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Part 4 Directional Derivatives Printed version of the lecture Differential Geometry on 13. September 2013 Tommy R. Jensen, Department of Mathematics, KNU 4.1 Overview Contents 1 Directional Derivatives 1 2 Operation of a Vector Field 4 3 Conclusion 5 4.2 1 Directional Derivatives Derivative of a function Definition Let f : R3 → R be a differentiable function. Let p ∈ R3 be a point. Let v p be a tangent vector to R3 in p, that is, v p ∈ Tp (R3 ). We write d f (p + tv)|t=0 . dt The number v p [ f ] is called the derivative of f with respect to v p . vp[ f ] = We also say that v p [ f ] is a directional derivative. 1 The directional derivative v p [ f ] informs us about the change in the value of f as we move away from p in the direction of v p . 4.3 An example of computing a directional derivative Let f = xy2 z: R3 → R. Let p = (1, 1, 0) and v = (1, 0, −3). Then p + tv = (1, 1, 0) + t(1, 0, −3) = (1 + t, 1, −3t). Evaluate f in this new point: f (p + tv) = f (1 + t, 1, −3t) = (1 + t)(−3t) = −3t − 3t 2 . d Differentiate with respect to t : dt f (p + tv) = −3 − 6t. Setting t = 0, we get the answer v p [ f ] = −3. We deduce that f is decreasing when moving away from p in the direction of v. To express the answer differently we could write (1, 0, −3)(1,1,0) [xy2 z] = −3. But we don’t want to ask for trouble. . . 4.4 How to differentiate composite functions The chain rule Let f : R3 → R be a differentiable function. Let a : R → R,t 7→ a(t) b : R → R,t 7→ b(t) c : R → R,t 7→ c(t) be differentiable functions of one variable t. Then f (a, b, c) : R → R,t 7→ f (a(t), b(t), c(t)) is a differentiable function of one variable t, and d ∂f da ∂ f db ∂ f dc f (a, b, c) = (a, b, c) + (a, b, c) + (a, b, c) . dt ∂ x1 dt ∂ x2 dt ∂ x3 dt 4.5 How to compute directional derivatives more efficiently Lemma 3.2 Let v p = (v1 , v2 , v3 ) p be a tangent vector to R3 in p. Then v p [ f ] = ∑ vi ∂f (p). ∂ xi Proof of Lemma 3.2 Let p = (p1 , p2 , p3 ). Then p + tv = (p1 + tv1 , p2 + tv2 , p3 + tv3 ). Now use the chain rule: 2 vp[ f ] = df (p + tv)|t=0 dt ∂f d = ∑ ∂ xi (p + tv) dt (pi + tvi )|t=0 = ∑ ∂ xi (p)vi . ∂f 4.6 The same example as before Let f = xy2 z: R3 → R. Let p = (1, 1, 0) and v = (1, 0, −3). Now we first find the partial derivatives of f : ∂f = y2 z, ∂x ∂f = 2xyz, ∂y ∂f = xy2 . ∂z ∂f (p) = 0, ∂y ∂f (p) = 1. ∂z At the point p = (1, 1, 0) we calculate: ∂f (p) = 0, ∂x By Lemma 3.2: v p [ f ] = 0 + 0 + (−3) · 1 = −3. 4.7 The main properties of directional derivatives Theorem 3.3 Let f : R3 → R and g : R3 → R be functions, let v p and w p be tangent vectors to R3 in p, and let a, b ∈ R be numbers. Then (1) (av p + bw p )[ f ] = av p [ f ] + bw p [ f ], (2) v p [a f + bg] = av p [ f ] + bv p [g], (3) v p [ f g] = v p [ f ] · g(p) + f (p) · v p [g]. (1) and (2) mean that v p [ f ] is linear in both the tangent vector v p and the function f . (3) means that the derivative satisfies the usual product rule for differentiation (also called Leibniz rule). 4.8 Proof of Theorem 3.3 Proof of v p [a f + bg] = av p [ f ] + bv p [g] Let v = (v1 , v2 , v3 ). From Lemma 3.2 we have v p [a f + bg] = = = = = = ∂ (a f + bg) (p) ∂ xi ∂f ∂g ∑ vi (a ∂ xi (p) + b ∂ xi (p)) ∂f ∂g ∑(vi a ∂ xi (p) + vi b ∂ xi (p)) ∂f ∂g ∑ vi a ∂ xi (p) + ∑ vi b ∂ xi (p) ∂f ∂g a ∑ vi (p) + b ∑ vi (p) ∂ xi ∂ xi av p [ f ] + bv p [g]. ∑ vi 3 4.9 A picture It is confusing, but nicer to look at than a proof. 4.10 2 Operation of a Vector Field The operation of a vector field on a function Definition Let V be a vector field, and let f : R3 → R be a function. Let V [ f ] be the function defined by V [ f ] : R3 → R, p 7→ V (p)[ f ]. We explain V [ f ] as the function which to each point p in R3 gives us the derivative of f in p with respect to the tangent vector V (p). Example For the natural frame field U1 ,U2 ,U3 we get by Lemma 3.2 U1 [ f ](p) = U1 (p)[ f ] = (1, 0, 0) p [ f ] = 1· ∂f ∂f ∂f ∂f (p) + 0 · (p) + 0 · (p) = (p). ∂x ∂y ∂z ∂x 4.11 4 Basic properties of operations of vector fields Corollary 3.4 Let V and W be vector fields on R3 , let f , g, h : R3 → R be functions, and let a, b ∈ R be numbers. Then (1) ( fV + gW )[h] = fV [h] + gW [h], (2) V [a f + bg] = aV [ f ] + bV [g], (3) V [ f g] = V [ f ] · g + f ·V [g]. Proof of V [a f + bg] = aV [ f ] + bV [g] Let p be any point in R3 . Def. V [a f + bg](p) = V (p)[a f + bg] Theorem 3.3 = aV (p)[ f ] + bV (p)[g] Def. = aV [ f ](p) + bV [g](p) = (aV [ f ] + bV [g])(p) 4.12 Example Let V be the vector field V = x2U1 + yU3 , and let f be the function f = x2 y − y2 z. V[f] = (x2U1 + yU3 )[x2 y − y2 z] = x2U1 [x2 y] − x2U1 [y2 z] + yU3 [x2 y] − yU3 [y2 z] = x2 (2xy) − x2 · 0 + y · 0 − y(y2 ) = 2x3 y − y3 . 4.13 3 Conclusion Here it ends END OF THE LECTURE! 4.14 5 Next time: Curves in R3 4.15 4.16 6