MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable CalculusFall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.18.02 Lecture 8. – Tue, Sept 25, 2007 Functions of several variables. Recall: for a function of 1 variable, we can plot its graph, and the derivative is the slope of the tangent line to the graph. Plotting graphs of functions of 2 variables: examples z = −y, z = 1 − x2 − y2, using slices by the coordinate planes. (derived carefully). Contour plot: level curves f(x, y) = c. Amounts to slicing the graph by horizontal planes z = c. Showed 2 examples from “real life”: a topographical map, and a temperature map, then did the examples z = −y and z = 1 − x2 − y2 . Showed more examples of computer plots (z = x2 + y2 , z = y2 − x2, and another one). Contour plot gives some qualitative info about how f varies when we change x, y. (shown an example where increasing x leads f to increase). Partial derivatives. fx = ∂f = lim f(x0 + Δx, y0) − f(x0, y0); same for fy. ∂x Δx 0 Δx→Geometric interpretation: fx, fy are slope s of tangent lines of vertical slices of the graph of f (fixing y = y0; fixing x = x0). How to compute: treat x as variable, y as constant. Example: f(x, y) = x3y + y2, then fx = 3x2y, fy = x3 + 2y. 18.02 Lecture 9. – Thu, Sept 27, 2007 Handouts: PS3 solutions, PS4. Linear approximation Interpretation of fx, fy as slopes of slices of the graph by planes parallel to xz and yz planes. Linear approximation formula: Δf ≈ fxΔx + fyΔy. Justification: fx and fy give slopes of two lines tangent to the graph: y = y0, z = z0 + fx(x0, y0)(x − x0) and x = x0, z = z0 + fy(x0, y0)(y − y0). We can use this to get the equation of the tangent plane to the graph: z = z0 + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0). Approximation formula = the graph is close to its tangent plane. Min/max problems. At a local max or min, fx = 0 and fy = 0 (since (x0, y0) is a local max or min of the slice). Because 2 lines determine tangent plane, this is enough to ensure that tangent plane is horizontal (approximation formula: Δf � 0, or rather, |Δf | � |Δx|, |Δy|). Def of critical point: (x0, y0) where fx = 0 and fy = 0. A critical point may be a local min, local max, or saddle. Example: f(x, y) = x2 − 2xy + 3y2 + 2x − 2y. Critical point: fx = 2x − 2y + 2 = 0, fy = −2x + 6y − 2 = 0, gives (x0, y0) = (−1, 0) (only one critical point). 1� �� � � 2 Is it a max, min or saddle? (pictures shown of each type ). Systematic answer: next lecture. For today: observe f = (x − y)2 + 2y2 + 2x − 2y = (x − y + 1)2 + 2y2 − 1 ≥ −1, so minimum. Least squares. Set up problem: experimental data (xi, yi) (i = 1, . . . , n), want to find a best-fit line y = ax + b (the unknowns here are a, b, not x, y!) Deviations: yi − (axi + b); want to minimize the total square deviation D = (yi − (axi + b))2 .i∂D ∂D = 0 and = 0 leads to a 2 × 2 linear sys tem for a and b (done in detail as in Notes LS): ∂a ∂b ��� ��� � xi 2 a + xi b = xiyi xi a + nb = yi setup also works in other cases: e.g. exponential laws y = ceax (taking logarithms: ln y = ln c + ax, so setting b = ln c we reduce to linear case); or quadratic laws y = ax2 + bx + c (minimizing total square deviation leads to a 3 × 3 linear system for a, b, c). Example: Moore’s Law (number of transistors on a computer chip increases exponentially with time): showed interpolation line on a log plot. 18.02 Lecture 10. – Fri, Sept 28, 2007 Second derivative test. Recall critical points can be local min (w = x2 + y2), local max (w = −x2 − y2), saddle (w = y2 − x2); slides shown of each type . Goal: determine type of a critical point, and find the global min/max. Note: global min/max may be either at a critical point, or on the boundary of the domain/at infinity. We start with the case of w = ax2 + bxy + cy2, at (0, 0). Example from Tuesday: w = x2 −2xy+3y2: completing the square, w = (x−y)2 +2y2, minimum. b b b2 1 bIf a �= 0, then w = a(x 2+ axy)+cy 2 = a(x+2ay)2+(c− 4a )y 2 =4a (4a 2(x+2ay)2+(4ac−b2)y 2). 3 cases: if 4ac − b2 > 0, same signs, if a > 0 then minimum, if a < 0 then maximum; if 4ac − b2 < 0, opposite signs, saddle; if 4ac − b2 = 0, degenerate case. This is related to the quadratic formula: w = y 2� a(x)2 + b(x) + c � . y y If b2 −4ac < 0 then no roots, so at2 +bt+c has a constant sign, and w is either always nonnegative or always nonpositive (min or max). If b2 − 4ac > 0 then at2 + bt + c crosses zero and changes sign, so w can have both signs, saddle. General case: second derivative test. ∂2fWe look at second derivatives: fxx = , fxy, fyx, fyy. Fact: fxy = fyx. ∂x2 Given f and a critical point (x0, y0), set A = fxx(x0, y0), B = fxy(x0, y0), C = fyy(x0, y0), then: – if AC − B2 > 0 then: if A > 0 (or C), local min; if A < 0, local m ax. – if AC − B2 < 0 then saddle. Least-squares3 – if AC − B2 = 0 then can’t conclude. Checked quadratic case (fxx = 2a = A, fxy = b = B, fyy = 2c = C, then AC − B2 = 4ac − b2). General justification: quadratic approximation formula (Taylor series at order 2): Δf � fx (x − x0) + fy (y − y0) + 1 fxx (x − x0)2 + fxy (x − x0)(y − y0) + 1 fyy (y − y0)2 .2 2 At a critical point, Δf � A (x − x0)2 + B(x − x0)(y − y0) + C (y − y0)2 . In degenerate case, would 2 2 need higher order derivatives to conclude. NOTE: the global min/max of a function is not necessarily at a critical point! Need to check boundary / infinity. 1Example: f(x, y) = x + y + xy , for x > 0, y > 0. fx = 1 − 1 2= 0, fy = 1 − 1 2 = 0. So x2y = 1, …
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