there is no special reason why the value of y should be fixed at y = 0.3. Any y such that

Transcription

1 25. More on bivariate functions: partial erivatives integrals Although we sai that the graph of photosynthesis versus temperature in Lecture 16 is like a hill, in the real worl hills are three-imensional objects that can be climbe in several irections. So the sie of a real hill looks more like Figure 1(a), where the bivariate function P efine by P(x,y) y /2 is graphe on [0.1, 1] [0.1, 1]. As iscusse in Lecture 24, P(x, y) represents the power in kilowatts neee to keep an x-kilogram bir with a y-meter wing span in steay flight at the spee that minimizes fuel consumption over a given istance. In Figure 1, a possible path up the sie of the hill inclues KL, a possible path own inclues MN. The curve KL is the curve in which surface z P(x, y) meets vertical plane y 0., so KL has equation P(x, 0.) (0.) /2 0.21x 5/. Thus KL is the graph z f(x) of the orinary function f efine on [0.1, 1] by x 5/, which escribes how power requirement varies with mass in birs with a 0 cm wing span; see Figure 2(a). To etermine its graient f, we apply (22.1) with β 5/: f The graph of f also appears in Figure 2(a). Note that f is concave up f is concave own, in agreement with Exercise 2.2. Similarly, the curve MN is the curve in which surface z P(x, y) meets vertical plane x 0.8, so MN has equation P(0.8, y) y / y /2. Thus MN is the graph z g(y) of the orinary function g efine on [0.1, 1] by y /2, which escribes how power requirement varies with wing span in 800 gm birs; see Figure 2(b). To etermine its graient g, we apply (22.1) with β /2: g 250 (0.8)5/ y 5/ y 5/2. The graph of g also appears in Figure 2(b); again, in agreement with Exercise 2.2, g g are concave up concave own, respectively. Now, for climbing the sie of Figure 1's hill in a irection parallel to the x-axis, there is no special reason why the value of y shoul be fixe at y 0.. Any y such that 0.1 y 1 woul o, say y k. The corresponing curve KL woul be that in which surface z P(x, y) meets vertical plane y k, KL woul have equation 250 x5/ (25.1) z 250 x5/ (25.2) f(x) 250 (0.) /2 (25.) (x) (0.) /2 { x5/ } 250 (0.) /2 x x5/ (25.4) (0.) /2 x(5/ 1) 50 (0.) /2 x 2/ 0.65x 2/. z 250 (0.8)5/ (25.5) g(y) 250 (0.8)5/ (25.6) (y) 250 (0.8)5/ y y /2 { } (25.7) 2 y( /2 1) (0.8)5/

2 z P(x, k) M. Mesterton-Gibbons: Biocalculus, Lecture 25, Page x5/ k /2, (25.8) KL woul be the graph of the orinary function f efine on [0.1, 1] by f(x) 250 k /2 x 5/, (25.) instea of by (). Similarly, for escening in a irection parallel to the y-axis, there is no special reason why the value of x shoul be fixe at x 0.8. Any x such that 0.1 x 1 woul o, say x m. The corresponing curve MN woul be that in which surface z P(x, y) meets vertical plane x m, MN woul have equation z P(m, y) 250 m5/ y /2, (25.10) MN woul be the graph of the orinary function g efine on [0.1, 1] by g(y) 250 m5/ y /2, (25.11) instea of by (6). To obtain the graient along KL, all we nee o is to replace 0. by k in (4); to obtain the graient along MN, all we nee o is to replace 0.8 by m in (7). Thus we obtain in place of (4), f (x) 50 k /2 x 2/ g (y) m5/ y 5/2 (25.12) (25.1) in place of (7). Comparing (12)-(1) with (8)-(11), we observe that f (x) is z/x for KL g (y) is z/y for MN. Use of the symbol z/x implies that z epens only on x, whereas use of the symbol z/y implies that z epens only on y. Isn't this a contraiction? The answer is no, not really, because even though z epens on both x y in general, on KL the value of y is fixe, so that z epens only on x; whereas on MN the value of x is fixe, so z epens only on y. Nevertheless, there is potential for confusion, so we introuce some new notation. For a bivariate function P, we use z/x to enote the erivative of z P(x, y) with respect to x alone, i.e., with y hel fixe; we use z/y to enote the erivative of z P(x, y) with respect to y alone, i.e., with x hel fixe. That is, by analogy with (16.1b), we efine z x P(x + h,y) P(x,y) lim h 0 h (25.14) z y lim P(x,y + h) P(x,y). (25.15) h 0 We refer to z/x, i.e., the graient of z in a irection parallel to the x-axis, as the partial erivative of z with respect to x; to z/y, i.e., the graient of z in a irection parallel to the y-axis, as the partial erivative of z with respect to y. For example, in the case of Figure 1, where k enotes an arbitrary y-value m an arbitrary x-value, we can replace k by y in (12) to obtain z x we can replace m by x in (1) to obtain h 50 y /2 x 2/, (25.16)

3 M. Mesterton-Gibbons: Biocalculus, Lecture 25, Page z y x5/ y 5/2. (25.17) The partial erivatives of z P(x, y) yiel two new bivariate functions, say Q R. 1 That is, P(x + P(x,y) Q(x, y) lim h,y) h 0 h P(x,y + h) P(x,y) R(x, y) lim h 0 h For example, in the case of Figure 1, we have Q(x, y) 50 x2/ y /2 R(x,y) x5/ y 5/2 (25.18). (25.1) (25.20) (25.21) from (16)-(17). An important special case occurs when the function P is separable, i.e., when orinary functions F G exist such that P(x, y) F(x)G(y) (25.22) for any relevant x or y. Then (18) implies (1) implies or, in mixe notation, F(x + F(x)G(y) Q(x, y) lim h)g(y) h 0 h F(x + h) F(x) lim h 0 h F(x)G(y + F(x)G(y) R(x, y) lim G(yx + h) G(y) h 0 h) h 0 h F(x)lim h { x F(x)G(y) } F (x)g(y), 250 y /2 (25.2) G(y) F (x)g(y) F(x) G (y) (25.24) { y F(x)G(y) } F(x) G (y). (25.25) For example, P efine by (1) is separable; so, from z x 5/ y /2 /250, we calculate z x x x5/ 250 y /2 ( ) x x5/ 250 y /2 5 (25.26) x2/ 50 x2/ y /2 in agreement with (16), 1 If P has omain [, ] [a2, b2], then the omains of Q R are, respectively, [, ) [a2, b2] [, ] [a2, b2), by (1.25); i.e., x is, strictly speaking, exclue from the omain of Q, whereas y b2 is exclue from the omain of R. In practice, however, both omains can be extene to [, ] [a2, b2] by analogy with (1.26), i.e., by agreeing that Q(, y) is the limit of Q(x, y) as x R(x, b2) is the limit of R(x, y) as y b2.

4 z y y 250 x5/ 250 x5/ y /2 2 y 5/2 M. Mesterton-Gibbons: Biocalculus, Lecture 25, Page 4 ( ) 250 x5/ y y / x5/ y 5/2. (25.27) in agreement with (17). Furthermore, because Q is itself a bivariate function, it has its own partial erivatives, which are secon partial erivatives of z. We write 2 z x 2 x z x Q(x + h,y) Q(x,y) lim h 0 h 2 z z yx y x For example, in the case of Figure 1, (25)-(26) imply 2 z x 2 x 2 x 1/ whereas (25) (27) imply 2 z yx y 50 x2/ y /2 50 y /2 Q(x,y + h) Q(x,y) lim h 0 50 x2/ y /2 h { } x x2/ 50 y / x 1/ y /2, { } 50 x2/ y y /2 (25.28). (25.2) 50 x2/ 2 y 5/2 100 x2/ y 5/2. Likewise, because R is a bivariate function, we have secon partial erivatives 2 z xy x z y R(x + h,y) R(x,y) lim h 0 h 2 z 2 y z R(x,y + h) R(x,y) y y lim h 0 h In particular, in the case of Figure 1, (25)-(26) imply 2 z xy x x5/ y 5/2 x x5/ 5 x2/ y 5/2 whereas (25) (27) imply 2 z y 2 y x5/ y 5/ x5/ 5 2 y 7/2 by Exercise 1. Note that (1) (4) imply 2 z yx { } x2/ y 5/2, (25.0) (25.1) (25.2). (25.) 500 y 5/2 27 { } 500 x5/ y y 5/ x5/ y 7/2 (25.4) (25.5) 2 z xy. (25.6)

5 M. Mesterton-Gibbons: Biocalculus, Lecture 25, Page 5 This result hols not only for z in Figure 1, but also in general. 2 Thus, although in principle a bivariate function has four secon erivatives, efine by (28), (2), (2) (), in practice there are only three, because (6) makes (2) (2) equal. Now, from Lecture 24, if a bivariate function P has a global minimum on the interior of its omain at ( ˆx,ŷ), then ˆx must minimize the orinary function f efine by f(x) P(x, ŷ), whose graph is a vertical section through the graph of P along y ŷ; ŷmust minimize the orinary function g efine by g(y) P( ˆx, y), whose graph is a vertical section through the graph of P along x ˆx. Thus the partial erivative of P with respect to x along y ŷ must change sign from negative to positive at x ˆx, the partial erivate of P with respect to y along x ˆx must change sign from negative to positive at y ŷ. In other wors, Q( ˆx, ŷ) 0 R( ˆx, ŷ), (25.7) where Q R are efine by (18)-(1); or, in ifferential notation, z x x ˆx yŷ 0 z y x ˆx yŷ. (25.8) This pair of equations is sometimes useful for fining global minima. For example, in Appenix 2A we require the global minimum of S efine by S(α,β) α α β + 667αβ + 7β 2. (25.) Here, with z S(α, β), we have z α (4.26) 155 (α) α α α (α2 ) α z β β (β) α (αβ) + 7 α (β2 ) α β α + 667β (25.40) (4.26) 155 β (α) β (α2 ) β (β) β (αβ) + 7 β (β2 ) α + 7 2β α + 14β, (25.41) so that (8) yiels 601α + 667β α + 14β This pair of equations is reaily solve to yiel ˆα / ˆβ 26751/ , yieling a global minimum of S( ˆα, ˆβ) It is well to remember, however, that (8) isn't always so useful; for example, it wouln't have helpe much in Lecture 24. We conclue by broaching the concept of a partial integral, which is important in probability theory. Partial integration is the opposite of partial ifferentiation, in the sense that a partial erivative is the erivative of a bivariate function with respect to one variable, whereas a partial integral is the integral of a bivariate function with respect to one variable. Now, you know from Lecture 12 that the result of integrating P(x) between x x is inepenent of x, so the result of integrating P(x, y) between x x must also be inepenent of x. The ifference is simply that if P is orinary then Int(P, [, ]) epens only on, whereas if P is 2 More precisely, whenever z represents the height of a smooth surface, which is usually the case in practice. The result is sometimes calle Clairaut's Theorem. See Exercise 2 for an important special case.

6 M. Mesterton-Gibbons: Biocalculus, Lecture 25, Page 6 bivariate then Int(P, [, ]) can also epen on y. So enote it by W(y). Then, by analogy with (12.20), we efine the partial integral of P with respect to x by W(y) P(x, y)x lim P(x,y)δx δx 0. (25.42) [,b 1 ] Corresponingly, the partial integral of P with respect to y is efine by b 2 U(x) P(x, y)y lim P(x,y)δy δy 0. (25.4) a 2 [a 2,b2 ] Both U W are orinary functions, in Appenix 28C we will require an expression for their erivatives. It is simplest to obtain if we assume that P is separable. Then, by (22), orinary functions F G exist such that P(x, y) F(x)G(y). If we are integrating with respect to x, then G(y) behaves like a constant, so from (42) (12.25) with u F, k G(y) we have W (y) y F(x)G(y)x y G(y) F(x)x. (25.44) But if we are ifferentiating with respect to y, then Int(F, [, ]) behaves like a constant, so from (44) (16.17) with k Int(F, [, ]) we have W (y) G (y) F(x)x G (y)f(x)x (25.45) after applying (12.25) again, but this time with k G (y), which behaves like a constant if we are integrating with respect to x. Now, applying (25) to (45), we obtain W (y) F(x) G (y)x In other wors, on using (42), { y F(x)G(y) }x { y P(x,y) }x. (25.46) P(x, y)x y { y P(x,y) }x. (25.47a) A parallel argument establishes that b 2 P(x, y)y x x P(x,y) a (see Exercise 2). Inee (47a) (47b) are in essence the same result; although we have establishe it only for the case where P is separable, it can be shown to hol more { }y (25.47b) 2 generally. b 2 a 2 Exercises Establish (6) for the special case of a separable function, i.e., for z F(x)G(y) Establish (47b) for the special case where P is separable, i.e., P(x, y) F(x)G(y). Hint: Differentiate U in (4) with respect to x use (12.25) (16.17).