**Unformatted text preview: **Math 20C Multivariable Calculus Lecture 12 1 ' $ Partial derivatives Review: Limits and continuity. (Sec. 14.2) Definition of Partial derivatives. (Sec. 14.3) Higher derivatives. Examples of differential equations. Slide 1 & % ' $ We recall the definition of limit of f (x, y) Let f (x, y) be a scalar function defined for P = (x, y) near P0 = (x0 , y0 ). Let dP0 P = (x - x0 )2 + (y - y0 )2 be the distance between (x, y) and (x0 , y0 ). We write
(x,y)(x0 ,y0 ) Slide 2 lim f (x, y) = L, to mean that the values of f (x, y) approaches L as the distance dP0 P approaches zero.
& % Math 20C Multivariable Calculus Lecture 12 2 ' $ In this case lim(x,y)(x0 ,y0 ) f (x, y) exists
z Slide 3 L f(x,y) y (x0,y0 ) x & % ' In this case lim(x,y)(0,0) f (x, y) does not exist
z f(x,y) $ Slide 4
y C2 x C1 & Compute side limits along C1 and C2 % Math 20C Multivariable Calculus Lecture 12 3 ' $ Continuous functions have graphs without holes or jumps Definition 1 f (x, y) is continuous at (x0 , y0 ) if Slide 5
(x,y)(x0 ,y0 ) lim f (x, y) = f (x0 , y0 ). Polynomial functions are continuous in I 2 , for example R P2 (x, y) = a0 + b1 x + b2 y + c1 x2 + c2 xy + c3 y 2 .
& % ' $ More examples of continuous functions Rational functions are continuous on their domain, f (x, y) = Slide 6 for example, f (x, y) = x2 + 3y - x2 y 2 + y 4 , x2 - y 2 x = y. Pn (x, y) , Qm (x, y) Composition of continuous functions are continuous, example f (x, y) = cos(x2 + y 2 ).
& % Math 20C Multivariable Calculus Lecture 12 4 ' $ To compute partial derivative with respect to x keep y constant Definition 2 (x-partial derivative) Let f : D I 2 R I The partial derivative of f (x, y) R R. with respect to x at (a, b) D is denoted as fx (a, b) and is given by fx (a, b) = lim 1 [f (a + h, b) - f (a, b)] . h0 h Slide 7 & % ' $ To compute partial derivative with respect to y keep x constant Definition 3 (y-partial derivative) Let f : D I 2 R I The partial derivative of f (x, y) R R. with respect to y at (a, b) D is denoted as fy (a, b) and is given by fy (a, b) = lim 1 [f (a, b + h) - f (a, b)] . h0 h Slide 8 & % Slide 9 So, to compute the partial derivative of f (x, y) with respect to x at (a, b), one can do the following: First, evaluate the function at y = b, that is compute f (x, b); second, compute the usual derivative of single variable functions; evaluate the result at x = a, and the result is fx (a, b). Example: To compute the partial derivative of f (x, y) with respect to y at (a, b), one follows the same idea: First, evaluate the function at x = a, that is compute f (a, y); second, compute the usual derivative of single variable functions; evaluate the result at y = b,, and the result is fy (a, b). Example: Math 20C Multivariable Calculus Find the partial derivative of f (x, y) = x2 + y 2 /4 with respect to y at (1, 3). Find the partial derivative of f (x, y) = x2 + y 2 /4 with respect to x at (1, 3). 1. f (1, y) = 1 + y 2 /4; 1. f (x, 3) = x2 + 9/4; 2. fx (x, 3) = 2x; 3. fx (1, 3) = 2. & ' 2. fy (1, y) = y/2; 3. fy (1, 3) = 3/2. Partial derivatives are slopes of lines tangent to the graph of f (x, y) x z f(x,y) (a,b) Lecture 12 fx fy y % $ 5 Math 20C Multivariable Calculus Lecture 12 6 ' $ Partial derivatives define new functions Definition 4 Consider a function f : D I 2 R I The functions partial derivatives R R. of f (x, y) are denoted by fx (x, y) and fy (x, y), and are given by the expressions fx (x, y) = lim 1 [f (x + h, y) - f (x, y)] , h0 h 1 fy (x, y) = lim [f (x, y + h) - f (x, y)] . h0 h
% Slide 10 & ' $ The partial derivative functions of a paraboloid are planes f (x, y) = ax2 + by 2 + xy. Slide 11 fx (x, y) = 2ax + 0 + y, = 2ax + y. fy (x, y) = 0 + 2by + x, = 2by + x.
& % Math 20C Multivariable Calculus Lecture 12 7 ' $ The partial derivative functions of a paraboloid are planes
z f(x,y) z fy(x,y) f x(x,y) z Slide 12 y x x y x y & More examples: f (x, y) = x2 ln(y), fx (x, y) = 2x ln(y), x2 fy (x, y) = . y f (x, y) = x2 + fx (x, y) fy (x, y) f (x, y) = fx (x, y) = = = 2x - y , x + 2y 2(x + 2y) - (2x - y) , (x + 2y)2 2x + 4y - 2x + y , (x + 2y)2 5y . (x + 2y)2 = 2x, y = . 2 y2 , 4 % Math 20C Multivariable Calculus Lecture 12 -(x + 2y) - (2x - y)2 , (x + 2y)2 -5x . (x + 2y)2 8 fy (x, y) = = f (x, y) = x3 e2y + 3y, fx (x, y) = 3x2 e2y , fy (x, y) fyy (x, y) fyyy (x, y) fxy fyx = 2x3 e2y + 3, = 4x3 e2y , = 8x3 e2y , = 6x2 e2y , = 6x2 e2y . ' Higher derivatives of a function f (x, y) are partial derivatives of its partial derivatives For example, the second partial derivatives of f (x, y) are the following: $ Slide 13 fxx (x, y) = lim 1 [fx (x + h, y) - fx (x, y)] , h0 h 1 fyy (x, y) = lim [fy (x, y + h) - fy (x, y)] , h0 h 1 [fx (x + h, y) - fx (x, y)] , h0 h 1 fyx (x, y) = lim [fy (x, y + h) - fy (x, y)] . h0 h fxy (x, y) = lim & % Math 20C Multivariable Calculus Lecture 12 9 ' $ Higher partial derivatives sometimes commute Slide 14 Theorem 1 Consider a function f (x, y) in a domain D. Assume that fxy and fyx exists and are continuous in D. Then, fxy = fyx . & % ' $ Differential equations are equations where the unknown is a function For example, the Laplace equation: Find (x, y, z) : D I 3 I solution of R R xx + yy + zz = 0.
This equation describes the gravitational effects near a massive object. Slide 15 and where derivatives of the function enter into the equation
& % Math 20C Multivariable Calculus Lecture 12 10 ' $ More examples of differential equations Heat equation: Find a function T (t, x, y, z) : D I 4 I solution of R R Tt = Txx + Tyy + Tzz .
The heat on a metal is described by this equation. T is the temperature on that object. Slide 16 & % ' $ More examples of differential equations Wave equation: Find a function f (t, x, y, z) : D I 4 I solution of R R ftt = fxx + fyy + fzz .
The sound in the air is described by this equation. f is the air density. Slide 17 & % Math 20C Multivariable Calculus Lecture 13 11 Exercises: Verify that the function T (t, x) = e-t sin(x) satisfies the one-space dimensional heat equation Tt = Txx . Verify that the function f (t, x) = (t - x)3 satisfies the one-space dimensional wave equation Ttt = Txx . Verify that the function below satisfies Laplace Equation, (x, y, z) = 1 x2 + y2 + z 2 . ' $ Differentiable functions (Sec. 14.4) Slide 18 Definition of differentiable functions. Equation of the tangent plane. Linear approximation. (Differentials.) & % Math 20C Multivariable Calculus Lecture 13 12 ' A function can have partial derivatives at a point and be discontinuous at that point
fx(0,0) = fy(0,0) = 0 z f(x,y) $ Slide 19
y C2 x C1 & This is a very bad property for a definition of derivative
% ' $ Here is one of such functions, given explicitly Slide 20 f (x, y) = 2xy/(x2 + y 2 ) 0 (x, y) = (0, 0), (x, y) = (0, 0). fx (0, 0) = fy (0, 0) = 0, although f (x, y) is not continuous at (0, 0). & % Math 20C Multivariable Calculus Lecture 13 13 ' $ Recall the following property of the derivative of f (x) Theorem 2 If f (x) exists, then f (x) is continuous. Slide 21
lim [f (x + h) - f (x)] = = lim {[f (x + h) - f (x)]/h}h, lim f (x)h = 0. h0 h0 h0 The analogous claim "If fx (x, y) and fy (x, y) exists, then f (x, y) is continuous" is false
& % ' One has to define a notion of derivative having the continuity property discussed above
f(x,y) = z L(x,y) Linear approximation at (x0,y0) $ Slide 22
y x (x 0,y0) & New definition: A differentiable function must be approximated by a plane % Math 20C Multivariable Calculus Lecture 13 14 ' In the case f (x) this definition says: The function must be approximated by a line
f(x) = y L(x) $ Slide 23 x0 x Only for functions f (x) the derivative f (x) implies the existence of an approximating line L(x)
& % ' $ A function of two variables is differentiable at (x0 , y0 ) if two conditions hold: Slide 24 There exists the plane from its partial derivatives at (x0 , y0 ); This plane approximates the graph of f (x, y) near (x0 , y0 ). & % Math 20C Multivariable Calculus Lecture 13 15 ' $ Here is a rewording of the definition Definition 5 The function f (x, y) is differentiable at (x0 , y0 ) if Slide 25 f (x, y) = L(x0 ,y0 ) (x, y) + where and
i (x, y) 1 (x - x0 ) + 2 (y - y0 ), 0 when (x, y) (x0 , y0 ), for i = 1, 2, L(x0 ,y0 ) (x, y) = fx (x0 , y0 )(x-x0 )+fy (x0 , y0 )(y-y0 )+f (x0 , y0 )
& % ' This notion of differentiability has the continuity property $ Theorem 3 If f (x, y) is differentiable, then f (x, y) is continuous. Slide 26
f(x,y) = z L(x,y) Linear approximation at (x0,y0) y x (x 0,y0) If f (x, y) is differentiable, then L(x0 ,y0 ) (x, y) is called the linear & approximation of f (x, y) at (x0 , y0 ). % Math 20C Multivariable Calculus Lecture 13 16 ' $ The following result is useful to check the differentiability of a function Slide 27 Theorem 4 Consider a function f (x, y). Assume that its partial derivatives fx (x, y), fy (x, y) exist at (x0 , y0 ) and near (x0 , y0 ), and both are continuous functions at (x0 , y0 ). Then, f (x, y) is differentiable at (x0 , y0 ). & % ' Consider the following exercise: 1. Show that f (x, y) = arctan(x + 2y) is differentiable at (1, 0). 2. Find its linear approximation at (1, 0). $ Slide 28 fx (x, y) = 1 , 1 + (x + 2y)2 fy (x, y) = 2 . 1 + (x + 2y)2 These functions are continuous in I 2 , so f (x, y) is differentiable at every R point in I 2 . R L(1,0) (x, y) = fx (1, 0)(x - 1) + fy (1, 0)(y - 0) + f (1, 0), where f (1, 0) = arctan(1) = /4, fx (1, 0) = 1/2, fy (1, 0) = 1. Then, L(1,0) (x, y) = 1 (x - 1) + y + . 2 4 & % Math 20C Multivariable Calculus Lecture 13 17 ' $ Second exercise, on linear approximation Find the linear approximation of f (x, y) = 17 - x2 - 4y 2 at (2, 1). Slide 29
We need three numbers: f (2, 1), fx (2, 1), and fy (2, 1). Then, we compute the linear approximation by the formula L(2,1) (x, y) = fx (2, 1)(x - 2) + fy (2, 1)(y - 1) + f (2, 1). The result is: f (2, 1) = 3, fx (2, 1) = -2/3, and fy (2, 1) = -4/3. Then the plane is given by 4 2 L(2,1) (x, y) = - (x - 2) - (y - 1) + 3. 3 3 & % ' df is a special name for L(x0 ) (x) - f (x0 ) Single variable case: df (x) = Lx0 (x) - f (x0 ) = f (x0 )(x - x0 ) = f (x0 )dx. $ We called (x - x0 ) = dx. Slide 30
y f(x) L(x) f f(x0) x0 x dx = x df x & % Math 20C Multivariable Calculus Lecture 13 18 ' $ df (x, y) is a special name for L(x0 ,y0 ) (x, y) - f (x0 , y0 ) Functions of two variables: df (x, y) = L(x0 ,y0 ) (x, y) - f (x0 , y0 ), Slide 31 dx = x - x0 , dy = y - y0 . Then, the formula is easy to remember: df (x, y) = fx (x0 , y0 ) dx + fy (x0 , y0 ) dy.
Different names for the same idea: Compute the linear approximation of a differentiable function. & % ' An exercise on differentials Compute the df of f (x, y) = ln(1 + x2 + y 2 ) at (1, 1) for dx = 0.1, dy = 0.2. $ df (x0 , y0 ) = = Slide 32
Then, fx (x0 , y0 )dx + fy (x0 , y0 )dy, 2y0 2x0 dx + dy. 2 2 1 + x 2 + y0 1 + x 2 + y0 0 0 2 1 2 2 + , 3 10 3 10 2 3 , 3 10 1 . 5 df (1, 1) = = = & % Math 20C Multivariable Calculus Lecture 13 19 ' Another exercise on differentials Use differentials to estimate the amount of tin in a closed tin can with internal diameter of 8cm and height of 12cm if the tin is 0.04cm thick. $ Data of the problem: h0 = 12cm, r0 = 4cm, dr = 0.04cm and dh = 0.08cm. Draw a picture of the cylinder. Slide 33 The function to consider is the volume of the cylinder, V (r, h) = r 2 h. Then, dV (r0 , h0 ) = = = Vr (r0 , h0 )dr + Vh (r0 , h0 )dh,
2 2r0 h0 dr + r0 dh 16.1cm. & % ' $ Chain rule and directional derivatives Slide 34 Review: Linear approximations. (Sec. 14.4) Chain rule. (Sec. 14.5) Cases 1 and 2. Examples. & % Math 20C Multivariable Calculus Lecture 13 20 ' $ Recall the chain rule for f (x) Given f (x), and x(t) differentiable functions, introduce z(t) = f (x(t)). Then, z(t) is differentiable, and Slide 35 df dx dz = (x(t)) (t). dt dx dt Or, using the new notation, zt (t) = fx (x(t)) xt (t). & % ' $ There are many chain rules for f (x, y) Case 1: Given f (x, y) differentiable, and x(t), y(t) differentiable functions of a single variable, then z(t) = f (x(t), y(t)) is differentiable and dz dx dy = fx (x(t), y(t)) (t) + fy (x(t), y(t)) (t). dt dt dt Slide 36 & % Math 20C Multivariable Calculus Lecture 13 21 ' $ Example of Chain rule, case 1 f (x, y) = x2 + 2y 3 , Slide 37 x(t) = sin(t), y(t) = cos(2t). Let z(t) = f (x(t), y(t)). Then, dz dx dy = 2x(t) + 6[y(t)]2 , dt dt dt = 2x(t) cos(t) - 12[y(t)]2 sin(2t), = 2 sin(t) cos(t) - 12 cos2 (2t) sin(2t). & % ' Second case of chain rule for f (x, y) Case 2: Let f (x, y) be differentiable, and x(t, s), y(t, s) be also differentiable functions of a two variable. $ Then z(t, s) = f (x(t, s), y(t, s)) is differentiable and Slide 38 zt (t, s) = fx (x(t, s), y(t, s)) xt (t, s) + fy (x(t, s), y(t, s)) yt(t, s) zs (t, s) =
& fx (x(t, s), y(t, s)) xs(t, s) + fy (x(t, s), y(t, s)) ys(t, s) % Math 20C Multivariable Calculus Lecture 13 22 ' $ Second case of chain rule for f (x, y) again Case 2: Let f (x, y) be differentiable, and x(t, s), y(t, s) be also differentiable functions of a two variable. Slide 39 Then z(t, s) = f (x(t, s), y(t, s)) is differentiable and z t = f x xt + f y y t z s = f x xs + f y y s
& % ' $ Example of chain rule, case 2 A change of coordinates: Slide 40 Consider the function f (x, y) = x2 + ay 2 , with a I R. Introduce polar coordinates r, by the formula x(r, ) = r cos(), y(r, ) = r sin(). Let z(r, ) = f (x(r, ), y(r, )) be f in polar coordinates. & % Math 20C Multivariable Calculus Lecture 13 23 ' $ A change of coordinates Then, the chain rule, case 2, says that z r = f x xr + f y y r . Slide 41 Each term can be computed as follows, fx = 2x, xr = cos(), then one has zr = 2r cos2 () + 2ar sin2 ().
& % fy 2ay, yr = sin(), ...

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