Discussion 8 Solution Thursday, February 10th. Consider the function f(x, y) := y 2 x 2.

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1 Discussion 8 Solution Thursday, February 10th. 1. Consider the function f(x, y) := y 2 x 2. (a) This function is a mapping from R n to R m. Determine the values of n and m. The value of n is 2 corresponding to x and y. The value of m is 1 corresponding to the single value y 2 x 2. This function is a scalar function. (b) Sketch the domain and range of f as a mapping. The domain lies in the space of R 2 and the range lies in the space of R. We can evaluate the function value for any pairs of real numbers (x, y). Therefore the domain is R 2, or we denote the domain by D = {(x, y) x R, y R}. In order to figure out the range, we can observe the possible output values corresponding to x = 0. f(0, y) = y 2 tells us that the range contains [0, ). Similarly, f(x, 0) = x 2 shows that the range contains (, 0]. Therefore the range is all real numbers, or we denote the range by R = {z z R}. The sketch of the mapping is shown below. 2. Consider the surface defined by the equation z = f(x, y). (a) Sketch three traces of this surface parallel to the xz-plane. First we would want to pick three planes parallel to the xz-plane. The planes will all have (0,1,0) as normal vector. The choice here will be y = 2, y = 0, and y = 2. On y = 2, we are looking at z = f(x, 2) = 4 x 2. This corresponds to a parabola on the xz-plane. Similarly we get z = f(x, 0) = x 2 on y = 0 and z = f(x, 2) = 4 x 2 on y = 2. Let us first sketch each parabola in R 2.

2 Combining our results and placing it in R 3 will give us (b) On a new set of coordinate axes, sketch three traces of this surface parallel to the yz-plane. For planes parallel to the yz-plane, we pick x = 2, x = 0, and x = 2. This time we get the equations z = y 2 4, z = y 2, and z = y 2 4 respectively. Pictures we get :

3 (c) On a new set of coordinate axes, sketch the cross-sections of this surface in the planes z = 4, z = 0 and z = 4, each of which is parallel to the xy-plane. We will follow along the same method as above. Observe that on z = 4, we get the curve defined by y 2 x 2 = 4. On z = 0, we get the curve defined by y 2 x 2 = 0. And on z = 4 we get the curve defined by y 2 x 2 = 4. The three curves are shown below And again combining the results to get one single picture 3. (a) In R 2, sketch the level curves of the function f(x, y) defined by f(x, y) = 4,

4 f(x, y) = 0, and f(x, y) = 4. How are these level curves different from the cross-sections you sketched in 2c? The level curve defined by f(x, y) = 4 lie in the domain of f. More specifically, it is the set of points (x, y) that satisfies the equation. By our previous work, the sketch is shown below The level curve defined by f(x, y) = 0 is And the level curve defined by f(x, y) = 4 is Notice that the cross-sections lie in the space of R 3, while the level curves in R 2. This is because we think of level curves as subsets of the domain corresponding to a function value. The cross-sections refers to the intersection of the graph of the function with a given plane. It is important to understand where these objects lie. (b) Find a parameterization of one branch of the level curve defined by f(x, y) = 4. We start by picking one branch of the level curve.

5 In order to parameterize the curve, we want a single variable that can point us to a specific position on the curve. Let t = x because we can solve for the y-value for any given x-value. Then y = 4 + t 2 gives us a positive y-value. Hence the parameterization is (x, y) = (t, 4 + t 2 ), for all t R. (c) Is it possible for two level curves to intersect each other? Why not? It is impossible for two different level curves to intersect. The reasoning is related to the vertical line test for functions mapping R to R. Suppose that I have two level curves defined by f(x, y) = A and f(x, y) = B. The only possibility for them to intersect is when you can find a pair (x 0, y 0 ) in the domain such that f(x 0, y 0 ) is equal to A and B simultaneously. This only makes sense if either f is not a function or A and B are the same number, both of them not what we are working with. (d) The level curve defined by f(x, y) = 0 appears to intersect itself. Why does this not contradict your answer to (c)? In (c) we were talking about how two different level curves can never intersect. The intersection point being on the same level curves corresponds to the part A = B. Hence no contradiction occurs. 4. (a) Use what you learned in 2 to sketch a graph of the function f(x, y), that is, sketch the surface defined by the equation z = f(x, y). [Hint: It may be easiest to restrict the domain to the unit disk.] The pictures below are from the internet. The general shape of the graph is correct but the scaling might be different.

6 (b) Describe the surface you drew in 4a in words. Everyone has their own way of describing this. I think of the surface as linking two parabolas together at their vertex with them being perpendicular and opposite facing. Next we fill in the surface in the four parts in a smooth way. The resulting surface is called a saddle.