math soluion 2 - Homework due 31th January Problem(§2.1...

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Unformatted text preview: Homework due 31th January Problem (§2.1 Exercise 2b and 20). Describe the behavior, as c varies, of the level curve f (as, y) = c for each of these functions: (b) f(m,y)=1—m2—y2 and (C) f(m,y)=m3—r. Solution. (b) The level curve f (m,y) = c is {(9341) E R21m2+y2 =1—C}‘ If a _<_ 1 this is a circle with radius V1 — c, If a > 1 we get an ‘empty’ level curve (because the range of f (as, y) is (—00, 1]) (c) The level curve f (32,11) = c is {(93,34) ER2|w3—m——c=0}. Note that there is no restriction on 3/. Therefore the level curve consists of vertical lines cutting the ac—axis at each solution of the cubic equation 3:3 — m — c = 0. There Will be 1, 2 or 3 vertical lines, depending on c, As an example, the level curve of height a = O is Problem (§2.1 Exercise 14). Describe the graph of the function R2 ——> R m,y) 1-—-+ my f:( by computing some level curves and sections, Solution. The level curves for c = —2, —1,0, 1,2 are shown below. The sections a: = 0 and y = 0 don’t help, so look at y = a: and y = —a:: So the graph of f (m,y) = my is a saddle, with uphill along the line 3/ = as, and downhill along the line y = —m, My attempt is: (it is ok to describe the surface in words), III Problem (§2.1 Exercise 18). Describe the graph of the function R2 ———> R my) *——> maXflmtlit/l} fi( by computing some level curves and sections. Solution, The range of f (say) is [0, 00), and so we look at level curves with c 2 O. For example, the level curve f(w,y) = 1 is maxflml, |yl} = 1, and so either lml =1 and lyl _<_ 1, or lyl = 1 and lax! S 1. So the level curves are: The section m = 0 is and so the graph of f (ray) is an inverted square pyramid With apex at (0, 0, 0): \ X Problem (§2.2 Exercise 4). Show that the subset D = {(m,y) I a: 74— 0 and y 75 O} of R2 is open. Solution, The set D is sketched below: By the picture, if x E D then there exists a radius 7“ > 0 such that DT(x) C; D. Therefore D is open, Explicitly, we can take 1" = min{|m|, |y|} 4. Problem (Extra problem 1). For each part below, specify the boundary of the set 5' whose . points (m,y) satisfy the given conditions. Include a sketch of the set 8. Also state Whether the set is open, closed, or neither. Please include a little bit of explanation, but a proof is not expected. (a) wZOandy<0. (13):)32—l—y2 <1andm+y>‘1. (C) lml + lyl S 1. Solution. (a) The sketch is The boundary of S is for if (90,34) is such a point, then every disk centered at (any) contains both points in S and points not in S. The set is not open, since it contains boundary points, for example, (0, ~2). The set S is not closed, since the complement of .S’ is not open. (Equivalently, S is not closed since it does not contain its boundary.) (b) The sketch is The boundary is 5' is open. S is not closed. (c) The sketch is The boundary is S is not open. 8 is closed. Problem (Extra problem 2). (a) For a any constant, define the subset 1‘0 of R2 by To : {(53,211) I y = ((12 ’_ C)2}' Sketch these subsets I‘c for various values of c. (b) Now suppose that f (m,y) is a real valued function of 2 variables. For each constant k, the level curves Lk are defined as usual by ‘ Lk : {(m,y) l f(a7ay) '2 k}' Is there any choice of function f (as, 3;) so that for all constants 0, each of the sets PC in part (a) is actually a level curve Lk of f for some value of k? Can you come up with a property of level curves that makes the answer to this question clear? Solution. (a) The curves To for c = ~2, —1,0,1,2 are: n,” C=-( , C” 2 . r . \ , AA > (b) One striking difference between the sketches in (a) and all of the level curves we have sketched is that the curves in (a) intersect one another, while level curves seem not to. In fact, level curves f (m,y) : cl and f(m,y) = 02 with 01 7é c2 cannot intersect, for this would mean that the function has tWo heights above the intersection point, Therefore the curves PC are not the level curves of a function, El ...
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