The Gradient and Level Sets. Let f(x, y) = x + y. (a) Find the gradient f. Solution. f(x, y) = x, y. (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 a, b, 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 Chain Rule to find d dtf( r(t))? (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 Since f( r(t)) = k for all t, dt f( r(t)) = d dt k = 0. By the Chain Rule, we know that d dtf( r(t)) is also equal to f( r(t)) r (t), so the gradient f( r(t)) is always perpendicular to r (t). Recall that r (t) gives the direction of the tangent vector, so f is always perpendicular to the tangent vector.. Here is the level set diagram of a function f(x, y); the value of f on each level set is labeled. Imagine that f(x, y) represents temperature on the blackboard, and an ant is standing at the point (a, b), which is marked on the diagram.
4 0 What direction should the ant go to warm up most quickly? That is, in what direction should he go to experience the highest instantaneous rate of change of temperature (with respect to distance)? What direction should the ant go to cool down most quickly? That is, in what direction should he go to experience the lowest (most negative) instantaneous rate of change of temperature? Solution. Using the same reasoning as in #, the fly should fly in the direction of f(a, b) to warm up most quickly (marked in red below) and the direction of f(a, b) to cool down most quickly (marked in blue below). Remember that we always want our direction vectors to be unit vectors, so the direction of f(a, b) really means the vector f(a,b) f(a,b) the vector f(a,b) f(a,b)., and the direction of f(a, b) really means 4 0. A fly is flying around a room in which the temperature is given by T (x, y, z) = x + y 4 + z. The fly is at the point (,, ) and realizes that he s cold. In what direction should he fly to warm up most quickly? If he flies in this direction, what will be the instantaneous rate of change of his temperature? Solution. To answer the first question, we want to find the unit vector u which maximizes the directional derivative D u T (,, ). We know that this directional derivative is equal to T (,, ) u, which is in turn equal to T (,, ) u cos θ, where θ is the angle between T (,, ) and u. Since we are looking for a unit vector u, u =, so the directional derivative will just be T (,, ) cos θ. To maximize this, we need to make cos θ as large as possible. The largest that cos θ can be is, and this happens when θ = 0. So, we want the angle θ between T (,, ) and u to be 0, which means that we want u to go in the
same direction as T (,, ). We can calculate T (,, ) easily: T = x, 4y, 4z T (,, ) =, 4, 4 We want u to be a unit vector going in the same direction, which means we simply divide this vector by its length (which is + 4 + 4 = 6), so u =,,. The instantaneous rate of change of the fly s temperature when he flies in this direction is simply D u T (,, ) = T (,, ) u =, 4, 4,, = 6. 4. You re hiking a mountain which is the graph of f(x, y) = x xy y. You re standing at (,, 9). You wish to head in a direction which will maintain your elevation (so you want the instantaneous change in your elevation to be 0). How many possible directions are there for you to head? What are they? Solution. If you head in a direction given by a unit vector u, then the directional derivative will be f(, ) u. You want this to be 0, so you want u to be perpendicular to f(, ). There are two unit vectors in the plane which are perpendicular to f(, ). We calculate f = x y, x 6y, so f(, ) = 4, 8. The two unit vectors perpendicular to this are, and,. () () To find these, start with any vector that is perpendicular to 4, 8 ; one example is 8, 4 (we can tell it is perpendicular to 4, 8 because if we dot it with 4, 8, we get 0). To get a unit vector, divide this vector by its length (4 ) to get,. The other unit vector which is perpendicular to 4, 8 must be the negative of the one we have already found.
. Let f(x, y) = (x y) = x xy + y. (The graph and level set diagram of f are shown.) z y 4 9 x x y 0 4 9 Calculate the following directional derivatives of f. (a) D u f(, 0) where u =,. Solution. We ll use the formula D u f(, 0) = f(, 0) u. In this case, f = (x y), (x y), so f(, 0) =, and D u f(, 0) =,, = 0. (b) D u f(, 0) where u =,. Solution. D u f(, 0) = f(, 0) u =,, =. (c) D u f(0, ) where u =,. Solution. D u f(0, ) = f(0, ) u =,, =. (d) D u f(0, ) where u =,. Solution. D u f(0, ) = f(0, ) u =,, =. 6. Let S be the cylinder x + y = 4. Find the plane tangent to S at the point (,, ). Solution. If we let f(x, y, z) = x + y, then S is the level set f(x, y, z) = 4. Therefore, f(,, ) will be a normal vector for the tangent plane we want. f = x, y, 0, so f(,, ) =,, 0. Since (,, ) is a point on the tangent plane, the tangent plane has equation,, 0 x, y, z = 0, or (x ) + (y ) = 0. Note that this is exactly the same problem as # from the worksheet Tangent Planes and Linear Approximation ; this solution, however, is simpler than the one we came up with before. 7. Let S be the surface z = y sin x. Find the plane tangent to S at the point ( π 6,, ). 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 ( π 6,, ) will be a normal vector for the tangent plane we want. f = y cos x, sin x,, so f ( π 6,, ) =,,. Since ( π 6,, ) is a point on the tangent plane, the tangent plane has equation,, x π 6, y, z = 0, or ( ) x π 6 + (y ) (z ) = 0. 4
Note that this is exactly the same problem as # from the worksheet Tangent Planes and Linear Approximation ; this solution, however, is simpler than the one we came up with before. 8. Suppose that x + 4y z = 4 is the plane tangent to the graph of f(x, y) at the point (,, ). (a) Find f(, ). Solution. We are told that x + 4y z = 4 is the plane tangent to the surface z = f(x, y) at the point (,, ). Let s think about how we normally find the tangent plane to a graph. We could use the result of # 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 (,, ), 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(,, ) must be a normal vector for the tangent plane. From the equation we are given from the tangent plane, we can see that, 4, is also a normal vector for the tangent plane. Therefore, we know that g(,, ) must be parallel to, 4, ; in other words, g(,, ) must be a scalar multiple of, 4,. Of course, we re interested in f = f x, f y, 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 = f x, f y,, and g(,, ) = f x (, ), f y (, ),. So, we can conclude that f x (, ), f y (, ), is a scalar multiple of, 4,. What scalar? Well, looking at the last component of each vector, we can see that the scalar must be ; that is, f x (, ), f y (, ), =, 4,. So, f x(, ) = and f y(, ) = 4, which means that f(, ) =, 4. (b) Use linear approximation to approximate f(.,.9). Solution. Now that we know f x (, ) and f y (, ), we can write down the linearization L(x, y) of f(x, y) at (, ). () However, there is an even simpler way to do this problem. Remember that the graph of L(x, y) was exactly the tangent plane to f at (, ). So, z = L(x, y) is the same as x + 4y z = 4, or z = x + 4 y + 4. That is, L(x, y) = x + 4 y + 4. So, f(.,.9) L(.,.9) = (.) + 4 (.9) + 4 =.98. Challenging problem : If r(t) is a curve on a surface, what are the relations between the unit tangent vector, the unit normal vector, the binormal vector of the curve, and the tangent plane of the surface? () It is L(x, y) = f(, ) + f x(, )(x ) + f y(, )(y ) = + (x ) + 4 (y ).
Not so challenging problem: In Problem 8, suppose the plane given is the tangent plane of g(x, y, z) = k at the point (,, ). Can you find g(,, )? 6