f(x) = C. (1) f(x,y) = C implies that x 2 + y 2 = C 0. (2)

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1 Lecture 4 Level Sets/Contours (Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 14.1) You are no doubt familiar with the idea of contour plots from geography topographic maps of a region in which points of identical elevation are plotted to produce contour lines. Such plots are two-dimensional representations of the actual three-dimensional land region. Some examples are sketched in the textbook, Section There is a good deal of information that can be read from such topographical maps. We ll return to this idea later. Definition: Given a function f : R n R and a number C that lies in the range of f, we define the C-level set of f to be the set of points x R n that satisfy f(x) = C. (1) In the special case n = 2, which we shall examine below in more detail, C-level sets are generally curves in R 2, which is why they are often called level curves or contours. They represent the intersection of the graph of f(x,y) with the plane z = C. This is illustrated nicely in Figures 11 and 17 of the textbook. Examples: 1. The function f(x,y) = x 2 + y 2 examined above. Note that f(x,y) 0. Then f(x,y) = C implies that x 2 + y 2 = C 0. (2) The level sets are concentric circles centered at (0,0) with radius C. When C = 0, the level set is the point (0,0). When C = 1, the level set is the circle of points x 2 + y 2 = 1. Some level sets are sketched below. 2. The function f(x, y) = 1 x y studied in the previous lecture. Since f(x, y) can assume any real value we consider C R consider the set (x,y) R 2 for which f(x,y) = C. This implies 22

2 y C = 4 C = 1 C = x Some level sets of z = x 2 + y 2 that 1 x y = C, which implies that x + y = 1 C or y = x + 1 C. (3) This represents a family of straight lines in the xy plane with slope 1 and intercept 1 C. Some sample lines are sketched below. y x C = 1 C = 0 C = 2 C = 1 Some level sets of z = 1 x y 3. The function f(x,y) = x 2 studied earlier. Then f(x,y) = C implies that x 2 = C 0. For C = 0, the level set is x = 0. For C > 0, the level set is the set of two parallel lines x = ± C, as shown in the plot below. The level sets of a function of three variables f(x,y,z) = C are generally surfaces in R 3. For example, consider the function f(x,y,z) = x 2 + y 2 + z 2. We cannot sketch the graph of f, since we would need four dimensions. We can, however, sketch its level sets. For C 0, the set of points that satisfy x 2 + y 2 + z 2 = C 0 (4) 23

3 y x -1-2 C = 4 C = 1 C = 0 C = 1 C = 4 Some level sets of z = x 2 is a sphere of radius C centered at the origin (0,0,0). For C = 0, the level set is the single point (0,0,0). Special case - functions with spherical symmetry In this course, we shall be often encountering functions f : R 3 R with spherical symmetry, for example, the function given above, f(x,y,z) = x 2 + y 2 + z 2. (5) Another example is g(x,y,z) = 1 (6) x 2 + y 2 + z2. The spherical symmetry is reflected in the fact that the value of the function depends only on the distance r from the origin (0, 0, 0) to the point (x, y, z). This distance, often called the radial distance is, of course, r = r = x 2 + y 2 + z 2. (7) In general, spherically symmetric functions may be written in terms of r alone. For this reason, they are also known as radially symmetric functions. In the examples given above, f(x,y,z) = x 2 + y 2 + z 2 = r 2, (8) and g(x,y,z) = 1 x 2 + y 2 + z 2 = 1 r. (9) 24

4 Both of these functions are spherically symmetric. Qualitatively, however, they are quite different. As the radial distance r increases, i.e., as we move away from the origin (0, 0, 0), the function f increases in value, whereas g decreases in value. Partial Derivatives (Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 14.3) Before discussing derivatives of functions of several variables, one has to start with the concepts of limit and continuity of these functions, as was done for the single variable case in your first-year calculus course. We shall simply state these concepts here they are discussed in a little more detail in Section 12.2 of the textbook. For simplicity, we consider only functions of two variables f(x,y). The extension to more than two variables is straightforward. 1. The function f(x,y) has a limit at (a,b). This is written as lim f(x,y) = L (x,y) (a,b) and implies that f(x, y) approaches the value L as the pair of values (x, y) approaches, i.e., gets arbitrarily close to, (a,b) regardless of the direction of approach. 2. The function f(x,y) is continuous at (a,b). This implies that lim f(x,y) = f(a,b). (x,y) (a,b) In other words, f(a,b) is defined at (a,b) and its value there is the limit of f(x,y) as (x,y) (a,b). 3. The function f(x,y) is differentiable at (x,y). This is the subject of this section. Given a function of several variables, say f(x,y), it is useful to have a method of determining the rate of change of f with respect to changes in its variables. The problem is, of course, that we now have more than one variable. We ll see that everything can still be taken care of in terms of two straightforward methods of differentiating f(x,y): 25

5 1. Keep y constant and differentiate with respect to x, just as you would do with a function g(x): The result is called the partial derivative of f with respect to x and is denoted as x The notation f x is also commonly employed. (pronounced dye f by dye x ). 2. Keep x constant and differentiate with respect to y, just as you would do with a function h(y): The result is called the partial derivative of f with respect to y and is denoted as y The notation f y is also commonly employed. (pronounced dye f by dye y ). These are known as first order partial derivatives. If f is a function of more than two variables, say, f(x,y,z), then you differentiate f with respect to each variable, keeping the other two constant, to produce the three first order partial derivatives x, y, Sometimes, it is important to keep track of all independent variables as well as which ones are being kept constant in the partial differentiation. As a result, the following notation is often employed x ) y,z where the subscripts y and z indicate that y and z are being kept constant while the function f is being partially differentiated with respect to x. This is often done in some scientific disciplines, e.g., Thermodynamics (which is why I am mentioning it here)., z. Formal definition of partial derivatives In what follows, we consider a function f(x,y) of only two variables the extension to more than two variables is quite straightforward. As you ll see below, the partial derivative of f with respect to a given independent variable, say x, uses the limit definition of derivatives you saw in single-variable calculus, while keeping the other variable(s) constant. The partial derivatives of f(x,y) at the point (a,b) R 2 are defined as follows: x (a,b) = y (a,b) = lim f(a + h,b) f(a,b) h 0 h lim k 0 f(a,b + k) f(a,b) k, (note that y = b is fixed) (10), (note that x = a is fixed) 26

6 provided that these limits exist. Recall the geometric or graphical interpretation of the derivative f (a) of a single-variable function f(x) at a point a: It is the slope of the tangent to the graph of f(x), i.e., the curve y = f(x), at x = a. y y = f(x) f(a) tangent to y = f(x) at x = a slope f (a) O a x Let us now determine the corresponding graphical interpretation for the function f(x, y) of two variables. Recall from a previous lecture that the graph of f, defined by z = f(x,y), is generally a surface in R 3. Let us now focus on the point (x,y) = (a,b) at which f(x,y) will be evaluated. It lies on the xy-plane, the space of input variables to the function f(x,y), see below. Directly above this point (a,b,0) exists the point (a,b,f(a,b) which lies on the graph/surface z = f(x,y). Now Let us keep y = b fixed and start varying x, by moving along the line (x,a,0). Directly above these points will be a set of points that form a curve C 1 that lies on the surface z = f(x,y): z z = f(x, y) tangent to C 1 P. curve C 1 z = f(x, b) x O. (a, b, 0) line y = b, z = 0 y This curve can also be defined as the intersection of the plane z = b with the surface z = f(x,y). It is the set of points defined as z = f(x,b) We ll call this function g(x), (11) 27

7 The instantaneous rate of change of f with respect to x, keeping y = b fixed is g (a), which is also the slope of the tangent to curve C 1 at point (a,b,f(a,b)). By definition, however, this rate of change hence the slope of the tangent is also the partial derivative x (a,b). Let us now go back to the input point (a,b,0) on the xy-plane and keep x = a fixed, varying y. This will produce a curve C 2 that is the intersection of the plane x = a with the graph z = f(x,y). The tangent to this curve at (a,b,f(a,b)) is the partial derivative (a,b), see below. y z z = f(x, y) P. curve C 2 curve C 1 x O. (a, b, 0) y This procedure has produced two tangent lines in the shape of a tire-iron that are both tangent to the surface z = f(x,y) at (x,y) = (a,b). We ll see below that they will uniquely determine an important plane that passes through the point (a,b,f(a,b)): the so-called linearization of f(x,y) at (a,b). Linearization, linear approximation and differentials (Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 14.4) Single-variable case f(x) Let us first go back to single-variable calculus and recall that the linearization of a function f(x) at the point x = a is given by L a (x) = f(a) + f (a)(x a). (12) (The right hand side is composed of the first two terms in the Taylor series of f about a.) This is the equation of a straight line that passes through the point (a,f(a)) with slope f (a). In other words, it matches both the value of the function f(a) and its derivative f (a) at the reference point x = a. 28

8 The linearization L a (x) provides the basis of the linear approximation of f at a: f(x) f(a) + f (a)(x a) for x near a. (13) Some simple examples: 1. sin x x for x near 0, 2. cos x 1 for x near 0, x x for x near 0. 2 Two-variable case f(x, y) With reference to the preceeding figure, we now wish to find the equation of the plane that is determined by the tire-iron at point (a,b,f(a,b)). We simply state the result, z = f(a,b) + (a,b)(x a) + (a,b)(y b), (14) x y and refer the reader to the brief discussion in the textbook, cf. pp Note that if we keep y = b fixed, we obtain the equation of a line passing through the point (a, b, f(a, b)) with slope / x(a, b). This is the tangent line to curve C 1. Likewise, if we keep x = a fixed, we obtain the equation of the tangent line to curve C 2. As such, this plane, sketched schematically below, is tangent the graph of f(x,y) at the point (a,b,f(a,b)). It is the two-variable analogue of the single-variable case, i.e., the tangent line to a function f(x). z z = f(x, y) z = L (a,b) (x, y) tangent plane P. x O. (a, b,0) y Linearization L (a,b) (x, y) of f(x, y) at the point (a, b). 29

9 As in the one-variable case, the tangent plane defines the linearization of the function f(x, y) at the point (a,b), denoted as follows, By construction, L (a,b) (x,y) = f(a,b) + (a,b)(x a) + (a,b)(y b). (15) x y The linear approximation associated with this linearization is L (a,b) (a,b) = f(a,b). (16) f(x,y) L (a,b) (x,y) for (x,y) near (a,b). (17) We ll consider some examples in the next lecture. 30

10 Lecture 5 Linearization, linear approximation and differentials (cont d) Let us now consider an example of linearization and linear approximation in the two-variable case. Example: Compute the linearization of the function f(x,y) = [1+2x+3y] 1/2 about (0,0). We must compute the appropriate value and derivatives of f at (0,0): 1. f(0,0) = We first compute and evaluate this derivative at (0,0): 3. Then compute and evaluate this derivative at (0,0): The linearization at (0,0) is thus given by x = 1 2 [1 + 2x + 3y] 1/2 (2) = (0,0) = 1. x y = 1 2 [1 + 2x + 3y] 1/2 (3) = x (0,0) = [1 + 2x + 3y] 1/2, 3 2[1 + 2x + 3y] 1/2. L (0,0) (x,y) = 1 + x + 3 y. (18) 2 Let us now examine the linear approximation afforded by this result. We seek to approximate f(x,y) at the point (x,y) = (0.1,0.02), which is quite close to (0,0). We use the linear approximation f(0.1,0.02) L (0,0) (0.1,0.02) (19) ( ) 3 = = =

11 The exact value of f(0.1,0.02), to five decimal digits is The error of this approximation is L (0,0) (0.1,0.02) f(0.1,0.002) (20) The relative error of an approximation is defined as follows, relative error = error of approximation. (21) actual value In this case, the relative error of the linear approximation is given by L (0,0) (0.1,0.02) f(0.1,0.002) f(0.1, 0.002) (22) The percentage relative error is 100 times the relative error, or roughly 1%. Incremental form of linear approximation and differentials Single-variable case f(x) Recall that the linearization L a (x) of a function f(x) of a single variable defines the linear approximation of f at a: f(x) f(a) + f (a)(x a) for x near a. (23) Now rewrite this relation by subtracting f(a) from both sides: f(x) f(a) f (a)(x a) for x near a. (24) The left side is the change in f(x) which is produced by a change in the independent variable as we move from x to x a. We may rewrite the above result as f f (a) x for x near a. (25) This is called the incremental form of the linear approximation to f at a. Note that it focuses on the change f in f produced by the change x = x a in the independent variable, as opposed to the actual values f(x) and f(a). In many applications, one is more interested in the change of a physical quantity, e.g., temperature, pressure, as opposed to the actual value of the quantity. As x gets smaller, we expect this approximation to be better. We can divide both sides by x 0 to give f x f (a). (26) 32

12 Once again, as x gets smaller, we expect the left side to provide better approximations of the derivative f (a). In the limit x 0, we write df dx = f (a). (27) provided that the limit of the quotient exists. (Note that the approximation sign has turned into an equality sign =.) We may rewrite this result as df = f (a)dx, (28) which is called the differential of f. It is a kind of infinitesimal limit of Eq. (25), where the changes f and x become infinitesimally small. We write f df as x dx, (29) The result df = f (a)dx relates the instantaneous change of f with respect to that of x. Our goal is now to obtain the differential form for functions of more than one variable. Two-variable case f(x,y) We now consider the linear approximation to a function f(x,y) yielded by its linearization at the point (a,b): This takes the form f(x,y) L (a,b) (x,y) for (x,y) close to (a,b). (30) f(x,y) f(a,b) + (a,b)(x a) + (a,b)(y b). (31) x y Subtracting f(a, b) from both sides of this relation yields the the following incremental form of the linear approximation, where f (a,b) x + (a,b) y, (32) x y f = f(x,y) f(a,b), x = x a, y = y b. (33) The incremental form of the linear approximation approximates f, the change in f that is produced by changes x and y in x and y, respectively. As in the one-variable case above, there is a differential of f(x,y) that is a kind of infinitesimal limit of f in (33) above, produced by letting changes in the two independent variables, x and y, 33

13 become infinitesimally small, i.e., x dx and y dy, respectively. The resulting differential of f(x,y) is df = dx + dy, (34) x y where it is understood that we must evaluate the partial derivatives at the reference point (a,b). Differentials and relative change In applications, one is often interested in the relative change of a property f which is given by f f, as opposed to f, the change in f. And often, the relative change is expressed as a percentage, e.g. 7.2%. The percentage relative change is given by 100 f. We show below that differentials are very f useful in this regard. Example 1: We start with a one-variable warmup example. The volume of a solid sphere of radius R is given by V (R) = 4 3 πr3. The differential of V (R) is given by dv = V (R)dR = 4πR 2 dr. (35) The change in volume of a sphere of radius R due to a change of R in its radius is therefore V 4πR 2 R. (36) You may know that 4πR 2 is the surface area of a sphere of radius R. Therefore Eq. (36) states that the change in volume V is roughly the surface area A(R) of the sphere multiplied by R. (You are basically flattening the thin onion skin spherical shell of radius R and thickness R to produce a flat carpet of area A(R) and thickness R.) Now suppose we are interested in the relative change in the volume of a sphere of radius R. We may calculate it as follows: V V 4πR2 R V = 4πR2 R 4 3 πr3 = 3 R R. (37) In other words, the relative change in volume of the sphere is three times the relative change in its radius. 34

14 Example 2: The area A of a rectangle of length L and width W is A = LW. The differential of the area function A(L,W) is therefore da = A A dl + dw (38) L W = WdL + LdW. The associated linear approximation to the change in area is A W L + L W. (39) Given values of L,W, W and L, we can compute the change in area A. If we divide A by A = LW, we obtain A A L L + W W. (40) Thus, the relative change in area is given approximately by the sum of the relative changes in L and W. As a result, we need only the latter two pieces of information to compute the former. Example 2: The pressure P, volume V and temperature T of one mole of an ideal gas obey the following ideal gas law equation PV = RT, (41) where R denotes the perfect gas constant. (Here, n = 1 in the ideal gas equation PV = nrt.) Let us rewrite Eq. (41) to define pressure in terms of temperature and volume, i.e. P(T,V ) = R T V. (42) Note that P increases with T and decreases with V. It remains to determine what happens to P when both T and V are changed simultaneously. The total differential will help here: dp = P P dt + dv (43) T V = R RT dt V V 2 dv. There is still a problem: How do we incorporate the given information, i.e., the relative changes in T and V into the above relation? Note that from the ideal gas law, we have: R V = P T and 35 RT V 2 = P V, (44)

15 so that the differential in (43) becomes dp = P T dt P dv (45) V or, dividing by P, dp P = dt T dv V. (46) This is the desired relation that relates the relative infinitesimal change in P to relative changes in T and V. The associated approximation in relative changes due to noninfinitesimal changes in T and V is P P T T V V. (47) 36

16 Lecture 6 Linearization (cont d) Extension to higher dimensions Let us return to the linearization of a two-variable function, f(x,y), L (a,b) (x,y) = f(a,b) + (a,b)(x a) + (a,b)(y b). (48) x y It will be useful to write this result in a more compact fashion. If we write a = (a,b), x = (x,y), (49) then where f(a) denotes the so-called gradient (vector) of f at a: L a (x) = f(a) + f(a) (x a), (50) f(a) = (a)i + (a)j. (51) x y The incremental form of the linear approximation may then be written as f = f(a) x, (52) where x = x a. The ideas of linearization, linear approximation and differentials are easily extended to scalarvalued functions of more variables, i.e., f : R n R, where n = 3,4,. And the compact notation introduced above is very helpful when we consider functions of many variables. We shall be particularly interested in the case n = 3 because of its application to three-dimensional physical space. The graph of a function f(x,y,z) will be a surface w = f(x,y,z) in four dimensions. In many cases, its behaviour will have to be represented in terms of sketches of its level sets f(x,y,z) = C, which will be surfaces in R 3. More on this later. Given a function f(x,y,z) of three variables, its gradient vector at a point (x,y,z) is defined as where the derivatives are evaluated at (x,y,z). f(x,y,z) = x i + y j + z k, (53) 37

17 The linearization of f(x,y,z) at the point a = (a,b,c) is defined as L (a,b,c) (x,y,z) = f(a,b,c) + x (a,b,c)(x a) + y This equation can be written in the compact form of Eq. (50), i.e., (a,b,c)(y b) (a,b,c)(z c). (54) z L a (x) = f(a) + f(a) (x a), (55) where a = (a,b,c) and x = (x,y,z) now have three components, as does the gradient vector f. And once again, the linearization of f furnishes its linear approximation at a: f(x) L a (x), for x near a. (56) In general, for functions f : R n R, the linearization at a point a = (a 1,a 2,,a n ) will have the same compact form as in Eq. (50), L a (x) = f(a) + f(a) (x a), (57) Here, the gradient vector has the general form f = x 1 e 1 + x 2 e x n e n, (58) where e k is the unit vector in the kth direction corresponding to the independent variable x k. This implies that the incremental form of the linearization of f at a may be written as follows, f f(a) x, (59) where x = x a. (60) We have now come to an important idea of this course: Associated with the scalar field f : R n R is a gradient vector field f. The question that remains: What does the gradient field f tell us about f? 38

18 The gradient vector and the directional derivative (Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 14.6) For simplicity, we examine in more detail the gradient vector associated with a function f(x,y): Examples: 1. f(x,y) = 50 x 2 2y 2. Then f(x,y) = 2xi 4yj. 2. f(x,y) = (x 2 + y 2 ) 1/2. Then f(x,y) = undefined at (0, 0). f or grad f = x i + j. (61) y x (x 2 + y 2 ) 1/2i + y (x 2 + y 2 ) 1/2j. Note that f(x,y) is We now consider the problem of finding the rate of change of a function f : R n R with respect to changes in its independent variables x i. The problem is that when n is greater than one, there are many directions that one can take. (When n = 1, there was only one direction to take along the x-axis.) In the following discussion, we consider the special case n = 2, i.e., functions f(x,y) of two variables, since it can be illustrated graphically. The results obtained below will generalize to the case of arbitrary n. We sketch the situation schematically in the figure below. We start at point P, with coordinates a = (a 1,a 2 ), where f assumes the value f(a) = f(a 1,a 2 ). We now move away from P in the direction of a unit vector û = (u 1,u 2 ) and wish to know the instantaneous rate of change of f in that direction. The rate of change is f(q) f(p) lim Q P, (62) PQ where Q is restricted to the line segment starting at P and having direction û. The rate of change of f in the direction of û is called the directional derivative of f and is denoted as Dûf(a). We already know two special cases of this derivative: 1. û = i, i.e., motion in the x-direction. Then by definition, the rate of change of f is D i f = x. 2. û = j, i.e., motion in the y-direction. Then by definition, the rate of change of f is D j f = y. What about all other directions û = (u 1,u 2 )? Let us derive the result. 39

19 z z = f(x, y). f(p). f(q) x. P û O. Q y We let the point Q be given by b = (b 1,b 2 ), where In component form, this becomes b = a + sû, s 0. (63) (b 1,b 2 ) = (a 1,a 2 ) + s(u 1,u 2 ). (64) Just to backtrack a bit: This construction ensures that point Q lies in the direction û from point P. By letting s approach zero, Q will approach P. Now let s go back to the rate of change/directional derivative in Eq. (62). We need to translate both numerator and denominator into workable quantities. First, note that the vector PQ represents the displacement x = ( x, y) from the reference point P to Q: In components, this implies that x = PQ = b a = sû. (65) ( x, y) = s(u 1,u 2 ) = (su 1,su 2 ). (66) Furthermore, note that PQ = x = s û = s. (67) Here, we have used the fact that s is nonnegative and û is a unit vector. We now have to take care of the numerator in Eq. (62), i.e., f(q) f(p) = f(b) f(a) = f, (68) 40

20 where f denotes the change in f produced by moving from P to Q. We assume that Q is close to P, i.e., s is very small, and use the incremental form of the linear approximation to f at a: f x (a) x + y (a) y x (a)su 1 + y (a)su 2. (69) We now have all of the ingredients for the rate of change expression in Eq. (62): f(q) f(p) PQ = f s x (a)u 1 + y (a)u 2. (70) Note the appearance of the approximation sign, because we are using the linear approximation. We claim here, without proof, that in the limit s 0, this approximation becomes an equality, i.e. f(q) f(p) lim Q P PQ f = lim s 0 s = x (a)u 1 + y (a)u 2 = f(a) û. (71) In other words, the desired rate of change the directional derivative of f at a in the direction û is simply the dot product of the gradient vector f(a) with the unit direction vector û: Dûf(a) = f(a) û = x (a)u 1 + y (a)u 2. (72) It turns out that the gradient vector f(a) contains all the information necessary to compute the directional derivative of f at a in any direction! Note: From Eq. (72), we see that the directional derivative Dûf(a) is the projection of the gradient vector f(a) in the direction of the unit vector û. Example: Consider the function f(x,y) = 50 x 2 2y 2. We shall assume that it represents the temperature at a point (x,y) on the surface of a hotplate for a suitable range of values of x and y. The level sets of f(x,y) are ellipses that are centered about the origin, as sketched roughly below. The maximum temperature of the hotplate is at (0, 0). As we move away from the origin, the temperature decreases the level-set values decrease. 41

21 y C = x Some level-set curves f(x, y) = C of the hotplate function f(x, y) = 50 x 2 2y 2. Now suppose that an ant is located at point (1, 1) where the temperature is f(1, 1) = 47. The ant would like to travel away from this point in an effort to cool itself. Let s compute the instantaneous rate of change of f in several directions to see which ones would be wise or unwise choices. This amounts to computing the directional derivative Dûf(1, 1). Since f = 2xi 4yj, (73) we have f(1, 1) = 2i + 4j = ( 2,4). (74) Here are the directions from point (1, 1) which we shall investigate: y x Directions from (1, 1) investigated in the ant-hotplate problem. 1. Direction û = i = (1,0). Then Dûf(1, 1) = f(1, 1) (1,0) = ( 2,4) (1,0) = 2. 42

22 The rate of change is negative, implying cooling, which is a good sign. 2. Direction û = i = ( 1,0). Then Dûf(1, 1) = f(1, 1) ( 1,0) = ( 2,4) ( 1,0) = 2. The rate of change is positive, implying warming, which is not a good sign. 3. Direction û = j = (0,1). Then Dûf(1, 1) = f(1, 1) (0,1) = ( 2,4) (0,1) = 4. Warming again, even worse than Case 2! 4. Direction û = j = (0, 1). Then Dûf(1, 1) = f(1, 1) (0, 1) = ( 2,4) (0, 1) = 4. Cooling, even better than Case 1! Of the four directions examined above, Cases 2 and 4 corresponded to cooling, with Case 4 being the better of the two. Recall that the hottest point of the hotplate is at (0,0). In both of these cases, the ant is moving farther away from (0,0). Is Case 4 the best of all possible directions û from point (1, 1)? We don t know. Let s examine another possible contender: the direction from point (1, 1) that points directly away from the ( 1 origin, i.e., the direction vector u = (1, 1) with associated unit vector û = 2, 1 ) Direction û = ( 1 2, 1 2 ). ( Dûf(1, 1) = f(1, 1 1) 2, 1 ) 2 43

23 ( 1 = ( 2,4) 2, 1 ) 2 = 6 2 = The rate of cooling is even greater than that of direction 4 above. But is this direction the best of all? In order to answer this question, we need to return to Eq. (72). 44