Therefore, Duf(a) f(a + hu) − f(a) = lim h!0 h Directional derivatives, steepest g(a + hu) − f(a) = lim ascent, tangent planes h!0 h Math 131 Multivariate Calculus f (a)hu + f (a)hu + ··· + f (a)hu = lim x1 1 x2 2 xn n D Joyce, Spring 2014 h!0 h = fx1 (a)u1 + fx2 (a)u2 + ··· + fxn (a)un Directional derivatives. Consider a scalar field In other notation, the directional derivative is the f : Rn ! R on Rn. So far we have only considered dot product of the gradient and the direction the partial derivatives in the directions of the axes. @f For instance gives the rate of change along a Duf(a) = rf(a) · u @x line parallel to the x-axis. What if we want the We can interpret this as saying that the gradient, rate of change in a direction which is not parallel rf(a), has enough information to find the deriva- to an axis? tive in any direction. First, we can identify directions as unit vectors, those vectors whose lengths equal 1. Let u be such Steepest ascent. The gradient rf(a) is a vector a unit vector, kuk = 1. Then we define the direc- in a certain direction. Let u be any direction, that tional derivative of f in the direction u as being the is, any unit vector, and let θ be the angle between limit the vectors rf(a) and u. Now, we may conclude f(a + hu) − f(a) that the directional derivative Duf(a) = lim : h!0 h Duf(a) = rf(a) · u = krf(a)k cos θ This is the rate of change as x ! a in the direction since, in general, the dot product of two vectors b u. When u is the standard unit vector e , then, i and c is as expected, this directional derivative is the ith b · c = kbk kck cos θ partial derivative, that is, Dei f(a) = fxi (a). These directional derivatives are linear combi- but in our case, u is a unit vector. But cos θ is nations of the partial derivatives, at least when between −1 and 1, so the largest the directional f is differentiable. Note that the direction u = derivative Duf(a) can be is when θ is 0, that is (u1; u2; : : : ; un) is a linear combination of the stan- when u is the direction of the gradient rf(a). dard unit vectors: In other words, the gradient rf(a) points in the direction of the greatest increase of f, that is, the direction of steepest ascent. Of course, the oppo- u = u1e1 + u2e2 + ··· + unen: site direction, −∇f(a), is the direction of steepest And, when f is differentiable, it is well- descent. approximated by the linear function g that de- Example 1. Find the curves of steepest descent scribes the tangent plane, that is, by g(x) = for the ellipsoid 2 2 2 f(a) + fx1 (a)(x1 − a1) + ··· + fxn (a)(xn − an): 4x + y + 4z = 16 for z ≥ 0: 1 If we can describe the projections of the curves in Tangent planes. We can, of course, use gradi- the (x; y)-plane, that's enough. This ellipsoid is the ents to find equations for planes tangent to surfaces. graph of a function f : R2 ! R given by A typical surface in R3 is given by an equation p f(x; y) = 1 16 − 4x2 − y2: 2 f(x; y; z) = c: The gradient of this function is @f @f That is to say, a surface is a level set of a scalar- rf = ; @x @y valued function f : R3 ! R. More generally, a typ- ! ical hypersurface in Rn+1 is a level set of a function −2x −y = ; f : Rn ! R. p 2 2 p 2 2 16 − 4x − y 2 16 − 4x − y Now, the gradient rf(a) of f points in the di- The curve of steepest descent will be in the opposite rection of the greatest change of f, and vectors or- direction, −∇f. thogonal to rf(a) point in directions of 0 change So, we're looking for a path x(t) = (x(t); y(t)) of f, that is to say, they lie on the tangent plane. whose derivative is −∇f. In other words, we need Another way of saying that is that rf(a) is a vector two functions x(t) and y(t) such that normal to the surface. If x is any point in R3, then 0 2x x (t) = ; rf(a) · (a − x) = 0 p16 − 4x2 − y2 y y0(t) = : says that the vector a − x is orthogonal to rf(a), 2p16 − 4x2 − y2 and therefore lies in the tangent plane, and so x is Each is a differential equation with independent a point on that plane. variable t. We can eliminate t from the discussion since dy dy .dx y = = : dx dt dt 2x A common method to solve differential equations is separation of variables, which we can use here. From the last equation, we get dy dx = y 4x and, then integrating, Z dy Z dx = ; y 4x so 1 ln jyj = 4 ln jxj + C; C which gives us, writing A for e , Example 2 (Continuous, nondifferentiable func- jyj = Apjxj: tion). You're familiar with functions of one vari- able that not continuous everywhere. For example, That describes the curves of steepest descent as a f(x) = jxj is continuous, and it's differentiable ev- family of curves parameterized by the real constant erywhere except at x = 0. The left derivative is −1 A (different from the last constant A) there, but the right derivative is 1. 4 x = Ay : Things like that can happen for functions of more 2 than one variable. Consider the function 8 0 if x = y = 0 < xy f(x) = otherwise : px2 + y2 This function is continuous everywhere, but it's not differentiable at (x; y) = (0; 0). The graph z = f(x; y) has no tangent plane there. There are directional derivatives in two directions, namely, along the x-axis the function is constantly 0, so the df partial derivative is 0; likewise along the y-axis, dx df and is 0. dy But in all other directions, the directional deriva- tive does not exist. For instance,p along the line y = x the function is f(x; x) = jxj= 2, which has no derivative at x = 0. Math 131 Home Page at http://math.clarku.edu/~djoyce/ma131/ 3.