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1 Chain Rules .2 Directional Derivative .3 Gradient Vector Field .4 1. Chain rules 2. Directional derivative 3. Gradient Vector Field 4. Most Rapid Increase 5. Implicit Function Theorem, Implicit Differentiation 6. Lagrange Multiplier 7. Second Derivative Test . Matb 210 in 2012 . Theorem. Suppose that w = f (x, y, z) is a differentiable function, where x = x(u, v), y = y(u, v), z = z(u, v), where the coordinate functions are parameterized by differentiable functions. Then the composite function w(u, v) = f ( x(u, v), (u, v), (u, v)) is a differentiable function in u and v, such that the partial functions are given by ¶w ¶w ¶x ¶w ¶y ¶w ¶z = · + · + · ; ¶u ¶x ¶u ¶y ¶u ¶z ¶u ¶w ¶w ¶x ¶w ¶y ¶w ¶z = · + · + · . ¶v ¶x ¶v ¶y ¶v ¶z ¶v . Remark. The formula stated above is very important in the theory of .surface integral. Matb 210 in 2012 . Theorem (Chain Rule for Coordinate Changes). Suppose that s = f (x, y, z) is a differentiable function, where x = x(u, v, w), y = y(u, v, w), z = z(u, v, w), where the coordinate functions are parameterized by differentiable functions in variables u, v and w. Then the composite function S(u, v, w) = f ( x(u, v, w), (u, v, w), (u, v, w)) is a differentiable function in u, v and w, such that the partial functions are given by ¶S ¶S ¶x ¶S ¶y ¶S ¶z = · + · + · ; ¶u ¶x ¶u ¶y ¶u ¶z ¶u ¶S ¶S ¶x ¶S ¶y ¶S ¶z = · + · + · ; ¶v ¶x ¶v ¶y ¶v ¶z ¶v ¶ ¶ ¶ ¶ ¶ ¶ ¶ S S · x S · y S · z ¶ = ¶ ¶ + ¶ ¶ + ¶ ¶ . w x w y w z w . Remark. The formula stated above is very important in the theory of .inverse function theory and integration theory. Matb 210 in 2012 . Example. In spherical coordinates, we have the parameters (r, q, f) to represent (x, y, z) as follows: x = r sin f cos q, y = r sin f sin q, z = r cos f, with r ≥ 0,p 0 ≤ q ≤ 2p, and 0 ≤ f ≤ p. Define 2 2 2 ¶S S(x, y, z) = x + y + z . Evaluate the partial derivative ¶r in two .ways. ¶S Solution. (i) Since S(r sin f cos q, r sin f sin q, r cos f) = r, so ¶r = 1 for any choices of parameters involved. (ii) The second method is to apply chain rule. ¶S p x ¶x ¶ ¶ = = sin f cos q, ¶r = ¶r (r sin f cos q) = sin f cos q, x x2+y2+z2 ¶S p y ¶y ¶ ¶ = = sin f sin q, ¶r = ¶r (r sin f sin q) = sin f sin q, y x2+y2+z2 and ¶S p z ¶z ¶ ¶ = = cos f, ¶r = ¶r (r cos f) = cos f. z x2+y2+z2 ¶S ¶S · ¶x ¶S · ¶y ¶S · ¶z And ¶r = ¶x ¶r + ¶y ¶r + ¶z ¶r = (sin f cos q)2 + (sin f sin q)2 + (cos f)2 = (sin2 f)(cos2 q + sin2 q ) + cos2 f = 1. Matb 210 in 2012 . Theorem. (Chain Rule of 2-variables) Suppose that f (x, y)nd is a real valued function defined on the planar domain D, and that r(t) = x(t)i + y(t)j is a curve in the domain D. Then we obtain a real-valued function g(t) = f (x(t), y(t)) which is a function of t. Then the derivative of g is given by 0 d ¶f dx ¶f dy g (t) = ( f (x(t), y(t)) ) = (r(t)) · + (r(t)) · = dt ¶x dt ¶y dt 0 0 .fx(r(t))x (t) + fy(r(t))y (t). Matb 210 in 2012 . Theorem. Chain Rule of 3-variables Suppose that f (x, y, z) is real valued function defined on the domain D which is part of R3, and that x = x(t), y = y(t) and z = z(t) is a curve in the domain D. One can think of the a particle moving in domain D, and its position is given by (x(t), y(t), z(t)) changing with respect to t, so it traces out a path in domain D given by r(t) = x(t)i + y(t)j + z(t)k. Then we obtain a real-valued function g(t) = f (x(t), y(t), z(t)). Then the chain rule tells us that 0 d g (t) = ( f (x(t), y(t), z(t)) ) dt ¶f · dx ¶f · dy ¶f · dz = ¶x (r(t)) dt + ¶y (r(t)) dt + ¶z (r(t)) dt (chain rule) 0 0 0 . = fx(r(t))x (t) + fy(r(t))y (t) + fz(r(t))z (t). Matb 210 in 2012 . Partial Derivatives . Suppose that z = f (x, y) is a function defined in a domain D, and P(a, b) is a point in D. Recall that the partial derivatives f (a + h, b) − f (a, b) fx(a, b) = lim , and h!0 h f (a, b + k) − f (a, b) fy(a, b) = lim . k!0 k .The limits are taken along the coordinate axes. Matb 210 in 2012 . Directional Derivative . Through the point P(a, b) we choose any direction u = (h, k) = hi + kj, then we consider the straight line through the point P along the direction u given by r(t) = (a + ht, b + kt), and the rate of change of g(t) = f (r(t)) at t = 0 is 0 ( ( ))− ( ( )) ( + + )− ( ) g (0) = lim f r t f r 0 = lim f a th,b tk f a,b . t!0 t t!0 t z T P(x¸, y¸, z¸) y 3 x . Matb 210 in 2012 . Directional Derivative . Through the point P(a, b) we choose any direction u = (h, k) = hi + kj, then we consider the straight line through the point P along the direction u given by r(t) = (a + ht, b + kt), and the rate of change of g(t) = f (r(t)) at t = 0 is 0 ( ( ))− ( ( )) ( + + )− ( ) g (0) = lim f r t f r 0 = lim f a th,b tk f a,b . t!0 t t!0 t . Suppose that f is differentiable, then it follows from the (multivariate) chain rule that 0 ¶f dx ¶f dy ¶f ¶f g (0) = + = (P)h + (P)k (¶x dt ¶y dt ¶)x ¶y = fx(a, b)i + fy(a, b)j · (hi + kj) = rf (a, b) · u, where rf is the vector-valued function fxi + fyj, called the gradient of 0 f at the point (x, y). Note g (0) only depends of the choice of the 0 .curve through P(a, b) with tangent direction r (0) only. Matb 210 in 2012 In order to simplify the notation more, one requires the directional vector u to be an unit vector. 0 Definition (Directional derivative) The resulting derivative g (0) is called the directional derivative Duf of f along the direction u, and hence we write Duf = rf · u to represent the rate of the change of f .in the unit direction u. ··· Remark. In general, if f = f (x1, , xn) is a function of n variables, one define ¶f ¶f (i) the gradient of f to be rf = ( ¶ , ··· , ¶ ), and x1 xn (ii) the directional derivative Duf by Duf = rf · u. Matb 210 in 2012 . Example. Evaluate the directional derivative y f (x, y) = xe + cos(xy) at the point P0(2, 0) in − .the direction 3i 4j. Remark. The blue curve is the level curve of f at different values. − Solution. Let u = p3i 4i = 3 i − 4 j. And rf (2, 0) = 32+42 5 5 y − y − j (fx(2, 0), fy(2, 0)) = (e y sin(xy), xe x sin(xy)) (x,y)=(2,0) = (1, 2). r · 3 − 8 − It follows that Duf (2, 0) = f (2, 0) u = 5 5 = 1. Matb 210 in 2012 . Proposition. The greatest rate of change of a scalar function f , i.e., the maximum directional derivative, takes place in the direction of, r .and has the magnitude of, the vector f . Proof. For any direction v, the directional derivative of f along the direction v at a point P in the domain of f , is given by ( ) = hr ( ) v i = kr k q where q is the angle between Dv P : f P , kvk f cos , the vectors rf (P) and v. Hence Dv(P) attains maximum (minimum) value if and only if cos q = 1 (−1), if and only if rf (P) ( −rf (P) ) is parallel to v. In this case, we have Dv(P) = krf k ( −krf k ). Matb 210 in 2012 . Example. (a) Find the directional derivative of f (x, y, z) = 2x3y − 3y2z at P(1, 2, −1) in a direction v = 2i − 3j + 6k. (b) In what direction from P is the directional derivative a maximum? .(c) What is the magnitude of the maximum directional derivative? Solution. r 2 3 − − 2 j − (a) f (P) = 6x yi + (2x 6yz)j 3y k (1,2,−1) = 12i + 14j 12k at P. Then the directional derivative of f along the direction v is v p2i−3j+6k 90 given by Dvf = rf · = h12i + 14j − 12k, i = − . kvk 22+32+62 7 (b) Dvf (P) is maximum(minimum) () v (−v) is parallel to rf (P) = 12i + 14j − 12k. (c) The maximum magnitude of Dvf (P) is given by r ( ) kr ( )k2 rf · f P = f P = krf (P)k = k12i + 14j − 12kk = p krf (P)k krf (P)k 144 + 196 + 144 = 22. Matb 210 in 2012 . Proposition. Let C : r(t) = x(t)i + y(t)j + z(t)k be a curve lying on the level surface S : f (x, y, z) = c for some c, i.e.
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