T. (13) Take the cross product of (12) and (13) and use the fact the T T = : ( r) ( r) = ds d 2 s T dt dt T + 2 ( ds ) 2 T T = dt So plugging into the previous formula we get ( ds ) 2 T T = ( r) 2 T T dt Proposition 2.2. The curvature κ of a curve with parametrization r(t) satisfies κ = T T (r) = (r) (r) (r) 3 Example 16. 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) + 1 = 2 We saw before that r = cos 2 (t) + sin 2 (t) = 2, so κ = r r r 3 = 2 ( 2) 3 = 1 2 We will explore further the geometric meaning of κ, but first 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 : 26

27 N := T This is of course 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). 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. Example 17. Find the principal unit normal and binormal vectors to the curve r(t) = (sin(t), cos(t), t), and describe the osculating circle to 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,, 1 > and N() =<, 1, >. The curvature is κ = 1/2, so the osculating circle has radius 2 and is centred at 2 N() + r() = 2 <, 1, > +(, 1, ) = (, 1, ). 27

28 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. We first consider functions of two variables, f(x, y), because they are easier to visualize. Draw a diagram of a two variable function: Example 18. The function f(x, y) = 2 + x xy + x 3 is a function of two variables. 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). 28

29 Example 19. 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 2. 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 >. Example 21. 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. 29

30 Example 22. 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 23. 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 24. 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. 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 25. 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

31 3.1 Limits and Continuity 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 Example 26. 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 31

32 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 F (f(t), g(t), h(t)) = L. t t Example 27. 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 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. 32

33 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 28. 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 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 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. 33

34 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 29. 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 2 1 Example 3. 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. 34

35 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 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 (f x ) y = f xy = y ( f ) = 2 f 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 = 35 3 f y( x) 2

36 Example 31. 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. 3.3 Partial Differential Equations This subsection is to motivate some of these concepts and will not be tested Harmonic Functions and the heat equation 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 36

37 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 32. 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 Wave equation Now suppose that u(x, t) describes the motion of a vibrating string 37

38 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. 3.4 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 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 ). 38

39 Example 33. Let f(x, y) = x 2 +2xy 3 +y. Calculate the equation of 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 = 2(x 1) + y. The linear approximation is p(x, y) = z = 2(x 1) + y + 1 so f(1.1,.1) p(1.1,.1) = 2(.1) = 1.3 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. 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 34. 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

40 Example 35. 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. 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). 4

41 3.5 Chain Rule 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 Proof. We use the limit definition for derivatives: f(a, b) dy(t ) y dt df dt f(x(t + h), y(t + h)) f(x(t ), y(t )) t=t = 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 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 36. 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) 41

42 At t = we have x = sin() = and y = cos() = 1, so dz dt t= = = 3 Example 37. 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 = lim 1 = 1 h h h 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 = = 1 1/2 which is a contradiction, so f is not differentiable at (, ). 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 f, y s t = f x x t + f y One way of remembering these formulas is using a tree diagram: y t 42