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.