3 Applications of partial differentiation

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1 Advanced Calculus Chapter 3 Applications of partial differentiation 37 3 Applications of partial differentiation 3.1 Stationary points Higher derivatives Let U R 2 and f : U R. The partial derivatives f x and f y are functions of x and y and so we can find their partial derivatives. We write f xy to denote f y differentiated with respect to x. Similarly f xx denotes f x differentiated with respect to x, and f yx and f yy denote f x and f y, respectively, differentiated with respect to y. The functions f xx, f xy, f yx and f yy are the second partial derivatives of f. For the other notation for partial derivatives, we write f xy, f yy. ( ) or just 2 f ( ) 2 or just for f xx, 2 f for f yx and ( ) or just ( ) 2 f for or just 2 f 2 We can also differentiate the second partial derivatives to get the third partial derivatives, and so on. For example, f xyy, or 3 f, is 2 the third partial derivative obtained from differentiating f yy with respect to x. for Example 1 Let f(x, y) = 3x 3 y + 2xy 2 4x 2 y. Then f x (x, y) = 9x 2 y + 2y 2 8xy; f y (x, y) = 3x 3 + 4xy 4x 2 ; f xx (x, y) = 18xy 8y; f yx (x, y) = 9x 2 + 4y 8x; f yy (x, y) = 4x; f xy (x, y) = 9x 2 + 4y 8x. Note that, in Example 1, f yx = f xy. This is true for all well-behaved functions. In particular, if f xy and f yx are both continuous on R 2 (or an open subset U of R 2 ) then f yx = f xy on R 2 (U).

2 Advanced Calculus Chapter 3 Applications of partial differentiation 38 Example 2 Let z = x cos 2y. Then z = cos 2y, z = 2x sin 2y, 2 z 2 = 0, 2 z = 2 sin 2y, 2 z = 2 sin 2y, 2 z 2 = 4x cos 2y. Example 3 Let z = xe2x. Find all the possible values of n given that y n 3x 2 z 2 xy2 2 z 2 = 12z. For z = xe2x y n we have z = e2x + 2xe 2x y n, = e2x (1 + 2x), y n z = nxe2x y, n+1 2 z = 2e2x (1 + 2x) + e 2x 2, 2 y n 2 z 2 = = 4e2x (x + 1) y n, n(n + 1)xe2x y n+2. Thus 3x 2 z 2 xy2 2 z = 12xe2x (x + 1) 2 y n n(n + 1)x2 e 2x y n = xe2x (12(x + 1) n(n + 1)x), yn

3 Advanced Calculus Chapter 3 Applications of partial differentiation 39 and so 3x 2 z 2 z xe2x 12xe2x 2 xy2 = 12z (12(x + 1) n(n + 1)x) =, 2 y n y n 12(x + 1) n(n + 1)x = 12, (Since the result is true for all x and y.) 12x + 12 n(n + 1)x = 12, 12 n(n + 1) = 0, (Since the result is true for all x.) n 2 + n 12 = 0, (n + 4)(n 3) = 0, n = 4 or n = 3. Stationary points Let U R 2, and let f : U R. Then (a, b) is a stationary point of f if the tangent plane at (a, b) is horizontal. That is, the tangent plane is parallel to the (x, y)-plane, and so has the form z = c for some constant c. Since the tangent plane at (a, b) passes through the point (a, b, f(a, b)), then if it is horizontal it has the form z = f(a, b). Recall from Chapter 1 that the equation of the tangent plane at (a, b) is f(a, b) z + f x (a, b)(x a) + f y (a, b)(y b) = 0. Now, if (a, b) is a stationary point then f(a, b) z = 0, and so f x (a, b)(x a) + f y (a, b)(y b) = 0. The last equation is true for all x and y. Putting x = a and y any value other than b into this equation gives f y (a, b) = 0. Similarly, putting y = b and x any value other than a into this equation gives f x (a, b) = 0. Hence, if (a, b) is a stationary point of f then f x (a, b) = 0 and f y (a, b) = 0. It is straightforward to establish the converse of this result. Thus, to find the stationary points of f, we have to find the solutions of the equations f x (a, b) = 0, f y (a, b) = 0. Example 4 Find all of the stationary points of f(x, y) = x 2 y + 3xy 2 3xy.

4 Advanced Calculus Chapter 3 Applications of partial differentiation 40 The partial derivative of f are f x (x, y) = 2xy + 3y 2 3y = y(2x + 3y 3); f y (x, y) = x 2 + 6xy 3x = x(x + 6y 3). Putting f x (x, y) = f y (x, y) = 0 gives y(2x + 3y 3) = 0, (1) x(x + 6y 3) = 0. (2) From equation (1) either y = 0 or 2x + 3y = 3. If y = 0 then equation 2 gives x(x 3) = 0, and so x = 0, 3. If 2x + 3y = 3 then 6y = 6 4x. Thus, in this case, equation 2 gives x(3 3x) = 0, and so x = 0, 1 and y = 1, 1, respectively. Thus the stationary points 3 of f are (0, 0), (0, 3), (0, 1) and (1, 1). 3 Types of stationary points There are three main types of stationary point: a local minimum, a local maximum and a saddle point. The following diagrams show a typical graph and contour-plot of these three types. A local minimum y x y 1 x A local maximum xy y x

5 Advanced Calculus Chapter 3 Applications of partial differentiation 41 A saddle point x y y x Classifying stationary points To determine the nature of a stationary point (a, b) of f we first find the Hessian matrix, ( ) fxx f xy. Here we are assuming that f xy f yy f xy = f yx. We then evaluate the determinant of the Hessian matrix at the point at (a, b). That is, we find the value, at (a, b), of ( ) fxx f = det xy = f xx f yy (f xy ) 2. f xy f yy If > 0 and f xx > 0 then (a, b) is a local minimum; If > 0 and f xx < 0 then (a, b) is a local maximum; If < 0 then (a, b) is a saddle point. If = 0 then this test provides no information about the nature of the stationary point. Other methods are required to classify the stationary point in this case. Example 5 Find all of the stationary points of and determine their nature. f(x, y) = x 2 y + 3xy 2 3xy, From Example 4 the stationary points of f are (0, 0), (0, 3), (0, 1) and (1, 1 3 ), f x(x, y) = 2xy + 3y 2 3y = y(2x + 3y 3) and

6 Advanced Calculus Chapter 3 Applications of partial differentiation 42 f y (x, y) = x 2 + 6xy 3x = x(x + 6y 3). Now f xx = 2y, f xy = 2x + 6y 3(= f yx ), f yy = 6x. Thus ( = det 2y 2x + 6y 3 2x + 6y 3 6x ) = 12xy (2x + 6y 3) 2. When x = y = 0, = 9 < 0, and so (0, 0) is a saddle point. When x = 3 and y = 0, = 9 < 0, and so (3, 0) is a saddle point. When x = 0 and y = 1, = 9 < 0, and so (0, 1) is a saddle point. When x = 1 and y = 1 3, = 3 > 0, f xx = 2 3 > 0, and so (1, 1 3 ) is a local minimum. Example 6 Find and classify the stationary points of f(x, y) = xye x+y. First we find the partial derivatives of f. f x (x, y) = ye x+y + xye x+y = y(x + 1)e x+y, f y (x, y) = x(y + 1)e x+y. (Since f is symmetrical in x and y.) Putting f x (x, y) = f y (x, y) = 0, and using the fact that e x+y 0, gives y(x + 1) = 0, (3) x(y + 1) = 0. (4) From equation (3) either y = 0 or x = 1. If y = 0 then equation (4) gives x = 0. If x = 1 then equation (3) gives y = 1. Thus the stationary points of f are (0, 0), and ( 1, 1). Now Thus f xx = ye x+1 + y(x + 1)e x+y = y(x + 2)e x+y, f xy = (y + 1)e x+y = x(y + 1)e x+y = (y + 1)(x + 1)e x+y, f yy = x(y + 2)e x+y. ( = det ) y(x + 2)e x+y (y + 1)(x + 1)e x+y (y + 1)(x + 1)e x+y x(y + 2)e x+y = e 2(x+y) ( xy(x + 2)(y + 2) (y + 1) 2 (x + 1) 2). When x = y = 0, = 1 < 0, and so (0, 0) is a saddle point. When x = y = 1, = e 4 > 0, f xx = e 2 < 0, and so ( 1, 1) is a local maximum.

7 Advanced Calculus Chapter 3 Applications of partial differentiation 43 Examples when = 0 We now consider the problem of classifying a stationary point (a, b) when = 0 at (a, b). There is no single method for determining the nature of (a, b) in this case, although it is often useful to find f(x, y) f(a, b) in order to determine the behaviour of f near (a, b); an alternative method is to consider how f varies along curves (normally straight lines) passing through (a, b). Example 7 Find and classify the stationary points of f(x, y) = x 2 + y First we find the partial derivatives of f. f x (x, y) = 2x, f y (x, y) = 4y 3. Putting f x (x, y) = f y (x, y) = 0, gives x = 0, y = 0. Thus the only stationary points of f is (0, 0). Now f xx = 2, f xy = 0, f yy = 12y 2. Thus ( 2 0 = det 0 12y 2 = 24y 2. ) Thus, at (0, 0), = 0, and so the Hessian gives us no information about the nature of this stationary point. However, for (x, y) (0, 0), note that f(x, y) f(0, 0) = x 4 + y , = x 2 + y 4 > 0. Thus f(x, y) > f(0, 0) for all (x, y) (0, 0). Hence (0, 0) is a local minimum. Example 8 Find and classify the stationary points of f(x, y) = xy 2 x 2 y 2 + x First we find the partial derivatives of f. = y2 2xy 2 + 4x 3, = 2xy 2x 2 y.

8 Advanced Calculus Chapter 3 Applications of partial differentiation 44 Putting = = 0 gives y 2 2xy 2 + 4x 3 = 0, (5) 2xy(1 x) = 0. (6) From equation (6) x = 0, y = 0 or x = 1. If either x = 0 or y = 0 then equation (5) gives y = 0 and x = 0, respectively. If x = 1 then equation (5) gives y 2 = 4, and so y = 2 or y = 2. Thus the stationary points of f are (0, 0), (1, 2) and (1, 2). Now 2 f 2 = 2y x 2, 2 f = 2y 4xy, 2 f 2 = 2x 2x 2. At the stationary point (0, 0), each of the second derivatives is zero. Thus = 0 at (0, 0), and so the Hessian gives no information about the nature of this stationary point. However, for (x, y) (0, 0), note that f(x, y) f(0, 0) = xy 2 x 2 y 2 + x 4. In particular, for x = y we get f(x, x) f(0, 0) = x 3 x 4 + x 4 = x 3. Since the sign of x 3 is the same as the sign of x, it follows that f(x, x) f(0, 0) > 0, when x > 0, and f(x, x) f(0, 0) < 0, when x < 0. Thus (0, 0) is a saddle point. The nature of the other stationary points of f can be determined using the Hessian. This is left as an exercise. Example 9 Find the stationary points of z = (x 2 + y 2 )e x2 y 2, and determine their nature. The partial derivatives of z are z z y2 = 2xe x2 + (x 2 + y 2 )( 2x)e x2 y 2 = 2xe x2 y 2 (1 x 2 y 2 ), = 2ye x2 y 2 (1 x 2 y 2 ). (Since z is symmetrical in x and y.) Putting z = z = 0, and using the fact that e x2 y 2 0 gives x(1 x 2 y 2 ) = 0, (7) y(1 x 2 y 2 ) = 0. (8)

9 Advanced Calculus Chapter 3 Applications of partial differentiation 45 From equation (7) either x = 0 or x 2 + y 2 = 1. If x = 0 then equation (8) gives y(1 y 2 ) = 0, and so y = 0, 1 or 1. Thus (0, 0), (0, 1) and (0, 1) are all stationary points of z. If x 2 + y 2 = 1 then equation (8) always holds. Thus every point on the circle x 2 +y 2 = 1 is stationary points of z. From equation (8) either y = 0 or x 2 + y 2 = 1. The second possibility has already been covered above; if y = 0 then equation (7) gives x(1 x 2 ) = 0, and so x = 0, 1 or 1. Thus (0, 0), (1, 0) and ( 1, 0) are all stationary points of z. Note that all of the points (0, 1), (0, 1), (1, 0) and ( 1, 0) lie on the circle x 2 + y 2 = 1. Thus the stationary points of z are (0, 0) and every point on the circle x 2 + y 2 = 1. Finding the second derivatives of z we get: 2 z x2 y2 = (2e + 2x( 2x)e x2 y 2 )(1 x 2 y 2 ) + 2xe x2 y 2 ( 2x), 2 ( = 2e x2 y 2 (1 2x 2 )(1 x 2 y 2 ) 2x 2), 2 z = 2e x2 y 2 ( 1 x 2 y 2 2x 2 + 2x 4 + 2x 2 y 2 2x 2), = 2e x2 y 2 ( 1 5x 2 y 2 + 2x 4 + 2x 2 y 2), = 2e x2 y 2 ( 1 5x 2 y 2 + 2x 4 + 2x 2 y 2), = 2e x2 y 2 ( 1 4x 2 (x 2 + y 2 ) + 2x 2 (x 2 + y 2 ) ), = ( 2y)2xe x2 y 2 (1 x 2 y 2 ) + 2xe x2 y 2 ( 2y), = 4xye x2 y 2 (2 x 2 y 2 ), 2 z 2 = ( 2e x2 y2 1 4y 2 (x 2 + y 2 ) + 2y 2 (x 2 + y 2 ) ), (Since z is symmetrical in x and y.) If x 2 + y 2 = 1 then 2 z 2 = 2e x2 y2 ( 1 4x 2 (x 2 + y 2 ) + 2x 2 (x 2 + y 2 ) ), = 2e 1 ( 1 4x x 2), = 4x 2 e 1, 2 z = 4xye x2 y 2 (2 x 2 y 2 ), = 4xye 1, 2 z 2 = ( 2e x2 y 2 1 4y 2 (x 2 + y 2 ) + 2y 2 (x 2 + y 2 ) ), = 2e 1 ( 1 4y y 2), = 4y 2 e 1.

10 Advanced Calculus Chapter 3 Applications of partial differentiation 46 Thus, in this case, = 2 z 2 2 z 2 ( ) 2 2 z, = ( 4x 2 e 1) ( 4y 2 e 1) ( 4xye 1) 2, = 16x 2 y 2 e 2 16x 2 y 2 e 2 = 0. Thus the Hessian gives us no information about the nature of the stationary points on the circle x 2 +y 2 = 1. To determine the nature of these stationary points, we consider the behaviour of z on the lines y = ax, where a R. On such a line z(x, y) = z(x, ax), = (x 2 + a 2 x 2 )e x2 a 2 x 2, = (a 2 + 1)x 2 e (a2 +1)x 2. Let g(x) = (a 2 +1)x 2 e (a2 +1)x 2. We make the following observations about g. Since g can be expressed as a function of x 2 it is symmetrical about x = 0. That is, g(x) = g( x). Since a > 0, x 2 0 and e (a2 +1)x 2 > 0, g(x) 0 for all x R. Furthermore, g(x) = 0 if and only if x = 0. As x ±, g(x) 0. From these observations we can deduce that g(x) takes its minimum value at x = 0. The value of g(x) then increases as we move away from x = 0 until g reaches a maximum; the value then decreases and approaches 0 as x approaches ±. The following diagram shows g for a typical value of a. From this we can see that every point on the circle x 2 + y 2 = 1 is a local maximum of z. We also get that (0, 0) is a local minimum.

11 Advanced Calculus Chapter 3 Applications of partial differentiation 47 The following diagram illustrates the graph of f that clearly shows the local maxima on the circle x 2 + y 2 = y x 2-2 Genralization to functions of several variables We can generalize the idea of a stationary point to a function of several variables. Let U R n, and let f : U R. Then (a 1, a 2,..., a n ) U is a stationary point of f if f x1 (a 1, a 2,..., a n ) = f x2 (a 1, a 2,..., a n ) = = f xn (a 1, a 2,..., a n ) = 0. It is also possible to classify the stationary points, but this is beyond the scope of the course. Example 10 Find the stationary points of f(x, y, z) = (x + y + z) 2 6xyz. The partial derivatives of f are f x (x, y, z) = 2(x + y + z) 6yz, f y (x, y, z) = 2(x + y + z) 6xz, f z (x, y, z) = 2(x + y + z) 6xy. Putting f x (x, y, z) = f y (x, y, z) = f z (x, y, z) gives x + y + z = 3yz, (9) x + y + z = 3xz, (10) x + y + z = 3xy. (11) Combining equations (9) and 10) gives 3yz = 3xz.

12 Advanced Calculus Chapter 3 Applications of partial differentiation 48 Thus either z = 0 or x = y. If z = 0 then equations (9) (11) reduce to x+y = 0 and x+y = 3xy. From this we get x = y = 0, and so (0, 0, 0) is a stationary point of f. If x = y then equations (9) (11) reduce to 2x + z = 3xz and 2x + z = 3x 2. From this we get xz = x 2, and so either x = 0 or x = z. If x = 0 we get x = y = z = 0, giving the stationary point found earlier. If x = z and x 0 then 2x+z = 3xz and 2x+z = 3x 2 reduce to x = x 2, and so x = 1. This gives x = y = z = 1, and so (1, 1, 1) is a stationary point of f. Thus the stationary points of f are (0, 0, 0) and (1, 1, 1). Exercises Let f(x, y) = 3x 3 y 2 5xy 3 + 3x 2 y 4 y 5. Find 2 f. 2. Determine all of the values of n such that z = 2xy + x n y 2n satisfies 2x 2 2 z 2 z 2 y2 + 18z = 36xy Find, and classify, the stationary points of g(x, y) = 1 2 x2 y 2xy y3. 4. Determine the nature of the non-zero stationary points of the function f given in Example Find the stationary points of f(x, y, z) = x 3 3x + y 3 3yz + 2z Lagrange multipliers Let U R 2 and let f, g : U R. We now consider the problem of finding the maximum and minimum values of f subject to the constraint g(x, y) = 0. Method 1 Suppose we can rewrite g(x, y) = 0 in the form y = h(x), for some function h. Then we can just find the maximum and minimum values of F (x) = f(x, h(x)) using the method for calculus of functions of one variable.

13 Advanced Calculus Chapter 3 Applications of partial differentiation 49 Example 11 Find the maximum and minimum values of f(x, y) = x 3 + 3xy 2 + 2xy, subject to the constraint x + y = 4. Here g(x, y) = x + y 4, and g(x, y) = 0 gives y = 4 x. (So h(x) = 4 x.) Thus F (x) = f(x, 4 x), = x 3 + 3x(4 x) 2 + 2x(4 x), = x 3 + 3x(16 8x + x 2 ) + 8x 2x 2, = 4x 3 26x x. Differentiating F with respect to x we get F (x) = 12x 2 52x + 56, and so F (x) = 0 12x 2 52x + 56 = 0, 3x 2 13x + 14 = 0, (3x 7)(x 2) = 0, x = 2, 7 3. For x = 2, y = 2 and F (2) = 40, and for x = 7, y = 5 and 3 3 F ( 7 25 ) = 39. Thus the maximum and minimum values of f(x, y) = 3 27 x 3 + 3xy 2 + 2xy, subject to the constraint x + y = 4, are 40 and 39 25, respectively. 27 Method 2 The local maximum and minimum of f(x, y) subject to the constraint g(x, y) = 0 correspond to the stationary points of L(x, y, λ) = f(x, y) λg(x, y). The variable λ is call a Lagrange multiplier. Example 12 Consider the problem from Example 11. Here f(x, y) = x 3 + 3xy 2 + 2xy and g(x, y) = x + y 4. So we let L(x, y, λ) = x 3 + 3xy 2 + 2xy λ(x + y 4). Now = 3x2 + 3y 2 + 2y λ, = 6xy + 2x λ, λ = (x + y 4).

14 Advanced Calculus Chapter 3 Applications of partial differentiation 50 Putting = = λ = 0 gives From equations (12) and (13) we get 3x 2 + 3y 2 + 2y = λ, (12) 6xy + 2x = λ, (13) x + y 4 = 0. (14) 3x 2 + 3y 2 + 2y = 6xy + 2x. We can now use equation (14) to eliminate y form the previous equation. After simplification this gives 12x 2 52x + 56 = 0 The solutions can now be found as in Example 11. Note that in Example 12 we did not need to find λ in order to find the solution to the problem. Lagrange multipliers are particularly useful when it is difficult (or just algebraically messy!) to rewrite g(x, y) = 0 in the form y = h(x). Example 13 Find the maximum and minimum values of f(x, y) = xy 3 on the circle x 2 + y 2 = a 2. Let g(x, y) = x 2 + y 2 a 2. Then we want to find the maximum and minimum values of f(x, y) subject to the constraint g(x, y) = 0. Let L(x, y, λ) = xy 3 λ(x 2 + y 2 a 2 ). Then Putting = = λ = 0 gives = y3 2λx, = 3xy 2 2λy, λ = x2 y 2 + a 2. y 3 2λx = 0, (15) 3xy 2 2λy = 0, (16) x 2 + y 2 = a 2. (17) Multiplying equation (15) by y, and equation (16) by x gives y 4 2λxy = 0, (18) 3x 2 y 2 2λxy = 0, (19)

15 Advanced Calculus Chapter 3 Applications of partial differentiation 51 and so subtracting equation (19) from equation (18) we get Thus and so either y = 0 or y 2 = 3x 2. y 4 3x 2 y 2 = 0. y 2 (y 2 3x 2 ) = 0, If y = 0 then equation (17) gives x 2 = a 2, and so x = ±a. If y 2 = 3x 2 then equation (17) gives 4x 2 = a 2. Thus x = ± a 2, and for each value of x, y = ± 3a 2. From the preceding calculations, the stationary points of L have ( a (x, y) = (a, 0), ( a, 0),, ) ( 3a a,, ) ( 3a, a, ) ( 3a or ( a Now f(a, 0) = f( a, 0) = 0, f, ) ( 3a = f a, ) 3a ( a and f, ) ( 3a = f a, ) 3a = a 4. a 2, 3a 2 = a 4, Thus the maximum and minimum values of f(x, y) = xy 3 on the circle x 2 + y 2 = a 2 are a 4 and a 4, respectively. Example 14 Find the point on the line 3x + 2y = 5 that is closest to the point (3, 1). The distance between a general point (x, y) and the point (3, 1) is (x 3)2 + (y 1) 2. ). We want to find the minimum value of this distance subject to the constraint 3x+2y 5 = 0. In fact, it is easier to minimize the square of the distance, and so we minimize f(x, y) = (x 3) 2 + (y 1) 2, subject to the given constraint. Let L(x, y, λ) = (x 3) 2 + (y 1) 2 λ(3x + 2y 5). Then λ = 2(x 3) + 3λ, = 2(y 1) + 2λ, = 3x 2y + 5. Putting = = λ = 0 gives 2(x 3) + 3λ = 0, (20) 2(y 1) + 2λ = 0, (21) 3x + 2y = 5. (22)

16 Advanced Calculus Chapter 3 Applications of partial differentiation 52 Multiplying equation (20) by 2, and equation (21) by 3 gives 4(x 3) + 6λ = 0, 6(y 1) + 6λ = 0. Thus 4(x 3) = 6(y 1). Multiplying out the brackets in this equation, and simplifying the result gives 2x 3y = 3. (23) Multiplying equation (22) by 3, and equation (23) by 2 gives 9x + 6y = 15, 4x 6y = 6. Adding the last two equations together we get 13x = 21, and so x = Using equation (23) with x = we get y =. Thus the point ( 21, 1 ) on the line on the line 3x + 2y = 5 is closest to the point (3, 1). Although not required, the distance of the point ( 21, 1 ) to the point (3, 1) is ( 21 f 13, 1 ) (21 ) 2 ( ) 2 1 = , 324 = = 169, 36 = 13 = Lagrange multipliers with more than one constraint Let U R n, and let f, g 1, g 2,..., g m : U R. To find the maximum and minimum values of f(x 1, x 2,..., x n ), subject to the constraints we find the stationary points of g 1 (x 1, x 2,..., x n ) = 0, g 2 (x 1, x 2,..., x n ) = 0, g m (x 1, x 2,..., x n ) = 0, L(x 1, x 2,..., x n, λ 1, λ 2,..., λ m ) = f(x 1, x 2,..., x n ) λ 1 g 1 (x 1, x 2,..., x n ). λ 2 g 2 (x 1, x 2,..., x n ) λ m g m (x 1, x 2,..., x n ). The introduced variables λ 1, λ 2,..., λ m are called Lagrange multipliers.

17 Advanced Calculus Chapter 3 Applications of partial differentiation 53 Example 15 Find the maximum and minimum values of x + z at the points where the plane 4x + y + z = 34 and the cylinder x 2 + y 2 = 40 intersect. Here we want to find the maximum and minimum values of f(x, y, z) = x + z, subject to the constraints g 1 (x, y, z) = 0 and g 2 (x, y, z) = 0, where g 1 (x, y, z) = 4x + y + z 34 and g 2 (x, y, z) = x 2 + y Let L(x, y, z, λ, µ) = x + z λ(4x + y + z 34) µ(x 2 + y 2 40). Then = 1 4λ 2µx, = λ 2µy, z = 1 λ, λ = (4x + y + z 34), µ = (x2 + y 2 40). Putting = = z = λ = µ = 0 gives 4λ + 2µx = 1, (24) λ + 2µy = 0, (25) λ = 1, (26) 4x + y + z = 34, (27) x 2 + y 2 = 40. (28) From equation (26), λ = 1. Substituting this value of λ into equations (24) and (25) gives 2µx = 3 and 2µy = 1, respectively. Multiplying the second of these equations by 3 gives 6µy = 3, and so 2µx = 3 = 6µy. Now µ 0 (otherwise equation (25) gives λ = 0, contrary to λ = 1) and so x = 3y. Substituting x = 3y into equation (28) gives 10y 2 = 40. Thus y 2 = 4, and so y = ±2. When y = 2, x = 6, and when y = 2, x = 6. From equation (27), z = 34 4x y. Thus when y = 2 and x = 6, z = 8, and when y = 2 and x = 6, z = 60. Now f(6, 2, 8) = 14 and f( 6, 2, 60) = 54. Thus the maximum and minimum values of x + z at the points where the plane 4x + y + z = 34 and the cylinder x 2 + y 2 = 40 intersect are 54 and 14, respectively.

18 Advanced Calculus Chapter 3 Applications of partial differentiation 54 Example 16 Find the maximum and minimum values of f(x, y, z) = xy + 4z subject to the constraints x + y + z = 0 and x 2 + y 2 + z 2 = 24. Let L(x, y, z, λ, µ) = xy + 4z λ(x + y + z) µ(x 2 + y 2 + z 2 24). Then = y λ 2µx, = x λ 2µy, z = 4 λ 2µz, λ = (x + y + z), µ = (x2 + y 2 + z 2 24). Putting = = z = λ = µ = 0 gives From equations (29) and (30) we get Thus y 2µx = λ, (29) x 2µy = λ, (30) 4 2µz = λ, (31) x + y + z = 0, (32) x 2 + y 2 + z 2 = 24. (33) y 2µx = λ = x 2µy. y x = 2µ(x y), and so either x = y or µ = 1. We now consider these two possibilities 2 separately. If x = y then equation (32) gives 2x + z = 0, and so z = 2x. Then equation (33) gives x 2 + x 2 + 4x 2 = 24, and so x 2 = 4. Hence x = ±2, and so x = ±2, y = ±2, z = 4. If µ = 1 then equations (29) and (31) give x+y = λ and z+4 = λ, 2 respectively. Thus x + y = λ = z + 4, and so x + y z = 4. (34)

19 Advanced Calculus Chapter 3 Applications of partial differentiation 55 Subtracting equation (34) from equation (32) gives 2z = 4, and so z = 2. Adding equations (34) and (32) gives 2(x + y) = 4, and so y = 2 x. Putting y = 2 x and z = 2 into equation (33) we get x 2 + (2 x) 2 + ( 2) 2 = 24 x 2 + (4 4x + x 2 ) + 4 = 24, x 2 2x 8 = 0, (x 4)(x + 2) = 0, x = 2, 4. When x = 2, y = 4 and z = 2, and when x = 4, y = 2 and z = 2. From the preceding calculations we have found four stationary points of L, namely (2, 2, 4), ( 2, 2, 4), ( 2, 4, 2) and (4, 2, 2). Now f(2, 2, 4) = 12, f( 2, 2, 4) = 20, f( 2, 4, 2) = 16 and f(4, 2, 2) = 16. Thus the maximum and minimum values of f(x, y, z), subject to the given constraints, are 20 and 16, respectively. Exercises Find the maximum and minimum values of f(x, y) = xy on the ellipse 18x 2 + 2y 2 = Find the maximum and minimum values of 2x + y on the ellipse x 2 + xy + 4y 2 + 2x + 16y + 7 = Find the maximum and minimum values of xy 2 on the circle x 2 + y 2 = Find the minimum distance from the origin to a point on the curve where the plane 2y + 4z = 15 and the surface z 2 = 4(x 2 + y 2 ) intersect. 3.3 The chain rule In this section we generalize the chain rule for functions of one variable to functions of more than one variable. First we extend our idea of a real function. Multivariable functions Let U R n. A (multivariable) function f : U R m determines for each point x = (x 1, x 2,..., x n ) of U a unique point y = (y 1, y 2,..., y m ) of R m. We write f(x 1, x 2,..., x n ) = (y 1, y 2,..., y m ) or f(x) = y.

20 Advanced Calculus Chapter 3 Applications of partial differentiation 56 Each coordinate of y is uniquely determined by (x 1, x 2,..., x n ). That is, each coordinate of y can be thought of as a function of (x 1, x 2,..., x n ). So there are functions f 1, f 2,..., f m : U R m such that f(x 1, x 2,..., x n ) = (f 1 (x 1, x 2,..., x n ), f 2 (x 1, x 2,..., x n ),..., f m (x 1, x 2,..., x n )), or f(x) = (f 1 (x), f 2 (x),..., f m (x)). Example 17 Let f : R 2 R 2, with f(x, y) = (x 2 y, x + 3y). Here f 1 (x, y) = x 2 y and f 2 (x, y) = x+3y. An alternative way of defining f is to write f(x, y) = (f 1, f 2 ) where f 1 (x, y) = x 2 y and f 2 (x, y) = x + 3y. Example 18 Let f : R R 3, with f(t) = (t, cos t, sin t). Functions of one variable from R to R m, like this one (that has m = 3), define a parametric curve in R m. If we plot the function f as t varies we obtain a spiral going round the x-axis, as illustrated in the following diagram. Example 19 Let f : R 3 R 2, with f(x, y, z) = (u, v), where u = x + y z and v = 2x y 2z. Let U R n, and let f : U R m with f(x) = f 1 (x), f 2 (x),..., f m (x)). Then the derivative of f at x, denoted by f (x) or df, is the m n dx matrix whose (i, j)th entry is i. That is, j f (x) = m 1. m 2 1 n 2 n. m n. Example 20 Let f : R 2 R 2, with f(x, y) = (x 2 y, x + 3y). Here f 1 (x, y) = x 2 y and f 2 (x, y) = x + 3y, and f (x, y) = ( ) ( 2xy x 2 = 1 3 Example 21 Let f : R R 3, with f(t) = (t, cos t, sin t). Then d f (t) dt 1 d (t) = (cos t) dt = sin t. (sin t) cos t d dt ).

21 Advanced Calculus Chapter 3 Applications of partial differentiation 57 Example 22 Let f : R 3 R 2, with f(x, y) = (u, v), where u = x + y z and v = 2x y 2z. Then ) ( ) f (x, y, z) = = ( u v u v u z v z Simple cases of the chain rule In this section we consider three simple cases of the chain rule, and illustrate them with an example. Case 1 For U R 2 and V R, let g : U R and f : V R and let F : U R with Then F (x, y) = (f g)(x, y) = f(g(x, y)). F = df g dg, F = df g dg. Example 23 Let F (x, y) = (x 2 + 3xy y 2 ) 4. Then F (x, y) = f(g) where g = x 2 + 3xy y 2 and f = g 4. Thus F = df g dg, = 4g 3 (2x + 3y), F = 4(x 2 + 3xy y 2 ) 3 (2x + 3y). = df g dg, = 4g 3 (3x 2y), = 4(x 2 + 3xy y 2 ) 3 (3x 2y). Example 24 Let f : R R and let z = f(x 2 + y 2 ). Show that y z = x z. Let g(x, y) = x 2 + y 2. Then z(x, y) = f(g(x, y)). Thus z = df g dg, = 2x df dg. z = df dg g, = 2y df dg.

22 Advanced Calculus Chapter 3 Applications of partial differentiation 58 Hence y z = 2xy df dg = x z. Case 2 For U R and V R 2, let g : U R 2, with g(t) = (u(t), v(t)), and f : V R and let F : U R with F (t) = (f g)(t) = f(g(t)). Then df dt = df du u t + df dv v t. Example 25 Let F (t) = u 2 + v 2, where u = t cos t and v = sin t. Then F (t) = f(g(t)) where g(t) = (u(t), v(t)) and f(u, v) = u 2 +v 2. Thus df dt = df u du t + df v dv t, = 2u(cos t t sin t) + 2v cos t, = 2t cos t(cos t t sin t) + 2 sin t cos t, = 2 cos t(t cos t t 2 sin t + 2 sin t). Case 3 For U R 2 and V R 2, let g : U R 2, with g(x, y) = (u(x, y), v(x, y)), and f : V R, and let F : U R with F (x, y) = (f g)(x, y) = f(g(x, y)). Then F = du F = du u + dv u + dv v, v. Example 26 Let u = x 2 y and v = xy 3 and let F (x, y) = 2u + v 2. Then F (x, y) = f(g(x, y)) where f(u, v) = 2u + v 2 and g(x, y) = (u, v) = (x 2 y, xy 3 ). Thus F = du F u + dv = 2(2xy) + 2vy 3, = 4xy + 2xy 6, = 2xy(2 + y 5 ). v, = u du + v dv, = 2x 2 + 2v(3xy 2 ), = 2x 2 + 6x 2 y 5, = 2x 2 (1 + 3y 5 ).

23 Advanced Calculus Chapter 3 Applications of partial differentiation 59 Two general cases of the chain rule Here we consider two general cases of the chain rule. General case 1 Suppose u 1, u 2,..., u n are functions of a single variable t, and g is a function of n variables. Let G(t) = g(u 1, u 2,..., u n ). Then G is a function of t and dg dt = g u 1 du 1 dt + G u 2 du 2 dt + + g u n du n dt. Example 27 Let u = t 2, v = t and w = 1, and let t G(t) = g(u, v, w) = uv. Then w dg dt = g du u dt + g dv v dt + g dw w dt, ( v ) ( u ) = (2t) + ( 1 ( 1 w w 2 t 2 ) + uv ) ( 1 ), w 2 t 2 = 2t 2 t t t 2 t, = 7 2 t 5 2. Note that in Example 27, G(t) = uv w = t2 t 1 t = t 3 t = t 7 2, and so dg = 7t 5 2. Thus, in this example, it is easier to find G as a function dt 2 of t and then differentiate it with respect to t, rather than use the chain rule. General case 2 Suppose u 1, u 2,..., u n are functions of m variables x 1, x 2,..., x n, and h is a function of n variables. Let H(x 1, x 2,..., x m ) = h(u 1, u 2,..., u n ). Then H is a function of x 1, x 2,..., x m and, for each i, with 1 i m, H i = h u 1 u 1 i + h u 2 u 2 i + + h u n u n i Example 28 Let h = uv + uw + vw, where u = y 2, v = x 2 + 2xy and w = x 2 2xy. Then, using the chain rule, we get h = h u u + h v v + h w w, = (v + w)(0) + (u + w)(2x + 2y) + (u + v)(2x 2y), = 2(x 2 2xy + y 2 )(x + y) + 2(x 2 + 2xy + y 2 )(x y), = 2(x y) 2 (x + y) + 2(x + y) 2 (x y), = 2(x y)(x + y)(x y + x + y), = 4x(x y)(x + y).

24 Advanced Calculus Chapter 3 Applications of partial differentiation 60 Similarly, h = h u u + h v v + h w w, = (v + w)(2y) + (u + w)(2x) + (u + v)( 2x), = (2x 2 )(2y) + (x 2 2xy + y 2 )(2x) + (x 2 + 2xy + y 2 )( 2x), = 4x 2 y + 2x(x y) 2 2x(x + y) 2, = 4x 2 y + 2x ( (x y) 2 (x + y) 2), = 4x 2 y + 2x ((2x)( 2y)), = 4x 2 y 8x 2 y, = 4x 2 y. Here we have used the identity: A 2 B 2 = (A + B)(A B). Example 29 Let f be a function of two variables x and y, with x = r cos θ and y = r sin θ. Find and in terms of r, θ, and r θ. Using the chain rule gives Similarly, Thus r = r + r θ = θ + θ r θ = cos θ + sin θ. = r sin θ + r cos θ. = cos θ + sin θ, (35) = r sin θ + r cos θ. (36) Multiplying equation (35) by r cos θ and equation (36) by sin θ gives r cos θ r sin θ θ = r cos 2 θ + r cos θ sin θ, (37) = r sin 2 θ + r cos θ sin θ. (38) Subtracting equation (38) from equation (37) gives r cos θ r Now, since cos 2 θ + sin 2 θ = 1, we get sin θ θ = r(cos2 θ + sin 2 θ). = cos θ r sin θ r θ. in either equa- Using a similar method, or by substituting for tion (35) or equation (36), we get = sin θ r + cos θ r θ.

25 Advanced Calculus Chapter 3 Applications of partial differentiation 61 Example 30 (Euler s Theorem on homogeneous functions.) A function f : R 2 R is homogeneous of degree n if, for all x, y, λ R, f(λx, λy) = λ n f(x, y). For example, f(x, y) = xy is homogeneous of degree 2 since f(λx, λy) = (λx)(λy) = λ 2 xy = λ 2 f(x, y). Suppose that g : R 2 R is homogeneous of degree n. For x, y R, let G : R R where u = λx and v = λy. By the chain rule G(λ) = g(λx, λy) = g(u, v), dg dλ = g du u dλ + g dv v dλ, = x g u + y g v. However, since g is homogeneous of degree n, G(λ) = λ n g(x, y). Thus dg dλ = nλn 1 g(x, y). Comparing the two expressions for dg dλ gives nλ n 1 g(x, y) = x g u + y g v. (39) Equation (39) is true for all λ. If we put λ = 1 then u = x, v = y, and equation (39) gives ng(x, y) = x g + y g. This result is known as Euler s Theorem on homogeneous functions. Exercises Let g(x, y) = x 2 + 2xy y 2, x = 3t 1 and y = t 2. Use the chain rule to find dg dt. 2. Let f : R R, and let z = f(xy). (a) Show that x z y z = 0. (b) Verify the result in part (a) for the particular case when f(t) = 2t 3 t 2 + 3t.

26 Advanced Calculus Chapter 3 Applications of partial differentiation Let F be a function of u and v, where u = x 2 +y 2 and v = xy. Find F F F F and in terms of x, y, and u v. 4. Let g(x, y) = 3xy3 2x 2 y 2. 4x + 5y (a) Show that g is homogeneous of degree 3. (b) Verify Euler s Theorem on homogeneous functions by evaluating x g + y g directly. 3.4 Taylor series Let U R, and let f : U R such that the derivative, and all the higher derivatives, of f exist on U. For h U, f(x) = f(h)+(x h)f (h)+ (x h)2 f (x h)n (h)+ + f (n) (h)+. 2! n! This is the Taylor series of f centred at h. If the series is truncated after n + 1 terms we get the Taylor polynomial or Taylor approximation of f, of degree n, centred at h. If we replace x with x + h throughout in the Taylor series of f, we get the following alternative way of expressing the Taylor series of f centred at h: f(x + h) = f(h) + xf (h) + x2 2! f (h) + + xn n! f (n) (h) +. The following are the Taylor series of some standard functions. The first three are centred at 0, and are valid for all x R; the other two are centred at 1, and are valid for x R with x < 1. e x = 1 + x + x2 2! + x3 3! + + xn n! +..., sin x = x x3 3! + x5 5! x7 7! + + ( 1)n x 2n+1 (2n + 1)! +..., cos x = 1 x2 2! + x4 4! x6 x2n + ( 1)n 6! (2n)! +..., ln(x + 1) = x x2 2 + x3 3 x4 xn + + ( 1)n+1 4 n +, (1 + x) α = α(α 1)x2 α(α 1)(α 2)x3 1 + αx ! 3! α(α 1)... (α n + 1)xn + +. n!

27 Advanced Calculus Chapter 3 Applications of partial differentiation 63 Now let U R 2, and let f : U R such that the partial derivatives, and all the higher partial derivatives, of f exist and are continuous on U. For (X, Y ) U, f(x + h, Y + k) = f(x, Y ) + (hf x (X, Y ) + kf y (X, Y )) ( h 2 f xx (X, Y ) + 2hkf xy (X, Y ) + k 2 f yy (X, Y ) ) + 2! + 1 ( h 3 f xxx (X, Y ) + 3h 2 kf xxy (X, Y ) + 3hk 2 f xyy (X, Y )+ 3! k 3 f yyy (X, Y ) ) n ( ) n h n i k i f (n i,i) (X, Y ) +, n! i i=0 ( ) n n! where = i i!(n i)!, and f (r,s) denotes the (r + s)th partial derivative of f found by differentiating r times with respect to x and s times with respect to y. This is the Taylor series of f centred at (X, Y ). If the series is truncated after n + 1 terms we get the Taylor polynomial or Taylor approximation of f, of degree n, centred at h. The notation f (r,s) is my own creation and maybe non-standard. I have not found any standard notation in any books! Example 31 Expand x 2 y + 3y 2 in powers of x 1 and y + 2. Let f(x, y) = x 2 y + 3y 2. Putting x = 1 + h and y = 2 + k we can obtain the required expansion by finding the Taylor series of f centred at (1, 2). Now f x (x, y) = 2xy, f y (x, y) = x 2 + 3, f xx (x, y) = 2y, f xy (x, y) = 2x, f yy (x, y) = 0, f xxx (x, y) = 0, f xxy (x, y) = 2, f xyy (x, y) = 0, f yyy (x, y) = 0,

28 Advanced Calculus Chapter 3 Applications of partial differentiation 64 and all higher derivatives are 0. Thus f(x, y) = f(1 + h, 2 + k), = f(1, 2) + hf x (1, 2) + kf y (1, 2) + 1 ( h 2 f xx (1, 2) + 2hkf xy (1, 2) + k 2 f yy (1, 2) ) 2! + 1 ( 3h 2 kf xxy (1 2) ), 3! = 10 4h + 4k + 1 ( 4h 2 + 4hk + 0k 2) (6h2 k), = 10 4h + 4k 2h 2 + 2hk + h 2 k, = 10 4(x 1) + 4(y + 2) 2(x 1) 2 + 2(x 1)(y + 2) + (x 1) 2 (y + 2), since h = x 1 and k = y + 2. Example 32 Find the Taylor approximation of degree 1 for f(x, y) = 12 x 2 + xy + y 2 centred at (2,2). Use your answer to estimate f(2.1, 1.8). The partial derivatives of f are f x (x, y) = 12(2x + y) (x 2 + xy + y 2 ), 2 f y (x, y) = 12(x + 2y) (x 2 + xy + y 2 ). 2 Now, f(2, 2) = 1, f x (2, 2) = 1 2 and f y(2, 2) = 1 2. Thus f(2 + h, 2 + k) = f(2, 2) + hf x (2, 2) + kf y (2, 2), = h 1 2 k. Using this Taylor approximation for f with h = 0.1 and k = 0.2 we get f(2.1, 1.8) = f(2 + h, 2 + k), ( 0.2), = Note that f(2.1, 1.8) , so the approximation given by the Taylor is quite good in this case. Exercises Find the Taylor polynomial of degree 2 of g(x, y) = xy + 1 centred at the point (0, 0). Use your answer to estimate g(0.1, x ).