Differentiable functions (Sec. 14.4)

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1 Math 20C Multivariable Calculus Lecture 3 Differentiable functions (Sec. 4.4) Review: Partial derivatives. Slide Partial derivatives and continuity. Equation of the tangent plane. Differentiable functions. Application: Differentials. (Linear approximation.) Review: Partial derivatives Slide 2 Definition Consider a function f : D IR 2 R IR. The functions partial derivatives of f(x, y) are denoted by f x (x, y) and f y (x, y), and are given by the expressions f x (x, y) = lim [f(x + h, y) f(x, y)], h 0 h f y (x, y) = lim [f(x, y + h) f(x, y)]. h 0 h

2 Math 20C Multivariable Calculus Lecture 3 2 Review: Higher derivatives Slide 3 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: f xx (x, y) = lim h 0 h [f x(x + h, y) f x (x, y)], f yy (x, y) = lim h 0 h [f y(x, y + h) f y (x, y)], f xy (x, y) = lim h 0 h [f x(x + h, y) f x (x, y)], f yx (x, y) = lim h 0 h [f y(x, y + h) f y (x, y)]. Higher derivatives Slide 4 Theorem (Partial derivatives commute) Consider a function f(x, y) in a domain D. Assume that f xy and f yx exists and are continuous in D. Then, f xy = f yx.

3 Math 20C Multivariable Calculus Lecture 3 3 Examples of differential equations Differential equations are equations where the unknown is a function, and where derivatives of the function enter into the equation. Examples: Laplace equation: Find φ(x, y, z) : D IR 3 IR solution of Slide 5 φ xx + φ yy + φ zz = 0. Heat equation: Find a function T (t, x, y, z) : D IR 4 IR solution of T t = T xx + T yy + T zz. Wave equation: Find a function f(t, x, y, z) : D IR 4 IR solution of f tt = f xx + f yy + f zz. Exercises: Verify that the function T (t, x) = e t sin(x) satisfies the one-space dimensional heat equation T t = T xx. Verify that the function f(t, x) = (t x) 3 satisfies the one-space dimensional wave equation T tt = T xx. Verify that the function below satisfies Laplace Equation, φ(x, y, z) = x2 + y 2 + z 2.

4 Math 20C Multivariable Calculus Lecture 3 4 Partial derivatives and continuity Partial derivatives generalize the idea of derivative from single variable functions, f(x) to functions f(x, y), as follows, Are the partial derivatives a faithful generalization? Slide 6 NO. Claim: If f (x) exists, then f(x) is continuous. True. (Proof: lim h 0 [f(x + h) f(x)] = lim h 0 {[f(x + h) f(x)]/h}h = lim h 0 f (x)h = 0.) Claim: If f x (x, y) and f y (x, y) exists, then f(x, y) i continuous. False. There is a counterexample: 2xy/(x 2 + y 2 ) (x, y) (0, 0), f(x, y) = 0 (x, y) = (0, 0). Slide 7 What is a faithful generalization of the concept of derivative to functions f(x, y)? The concept of linear approximation. If f (x 0 ) exists, then L(x) = f (x 0 )(x x 0 ) + f(x 0 ) approximates f(x) for x near x 0. What is the analog of L(x) in functions of two variables? The analog to the line L(x) is a plane L(x, y).

5 Math 20C Multivariable Calculus Lecture 3 5 Summary Slide 8 Consider a function f(x, y) such that f(x 0, y 0 ), f x (x 0, y 0 ), and f y (x 0, y 0 ) exist. Then, the plane L (x0,y 0)(x, y) = f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) + f(x 0, y 0 ) is well defined. If this plane approximates f(x, y) for (x, y) near (x 0, y 0 ), then we will say that f(x, y) is differentiable at (x 0, y 0 ). Differentiable functions of two variables Idea: A function f(x, y) is differentiable at (x 0, y 0 ) if there exists the plane from its partial derivatives at (x 0, y 0 ), Slide 9 AND this plane approximates the graph of f(x, y) near (x 0, y 0 ). Definition 2 (Differentiable functions) The function f(x, y) is differentiable at (x 0, y 0 ) if f(x, y) = L (x0,y 0)(x, y) + ɛ (x x 0 ) + ɛ 2 (y y 0 ), and ɛ i (x, y) 0 when (x, y) (x 0, y 0 ), for i =, 2.

6 Math 20C Multivariable Calculus Lecture 3 6 The following result is useful to check the differentiability of a function. Slide 0 Theorem 2 Consider a function f(x, y). Assume that its partial derivatives f x (x, y), f y (x, y) exist at (x 0, y 0 ) and near (x 0, y 0 ), and both are continuous functions at (x 0, y 0 ). Then, f(x, y) is differentiable at (x 0, y 0 ). Definition 3 (Linear approximation) If f(x, y) is differentiable, then L (x0,y 0)(x, y) is called the linear approximation of f(x, y) at (x 0, y 0 ). Differentials and chain rule Slide Review: Differentiable functions. (Sec. 4.4) Linear approximation and differentials. Chain rule. (Sec. 4.5)

7 Math 20C Multivariable Calculus Lecture 3 7 Review: Differentiable functions Let f(x, y) be a function defined in a neighborhood of (x 0, y 0 ) such that the partial derivatives f x (x 0, y 0 ), f y (x 0, y 0 ) exist. Slide 2 Consider the plane L (x0,y 0)(x, y) constructed with f(x 0, y 0 ) and with the partial derivatives f x (x 0, y 0 ), f y (x 0, y 0 ) given by L (x0,y 0)(x, y) = f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) + f(x 0, y 0 ). If this plane approximates the function f(x, y) near (x 0, y 0 ), then we call f(x, y) differentiable at (x 0, y 0 ). (Then, for differentiable functions, the plane is called the linear approximation of f(x, y) at (x 0, y 0 ).) Exercise: Differentiable functions Show that f(x, y) = arctan(x + 2y) is differentiable at (, 0). Find its linear approximation at (, 0). Slide 3 f x (x, y) = + (x + 2y) 2, f 2 y(x, y) = + (x + 2y) 2. These functions are continuous in IR 2, so f(x, y) is differentiable at every point in IR 2. L (,0) (x, y) = f x (, 0)(x ) + f y (, 0)(y 0) + f(, 0), where f(, 0) = arctan() = π/4, f x (, 0) = /2, f y (, 0) =. Then, L (,0) (x, y) = 2 (x ) + y + π 4.

8 Math 20C Multivariable Calculus Lecture 3 8 Exercise: Linear approximation Find the linear approximation of f(x, y) = 7 x 2 4y 2 at (2, ). Slide 4 We need three numbers: f(2, ), f x (2, ), and f y (2, ). Then, we compute the linear approximation by the formula L (2,) (x, y) = f x (2, )(x 2) + f y (2, )(y ) + f(2, ). The result is: f(2, ) = 3, f x (2, ) = 2/3, and f y (2, ) = 4/3. Then the plane is given by L (2,) (x, y) = 2 3 (x 2) 4 (y ) Differentials Different names for the same idea: Compute the linear approximation of a differentiable function. The differential is a special name for L (x0,y 0)(x, y) f(x 0, y 0 ). Single variable case: Slide 5 df(x) = L x0 (x) f(x 0 ) = f (x 0 )(x x 0 ) = f (x 0 )dx. We called (x x 0 ) = dx. Functions of two variables: df(x, y) = L (x0,y 0)(x, y) f(x 0, y 0 ), dx = x x 0, dy = y y 0. Then, the formula is easy to remember: df(x, y) = f x (x 0, y 0 )dx + f y (x 0, y 0 )dy.

9 Math 20C Multivariable Calculus Lecture 3 9 Exercise: Differentials Compute the df of f(x, y) = ln( + x 2 + y 2 ) at (, ) for dx = 0., dy = 0.2. df(x 0, y 0 ) = f x (x 0, y 0 )dx + f y (x 0, y 0 )dy, Slide 6 Then, = 2x 0 2y 0 + x dx + y2 0 + x dy. y2 0 df(, ) = = , = , Exercise: Differentials Use differentials to estimate the amount of tin in a closed tin can with internal diameter f 8cm and height of 2cm if the tin is 0.04cm thick. Slide 7 Data of the problem: h 0 = 2cm, r 0 = 4cm, dr = 0.04cm and dh = 0.08cm. Draw a picture of the cylinder. The function to consider is the volume of the cylinder, V (r, h) = πr 2 h. Then, dv (r 0, h 0 ) = V r (r 0, h 0 )dr + V h (r 0, h 0 )dh, = 2πr 0 h 0 dr + πr 2 0 dh = 6.cm.

10 Math 20C Multivariable Calculus Lecture 3 0 Chain rule Slide 8 Single variable case. Given f(x), and x(t) differentiable functions, introduce z(t) = f(x(t)). Then, z(t) is differentiable, and dz dt = df dx (x(t)) dx dt (t). Or, using the new notation, z t (t) = f x (x(t)) x t (t). Chain rule Case : 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 Slide 9 dz dt = f x(x(t), y(t)) dx dt (t) + f y(x(t), y(t)) dy dt (t). Example: f(x, y) = x 2 + 2y 3, x(t) = sin(t), y(t) = cos(2t). Let z(t) = f(x(t), y(t)). Then, dz dt = 2x(t) dx dy + 6[y(t)]2 dt dt, = 2x(t) cos(t) 2[y(t)] 2 sin(2t), = 2 sin(t) cos(t) 2 cos 2 (2t) sin(2t).

11 Math 20C Multivariable Calculus Lecture 3 Chain rule Slide 20 Case 2: Given f(x, y) differentiable, and x(t, s), y(t, s) differentiable functions of a two variable, then z(t, s) = f(x(t, s), y(t, s)) is differentiable and z t(t, s) = f x(x(t, s), y(t, s)) x t(t, s) + f y(x(t, s), y(t, s)) y t(t, s), z s(t, s) = f x(x(t, s), y(t, s)) x s(t, s) + f y(x(t, s), y(t, s)) y s(t, s). Example: Change of coordinates Consider the function f(x, y) = x 2 + ay 2, with a IR. Introduce polar coordinates r, θ by the formula x(r, θ) = r cos(θ), y(r, θ) = r sin(θ). Slide 2 Let z(r, θ) = f(x(r, θ), y(r, θ)). Then, the chain rule, case 2, says that z r = f x x r + f y y r. Each term can be computed as follows, f x = 2x, x r = cos(θ), f y 2ay, y r = sin(θ), then one has z r = 2r cos 2 (θ) + 2ar sin 2 (θ).

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