Math S21a: Multivariable calculus Oliver Knill, Summer 2011Lecture 12: GradientThe gradient of a function f (x, y) is defined as∇f(x, y) = hfx(x, y), fy(x, y)i .For functions of three dimensions, we define∇f(x, y, z) = hfx(x, y, z), fy(x, y, z), fz(x, y, z)i .The symbol ∇ is spelled ”Nabla” and named after an Egyptian har p. Here is a ver y import antfact:Gradients are orthogonal to level curves and level surfaces.Pro of. Every curve ~r(t) on the level curve or level surface satisfiesddtf(~r(t)) = 0. By the chainrule, ∇f (~r(t)) is perpendicular to t he tangent vector ~r′(t).Because ~n = ∇f (p, q) = ha, bi is perpendicular to the level curve f (x, y) = c t hrough (p, q), t heequation for the tangent line is ax + by = d, a = fx(p, q), b = fy(p, q), d = ap + bq. Compactlywritten, this is∇f(~x0) ·(~x −~x0) = 0and means that the gradient of f is perpendicular to any vector (~x −~x0) in the plane. It is one ofthe most important statements in multivariable calculus. since it provides a crucial link betweencalculus and geometry. The just mentioned gradient theorem is also useful. We can immediatelycompute tangent planes and tangent lines:1 Compute the tangent plane to the surface 3x2y + z2−4 = 0 at the point (1, 1, 1). Solution:∇f(x, y, z) = h6xy, 3x2, 2zi. And ∇f(1, 1, 1) = h6, 3, 2i. The plane is 6 x +3y +2z = d whered is a constant. We can find the constant d by plugging in a point and get 6x+3y+2z = 11.2 Problem: reflect the ray ~r(t) = h1 − t, −t, 1i at the surfacex4+ y2+ z6= 6 .Solution: ~r(t) hits the surface at the time t = 2 in the point (−1, −2, 1). The velocityvector in that ray is ~v = h−1, −1, 0i The normal vector at this point is ∇f (−1, −2, 1) =h−4, 4, 6i = ~n. The reflected vector isR(~v = 2Proj~n(~v) −~v .We have Proj~n(~v) = 8/68h−4, −4, 6i. Therefore, the reflected ray is ~w = (4/17)h−4, −4, 6i−h−1, −1, 0i.If f is a function of several variables and ~v is a unit vecto r then D~vf = ∇f · ~v iscalled the directional derivative of f in the direction ~v.The name directional derivat ive is related to the fact that every unit vector gives a direction. If~v is a unit vector, then the chain rule tells usddtD~vf =ddtf(x + t~v).The directional derivative tells us how the function changes when we move in a given direction.Assume for example that T (x, y, z) is the temperature at position (x, y, z). If we move with veloc-ity ~v through space, then D~vT tells us at which rate the temperature changes for us. If we movewith velocity ~v on a hilly surface of height h(x, y), then D~vh(x, y) gives us the slope we drive on.3 If ~r(t) is a curve with velocity ~r′(t) and the speed is 1, then D~r′(t)f = ∇f (~r(t)) ·~r′(t) is thetemperature change, one measures at ~r(t). The chain rule told us that this is d/dtf(~r (t)).4 For ~v = (1, 0, 0), then D~vf = ∇f · v = fx, the directional derivative is a generalization ofthe partial derivatives. It measures the rate of change of f , if we walk with unit speed intothat direction. But as with partial derivatives, it is a scalar.The directional derivative satisfies |D~vf| ≤ |∇f ||~v| because ∇f · ~v =|∇f||~v||cos(φ)| ≤ |∇f||~v|.The direction ~v = ∇f /|∇f| is the direction, where f increases most. It is thedirection of steepest ascent.If ~v = ∇f /|∇f|, then the directional derivative is ∇f · ∇f/|∇f | = |∇f|. Thismeans f increases, if we move into the direction of the gradient . The slope in thatdirection is |∇f|.5 You ar e on a trip in a air-ship over Cambridge at (1, 2) and you want to avoid a thunderstorm,a region of low pressure. The pressure is given by a function p(x, y) = x2+ 2y2. In whichdirection do you have to fly so that the pressure change is largest?Solution: The gradient ∇p(x, y) = h2x, 4yi at the point (1, 2) is h2, 8i. Normalize to getthe direction h1, 4i/√17.The directional derivative has the same properties than any derivative: Dv(λf) =λDv(f), Dv(f + g) = Dv(f) + Dv(g) and Dv(fg) = Dv(f)g + f Dv(g).We will see later that points with ∇f =~0 are candidates for local maxima or minima of f .Points (x, y), where ∇f(x, y) = (0, 0) are called critical points and help to understand the func-tion f .6 The Matterhorn is a 4’478 meter high mountain in Switzerland. It is quite easy to climbwith a guide because there are ropes and ladders at difficult places. Evenso there arequite many climbing accidents at the Matterhorn, this does not stop you from trying anascent. In suitable units on the ground, the height f(x, y) of the Matterhorn is approximatedby the function f (x, y) = 40 00 − x2− y2. At height f(−10 , 10) = 3800, at the point(−10, 10, 3800), you rest. The climbing route continues into the south-east direction v =h1, −1i/√2. Calculate the rate of change in that direction. We have ∇f(x, y) = h−2x, −2yi,so that h20, −20i · h1, −1i/√2 = 40/√2. This is a place, with a ladder, where you climb40/√2 meters up when advancing 1m forward.The rate of change in all directions is zero if a nd only if ∇f(x, y) = 0: if ∇f 6=~0, we canchoose ~v = ∇f/|∇f| and get D∇ff = |∇f |.7 Assume we know Dvf(1, 1) = 3/√5 and Dwf(1, 1) = 5/√5, where v = h1, 2i/√5 andw = h2, 1i/√5. Find the gradient of f . Not e that we do not know anything else about thefunction f .Solution: Let ∇f (1, 1) = ha, bi. We know a + 2b = 3 and 2a + b = 5. This allows us to geta = 7/3, b = 1/3.Homework1 A surface x2+ y2−z = 1 radiates light away. It can be parametrized as ~r(x, y) = hx, y, x2+y2−1i. Find the parametrization of the wave front which is distance 1 from the surface.2 Find the directional derivative D~vf(2, 1) = ∇f (2, 1) ·~v into the direction ~v = h−3, 4i/5 forthe function f(x, y) = x5y + y3+ x + y.3 Assume f (x, y) = 1 − x2+ y2. Compute the directional derivative D~v(x, y) at (0, 0) where~v = hcos(t), sin(t)i is a unit vector. Now computeDvDvf(x, y)at (0, 0), for any unit vector. Fo r which directions is this second directional derivativepositive?4 The Kitchen-Rosenberg formula gives the curvature of a level curve f(x, y) = c asκ =fxxf2y− 2fxyfxfy+ fyyf2x(f2x+ f2y)3/2Use this formula to find the curvature of the ellipsoid f(x, y) = x2+ 2y2= 1 at the point(1, 0).P.S. This formula is known since a hundred years at least but got revived in computer vision.If you want to derive the formula, you can check that the angleg(x, y) = arctan(fy/fx)of the g r adient vecto r has κ as the directional
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