The Gradient and Level Sets

1. Let f(x, y) = x2 + y2.

(a) Find the gradient ∇f. Solution. ∇f(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 ∇f(a, b)? Draw the vector ∇f(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 ∇f(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 ∇f(~r(t)) · ~r0(t), so the gradient ∇f(~r(t)) is always perpendicular to r~0(t). Recall that r~0(t) gives the direction of the tangent vector, so ∇f 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 ∇f(a, b). (Don’t worry about the magnitude of ∇f(a, b).)

1 1

4 3

2

0

Solution. Let’s look at the upper right point. We know that ∇f(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 ∇f(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 ∇f(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 √ 3. Let S be the cylinder x2 + y2 = 4. Find the plane tangent to S at the point (1, 3, 5).

2 √ Solution. If we let f(x, y, z) = x2 + y2, then S is the level set f(x, y, z) = 4. Therefore, ∇f(1, 3, 5) will be a normal vector for the tangent plane we want. √ √ √ ∇f = h2x, 2y, 0i, so ∇f(1, 3, 5) = h2, 2 3, 0i. Since (1, 3, 5) is a point on the tangent plane, the √ √ √ √ 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, ∇f 6 , 2, 1 will be a normal vector for the tangent plane we want. √ ∇f = hy cos x, sin x, −1i, so ∇f π , 2, 1
= 3, 1 , −1 . Since π , 2, 1
is a point on the tangent plane, √6 2 6 √ 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 ∇f(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, ∇g(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 ∇g(1, 2, 3) must be parallel to h3, 4, −5i; in other words, ∇g(1, 2, 3) must be a scalar multiple of h3, 4, −5i.

Of course, we’re interested in ∇f = hfx, fyi, and we had defined g in terms of f, so let’s try to write ∇g in terms of f. Since g(x, y, z) = f(x, y) − z, ∇g = hfx, fy, −1i, and ∇g(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 ∇f(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 .

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