14.6 Directional Derivatives

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1 CHAPTER 14. PARTIAL DERIVATIVES Directional Derivatives Comments. Recall that the partial derivatives can be interpreted as the derivatives along traces of f(x, y). We can reinterpret this in terms @y : rate of change of f in direction of x rate of change of f in direction of y It is possible to take the derivative in other directions too. Definition. The directional derivative of f in the direction of u, whereu ha, bi is a unit vector, is f(x + ha, y + hb) f(x, y) lim h!0 h if this limit exists. The notation for this derivative is Duf(x, y) or Duf. Note. Note that the usual partial derivatives are special cases of the directional derivative: Duf f x Duf f y with u h1, 0i i with u h0, 1i j Comments. Hopefully the definition of the directional derivative makes sense: you calculate the di erence quotient using x and y that are allowed to move along the direction of u. But we want to be able to calculate it without limits. The next result shows us how. Theorem. If f is a di erentiable function of x and y, and u any unit vector, then Duf hf x,f y i u (where is the dot product) If u ha, bi then we have Duf f x a + f y b. This is where we ended on Monday, February 25 Comments. The vector hf x,f y i in the theorem above gets a special name. Definition. Let f(x, y) be given. The gradient of f is rf hf x,f j. The combination rf is usually read as grad f or del f. Comments. Using this notation, and the previous theorem, we can say that Duf rf u, where u 1. Example 1. Let f(x, y) x 3 y 4x 2 and v 3i +2j. Find the directional derivative at the point (1, 2) in the direction of the vector v. rf 3x 2 y 8x, x 3

2 CHAPTER 14. PARTIAL DERIVATIVES 108 rf(1, 2) 3(1) 2 2 h 2, 1i v p 9+4 p use u v v D 3 8, 1 E p 2, p D Duf(1, 2) h 2, 1i 6 p + p 2 E p3 2, p 4 p Functions of 3 variables Definition. Let f(x, y, z) be given and u a unit vector. We define the directional derivative: f(x + hu) f(x) Duf lim (where x hx, y, zi) h!0 h and the gradient rf(x, y, z) hf x,f y,f k Theorem. If u 1then Duf rf u. Example 2. Let f(x, y, z) e y cos(xz) (a) Find rf (b) Find Duf(0, 1, 2) where u is the unit vector in the direction of v 2i j + k (a) rf h ze y sin(xz), e y cos(xz), xe y sin(xz)i rf(0, 1, 2) h0, e,0i (b) v p p 6 u v v 1 p 6 h2, 1, 1i

3 CHAPTER 14. PARTIAL DERIVATIVES 109 Duf(0, 1, 2) h0, e,0i e p 6 1 p 6 h2, 1, 1i This is where we ended on Wednesday, February Maximizing the Directional Derivative Theorem. Fix (a, b) in the domain of f. ThenDuf(a, b) can vary as di erent unit vectors u are used to calculate the directional derivative. The maximum value of Duf(a, b) is rf and it occurs when u is in the same direction as rf. Proof. Duf rf u rf u cos( ) rf cos( ) (recall u 1). Since cos( ) apple1, this shows that Du apple rf, and that they are equal when 0, i.e. there is no angle between u and rf. Example 3. Find the maximal directional derivative of f(x, y) ln(x 2 + y 2 ) at (1, 2) and also find the direction in which it occurs. rf(x, y) 2x/(x 2 + y 2 ), 2y/(x 2 + y 2 ) rf(1, 2) h2/5, 4/5i max derivative rf r p 5 5 Max rate of change is 2 p 5/5 and occurs in same direction as rf, so in the direction of h2/5, 4/5i Gradients are Normal to Level Curves Theorem. Let z f(x, y). At each point (a, b), the vector rf(a, b) is orthogonal to the level curves through that point. In other words, unique path that causes the maximum increase in z is a path orthogonal to the level curves. Comments. Here s an intuitive explanation of why this would be true. Suppose you fix a point (a, b) on a level curve, e.g. something like below

4 CHAPTER 14. PARTIAL DERIVATIVES 110 (1, 1) z 5 z 6 z 8 z 11 If you draw a path starting at (1, 1), on the z 6 level curve, and want to have the values of z increase as rapidly as possible on the path, which way should it go? It shouldn t go %, because that way z decreases to z 5. It shouldn t go - or & because z doesn t change much that way. It should go as directly as possible towards the z 8 level curve. In other words, it should go in the. direction, orthogonal to the z 6 level curve. In general, such a path should always go orthogonal to each level curve, that takes the values towards the next level curve as rapidly as possible. This explains the last part of the theorem. The first part is justified by remembering that the direction that changes the function the most is given by rf. Thus, the path that changes z the most is orthogonal to the level curves, and the direction of that path must be the same as rf. Proof. To prove this we first have to make the statement more precise. Fix a point P, fix the level curve C that contains P, and a line L tangent to C at P. We will show that rf orthogonal to L. Let r(t) be the function that parameterizes the curve C and write r(t) hx(t), y(t)i. Since r(t) gives the curve C, and since C is defined by f(x, y) k for some constant k, we have f(x(t),y(t)) k. Now we take f(x(t),y(t)) k d dt f(x(t),y(t)) d dt dt dt 0 dx dt, dy dt 0 rf r 0 (t) 0 rf is orthogonal to r 0 (t) rf is orthogonal to L rf is orthogonal to C Now we recall that the path with maximum directional derivative is the path that is in the same direction as rf. Thus, the path that changes z the most is the path in the same direction as rf, which we have just shown is orthogonal to C.

5 CHAPTER 14. PARTIAL DERIVATIVES 111 Example 4. Shown below is a contour map the big island of Hawaii, and in particular, the volcano of Mauna Loa. A B (a) You are sitting in your 4 wheel ATV at the position marked with A and there is a huge earthquake. You hear sirens warning of a large tsunami coming. Your goal is to get as high up the mountain as fast as possible. Draw a path. (b) After escaping the tsunami, you find out that during the earthquake, a large lava flow has started at the position marked with B. The lava will take weeks to make it down the mountain, but evacuations need to start soon, and so the path the lava will take needs to be predicted. The lava flows on a path that decreases elevation as fast as possible. Draw the path.

6 CHAPTER 14. PARTIAL DERIVATIVES 112 Both paths should be defined by being orthogonal to the contour lines. In theory this uniquely defines the path, but in practice, the gaps between the lines, and di erences in drawing and reading the maps will produce di erent results. Two possible solutions are shown below. You might want to compare the solution to part (b) with the image on the next page. The image is a composite photograph, and the brown and black lines show old lava flows. The one that looks like our solution is from 1859.

7

8 CHAPTER 14. PARTIAL DERIVATIVES 114 Extra Examples Example 5. Let f(x, y) 4x 2 +2xy +2y 3 +3x and u be the unit vector with angle /3. Then find Duf and Duf(0, 1). a 1 cos( /3) 1 2 b 1sin( /3) p 3 2 f x 8x +2y +3 f y 2x +6y 2 Duf f x a + f y b (8x +2y + 3) 1 p 3 2 +(2x +6y2 ) 2 4x + y +3/2+ p 3x +3 p 3y 2 (4+ p 3)x +3 p 3y 2 + y +3/2 Duf(0, 1) 3 p 3+5/2 Example 6. Find rf where f(x) e x y sin(xy). f x e x y cos(xy)y f y e x y ( 1) cos(xy)x rf hf x,f y i e x y y cos(xy), e x y x cos(xy)

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