3.1 Iterated Partial Derivatives

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1 3.1 Iterated Partial Derivatives Prof. Tesler Math 20C Fall 2018 Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

2 Higher Derivatives Take the partial derivative of f (x, y) = x 2 y 3 with respect to x: f x (x, y) = 2xy 3 This is also a function of x and y, and we can take another derivative with respect to either variable: The x derivative of f x (x, y) is ( f x ) x = f xx = 2y 3. The y derivative of f x (x, y) is ( f x ) y = f xy = 6xy 2. f xx and f xy are each an iterated partial derivative of second order. The y derivative of the x derivative can also be written: y (x2 y 3 ) = y (2xy3 ) = 6xy 2 or 2 y (x2 y 3 ) = 6xy 2 Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

3 Iterated Derivative Notations Let f (x, y) = x 2 y 3. There are two notations for partial derivatives, f x and f. Partial derivative of f with respect to x in each notation: f x = 2xy 3 f f (x, y) = = 2xy3 Partial derivative of that with respect to y: ( ) ( f x ) y = f xy, y f = 2 y f so f xy (x, y) = 6xy 2 = 2 f y = 6xy2 Notice derivatives are listed in the opposite order in each notation. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

4 Iterated Derivative Notations Notice derivatives are listed in the opposite order in each notation. In each notation, compute the derivatives in order from the one listed closest to f, to the one farthest from f : f xyzzy = y z z y f = 5 y z 2 y f = 5 f y z 2 y Both notations say to take derivatives in the order x, y, z, z, y. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

5 Mixed Partial Derivatives f (x, y) = x 2 y 3 f x = 2xy 3 f y = 3x 2 y 2 f xx = 2y 3 f yx = 6xy 2 f xy = 6xy 2 f yy = 6x 2 y A mixed partial derivative has derivatives with respect to two or more variables. f xy and f yx are mixed. f xx and f yy are not mixed. In this example, notice that f xy = f yx = 6xy 2. The order of the derivatives did not affect the result. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

6 Continuity Notation A function f (x, y, z) is in class C 1 if f and its first derivatives f x, f y, f z are defined and continuous. Class C 2 : f and all of its first derivatives (f x, f y, f z ) and second derivatives (f xx, f xy, f xz, f yx, f yy, f yz, f zx, f zy, f zz ) are defined and continuous. Class C n : f and all of its first, second,..., n th derivatives are defined and continuous. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

7 Clairaut s Theorem If f and its first and second derivatives are defined and continuous (that is, f is class C 2 ), then f xy = f yx. Example The function f (x, y) = x 2 y 3 is C 2 (in fact, C ), and f xy = f yx = 6xy 2. Example with higher order derivatives As long as f and all its derivatives are defined and continuous up to the required order, you can change the order of the derivatives. E.g., if f (x, y, z) is class C 5, then 5 f f xyzzy = f xyyzz y z 2 y = 5 f y 2 z 2 Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

8 Clairaut s Theorem Example Is there a C 2 function f (x, y) with f x = cos(x + y) and f y = ln(x + y)? If so, Clairaut s Theorem says f xy = f yx. f xy = ( f x ) y = y cos(x + y) = sin(x + y) f yx = ( f y ) x = 1 ln(x + y) = x+y These don t agree, so there is no such function. Example The book, p. 158, 3.1#32, has an example where f xy (a, b) f yx (a, b) at a point (a, b) that has discontinuous 2 nd derivatives. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

9 Differential Equations Many laws of nature in Physics and Chemistry are expressed using differential equations. Ordinary Differential Equations (ODEs): You re given an equation in x, y and derivatives y, y,.... The goal is to find a function y = f (x) satisfying the equation. ODEs were introduced in Math 20B. Solution methods will be covered in Math 20D. For now, we will just show how to verify a solution. Example: Solve y = 2y The answer turns out to be y = Ce 2x, where C is any constant. Verify this is a solution: Left side: y = C(2e 2x ) = 2Ce 2x Right side: 2y = 2Ce 2x. They re equal, so y = 2y. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

10 Wave Equation u=f(x,t) bump 1 A A sin(bkt) x=l/n x=2l/n bump 2 bump 3... bump n x=3l/n time t 2 time t x x=nl/n=l A partial differential equation (PDE) has a function of multiple variables, and partial derivatives. The wave equation describes motion of waves. We ll study the one dimensional wave equation: u tt = b 2 u xx (where b is constant) Goal is to solve for a function u = f (x, t) satisfying this equation. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

11 Wave Equation u=f(x,t) bump 1 A A sin(bkt) x=l/n x=2l/n bump 2 bump 3... bump n x x=3l/n time t 2 x=nl/n=l time t 1 Parameters (constant) A = amplitude L = length n = # bumps b = oscillation speed Variables x = horizontal position t = time u = f (x, t) = y-coordinate Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

12 Wave Equation u tt = b 2 u xx or 2 u t 2 = b2 2 u 2 The solution is u = A sin(bkt) sin(kx) where k = nπ/l. Verify the equation u tt = b 2 u xx : Left side: compute u tt Right side: compute b 2 u xx u t = Abk cos(bkt) sin(kx) u tt = A(bk) 2 sin(bkt) sin(kx) u x = Ak sin(bkt) cos(kx) u xx = Ak 2 sin(bkt) sin(kx) b 2 u xx = Ak 2 b 2 sin(bkt) sin(kx) The left and right sides are equal, so it s a solution. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

13 Wave Equation u=f(x,t) bump 1 A A sin(bkt) x=l/n x=2l/n bump 2 bump 3... bump n x=3l/n time t 2 time t x x=nl/n=l A sin(bkt) is the amplitude at time t. This varies between ±A, so the maximum amplitude is A. k = nπ nπx L so sin(kx) = sin( L ). So sin(kx) = 0 at x = 0, L n, 2L n,..., nl n. Hence those points are on the x-axis at all times t. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

14 Implicit Equations and Partial Derivatives z = 1 x 2 y 2 gives z = f (x, y) explicitly. x 2 + y 2 + z 2 = 1 gives z in terms of x and y implicitly. For each x, y, one can solve for the value(s) of z where it holds. sin(xyz) = x + 2y + 3z cannot be solved explicitly for z. Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

15 Implicit Equations and Partial Derivatives To compute z and z y with an implicit equation: Assume z = f (x, y). For, treat x as a variable y as a constant z as a function of x, y x 2 + y 2 + z 2 = 1. Find z/. Left side: (x2 + y 2 + z 2 ) = 2x z z Right side: (1) = 0 Combine: 2x z z = 0 Solve: z = x/z Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

16 Implicit Equations and Partial Derivatives x 2 + y 2 + z 2 = 1. Find z/ at (x, y) = (1/3, 2/3). At (x, y) = (1/3, 2/3): ( 1 3 )2 + ( 2 3 )2 + z 2 = 1 At (x, y, z) = (1/3, 2/3, 2/3): z 2 = = 4 9, so z = ± 2 3 z = x/z = 1/3 2/3 = 1 2 At (x, y, z) = (1/3, 2/3, 2/3): z = x/z = 1/3 2/3 = 1 2 Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

17 Implicit Equations and Partial Derivatives Find z in the above equation: of left side: cos(xyz) of right side: 1 Combined: Solve this for z/: sin(xyz) = x + 2y + 3z (yz + xy z z cos(xyz) (yz + xy z ) ) = z 1 yz cos(xyz) = z (xy cos(xyz) 3) z = 1 yz cos(xyz) xy cos(xyz) 3 Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

18 Implicit Equations and Partial Derivatives sin(xyz) = x + 2y + 3z Find z/ at (x, y) = (0, 0). At (x, y) = (0, 0), the equation becomes sin(0) = 0 + 2(0) + 3z so z = 0. Plug numerical values of x, y, z into the formula for z/: z = 1 yz cos(xyz) xy cos(xyz) 3 = 1 0 cos(0) 0 cos(0) 3 = 1 3 = 1 3 Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19

19 Implicit Equations and Partial Derivatives sin(xyz) = x + 2y + 3z Find z/ at (x, y) = (1,.1). sin((1)(.1)z) = 1 + 2(.1) + 3z, so sin( 0.1z) =.8 + 3z. Use a numerical solver to get z Plug x = 1, y =.1, z into formula for z/: z = 1 yz cos(xyz) xy cos(xyz) Prof. Tesler 3.1 Iterated Partial Derivatives Math 20C / Fall / 19