The Gradient and Level Sets 1. Let f(x; y) = x2 + y2. (a) Find the gradient rf. Solution. rf(x; y) = h2x; 2yi. (b) Pick your favorite positive number k, and let C be the curve f(x; y) = k. Draw the curve on the axes below. Now pick a point (a; b) on the curve C. What is the vector rf(a; b)? Draw the vector rf(a; b) with its tail at the point (a; b). What relationship does the vector have to the curve? Solution. The curve f(x; y) = k is a circle in the xy-plane centered at the origin. The vector rf(a; b) is equal to h2a; 2bi, and this vector is perpendicular to the circle (no matter what value of k you picked and what point (a; b) you picked). (c) Let ~r(t) be any parameterization of your curve C. What is f(~r(t))? What happens if you use the d Chain Rule to find dt f(~r(t))? Use this to explain your observation from (b). Solution. f(~r(t)) is always equal to k, since ~r(t) parameterizes the curve C, and the curve C is exactly the set of points where f(x; y) = k. d d d Since f(~r(t)) = k for all t, dt f(~r(t)) = dt k = 0. By the Chain Rule, we know that dt f(~r(t)) is also equal to rf(~r(t)) · ~r0(t), so the gradient rf(~r(t)) is always perpendicular to r~0(t). Recall that r~0(t) gives the direction of the tangent vector, so rf is always perpendicular to the tangent vector. 2. Here is the level set diagram (contour map) of a function f(x; y). The value of f(x; y) on each level set is labeled. For each of the three points (a; b) marked in the picture, draw a vector showing the direction of rf(a; b). (Don't worry about the magnitude of rf(a; b).) 1 1 4 3 2 0 Solution. Let's look at the upper right point. We know that rf(a; b) is perpendicular to the curve at (a; b), but there are still two possible directions: 1 4 3 2 0 To determine which is correct, we use the definition of rf(a; b): it is defined to be hfx(a; b); fy(a; b)i. We can tell from the original picture that fx(a; b) > 0 (see the worksheet \Partial Derivatives" for more explanation), so the vector rf(a; b) must point to the right. Using the same reasoning for the other two points, we get the following vectors: 1 4 3 2 0 p 3. Let S be the cylinder x2 + y2 = 4. Find the plane tangent to S at the point (1; 3; 5). 2 p Solution. If we let f(x; y; z) = x2 + y2, then S is the level set f(x; y; z) = 4. Therefore, rf(1; 3; 5) will be a normal vector for the tangent plane we want. p p p rf = h2x; 2y; 0i, so rf(1; 3; 5) = h2; 2 3; 0i. Since (1; 3; 5) is a point on the tangent plane, the p p p p tangent plane has equation h2; 2 3; 0i · hx − 1; y − 3; z − 5i = 0, or 2(x − 1) + 2 3(y − 3) = 0 . Note that this is exactly the same problem as #1 from the worksheet \Tangent Planes and Linear Approximation"; this solution, however, is simpler than the one we came up with before. π 4. Let S be the surface z = y sin x. Find the plane tangent to S at the point 6 ; 2; 1 . π Solution. If we let f(x; y; z) = y sin x−z, then S is the level set f(x; y; z) = 0. Therefore, rf 6 ; 2; 1 will be a normal vector for the tangent plane we want. p rf = hy cos x; sin x; −1i, so rf π ; 2; 1 = 3; 1 ; −1 . Since π ; 2; 1 is a point on the tangent plane, p6 2 6 p 1 π π 1 the tangent plane has equation 3; 2 ; −1 · x − 6 ; y − 2; z − 1 = 0, or 3 x − 6 + 2 (y − 2) − (z − 1) = 0 . Note that this is exactly the same problem as #2 from the worksheet \Tangent Planes and Linear Approximation"; this solution, however, is simpler than the one we came up with before. 5. Suppose that 3x + 4y − 5z = −4 is the plane tangent to the graph of f(x; y) at the point (1; 2; 3). (a) Find rf(1; 2). Solution. We are told that 3x + 4y − 5z = −4 is the plane tangent to the surface z = f(x; y) at the point (1; 2; 3). Let's think about how we normally find the tangent plane to a graph. We could use the result of #3 from the worksheet \Tangent Planes and Linear Approximation", or we could simply express the graph as a level surface. Let's take the second approach. To find the plane tangent to z = f(x; y) at the point (1; 2; 3), let's write g(x; y; z) = f(x; y) − z. Then, the surface z = f(x; y) can also be described as the level set g(x; y; z) = 0. Therefore, rg(1; 2; 3) must be a normal vector for the tangent plane. From the equation we are given from the tangent plane, we can see that h3; 4; −5i is also a normal vector for the tangent plane. Therefore, we know that rg(1; 2; 3) must be parallel to h3; 4; −5i; in other words, rg(1; 2; 3) must be a scalar multiple of h3; 4; −5i. Of course, we're interested in rf = hfx; fyi, and we had defined g in terms of f, so let's try to write rg in terms of f. Since g(x; y; z) = f(x; y) − z, rg = hfx; fy; −1i, and rg(1; 2; 3) = hfx(1; 2); fy(1; 2); −1i. So, we can conclude that hfx(1; 2); fy(1; 2); −1i is a scalar multiple of h3; 4; −5i. What scalar? 1 Well, looking at the last component of each vector, we can see that the scalar must be 5 ; that 1 3 4 is, hfx(1; 2); fy(1; 2); −1i = 5 h3; 4; −5i. So, fx(1; 2) = 5 and fy(1; 2) = 5 , which means that 3 4 rf(1; 2) = ; . 5 5 (b) Use linear approximation to approximate f(1:1; 1:9). Solution. Now that we know fx(1; 2) and fy(1; 2), we can write down the linearization L(x; y) of f(x; y) at (1; 2).(1) However, there is an even simpler way to do this problem. Remember (1) 3 4 It is L(x; y) = f(1; 2) + fx(1; 2)(x − 1) + fy(1; 2)(y − 2) = 3 + 5 (x − 1) + 5 (y − 2). 3 that the graph of L(x; y) was exactly the tangent plane to f at (1; 2). So, z = L(x; y) is the 3 4 4 3 4 4 same as 3x + 4y − 5z = −4, or z = 5 x + 5 y + 5 . That is, L(x; y) = 5 x + 5 y + 5 . So, 3 4 4 f(1:1; 1:9) ≈ L(1:1; 1:9) = 5 (1:1) + 5 (1:9) + 5 = 2:98 . 4.