Gradients Directional Derivatives

Gradients Directional Derivatives - Gradients...

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Unformatted text preview: Gradients http://wwwmautexas.edu/users/gilbert/aseiimath/‘Gradientshuni GRADEENTS, DIRECTIGNAL DERIVATiVES So you are back in Yosemite and feel like doing a hit of serious hiking. You've puiied up the contour map to the right on your iPhone and iocatect yourself at the black dot on the 9200' contour. One natural question to ask is for the slope of the terrain from where you in the particular direction you set off to walk. lfyou are feeling particularly energetic, you might even try to take the steepest path to the top of the mountain — we might call that the Path of Steepest Ascent‘. This will mean that at each point on your path you always take the direction of greatest slope. But if you are on the 9200‘ contour, that direction of greatest slope must he perpendicular to the contour, hecattse if you walk uphill in a direction that is not perpendicular, then you are 'cutting across‘ the mountain and not climbing uphill as fast as you can. Now, inathematicaliy, the terrain is the graph of z m f(x, y) and if (x(t), y(r)) is a curve in the xy~plane, then the graph of 1'0) '3 {36(1), W), 1'06th ND) is a space curve lying on the graph off. it’s the mathematical way of describing a path in Yosemite, say. Different curves provide different information. But first we make a crucial For a function f(x, )2) the GRADIENT off is the vector fimction Vftxa y) = Ji-(xa 30% 4" fvtxa JOi whose respective components are )3 and fi,. For example, when fix, y) = x2 “ yg, then (V f)(x, y) m 23: i~2 y j‘ Composition suggests the Chain Rule: 1,,321935 21952;. +. drftattltht‘D 5:: dr a}; dr 360).); NW}, which in turn suggests the dot product. For if we set (2(1) = (x(t), y(t)>, then c'(r) e" (36(1), y'(t)> and lof4 11/30/2009 4:57 PM Gradients 20M http:/r’www.ma.utexas.edu/users/gilbert/asciimath/Gradients.html d d :17"; f (0(1)) =Ef{X(f), M0} = xiiiflty'fl'm ” (V fXCUJ) ' 5(0- An example {iom a familiar setting wilt help. W WWWWWMWWNWWWWWMMWWWWMWWWWWMW Mum ‘ CURVE, 1 : set flx, y) = 2 + x2 w )9, C(t') 2 (cos t, sin z). The graph of c is a circle in the xy-plane, while the graph of f(c(t)) : f(cos 2‘, sin I) = 2 -+ 0052: is the intersection of a hyperboiic paraboloid and a circular cylinder as shown to the right. Now 61 ‘ . dt f(c(t)) ~ 2 $112!, while 62’0“} = (m sin t, cos i), V _f'{c{t)) : 2 cost i *2 sinrj. But then (V fl(c(t)) - c'(f) : ~— 4 cos tsint‘ = *2 seat, Wmfiwmmmmflmmmmwmhfifimm mm WWW-“WW J confirming that 31W!» = (VfXCGD - 6(2‘). Now the tangent vector to the space curve can be calcuiated: d 1"(1‘) = (X'tl'), fir), g; f (360’), ya)» = (XV), y'U), (V f)(C(I)) ' 6(0)}. It is shown in orange in the previous graphic. w. W WWE,WMM~WW- h. i. M...“ w A r m...» finition: wémfénéifie For a UNIT vector v "4* h i +kj the DIRECTIONAL SERIVATIVE of 2: m f (x, y) at (a, b) in the direction v is i- :11, b ~4~ fk} —- flat, 1)) va(a, b) 313315 ”I" r Of course, D f 1051, b) = jjia, b) and D f flex, 5)) m 13,027, b). Since HBO/2009 4:57 PM Gradients http:l/www.ma.utexas.edu/users/gtlbert/ascimath/Gradients.html "l" ., 13+ k ~ ,5 vamg 13) 3 E133 W = g;f(a+fl7; 33.14% {:0 ’ this suggests what the next choice of curve should be. m : CURVE 2. Let wwwmmwwwm C(f) = (a + th)i + (b +1k)j be the line through (a, b} in the direction ofa vector v = hi +1: j. When th ‘ surface to the right is the graph off, the graph of flew) = flaw/h, b+tk) : is the curve shown in orange on the surface. On the other hand, C(0) = (a, 5), 6(0) = h i +kj m v, 50 mmammwwww‘ ny(a, b) = Vfla, b) - v. g This tells us how to compute all directional derivatives D f v(a, b). Finally, we can see what the gradient vector has to do with contours. What's a contour? Well, its a curve c (t) = (x(t), 32(1)) in the xy—plane such that f {c(t)) m k where k is some constant. The graph to the Iefi: below shows one contour, while the one to the right shows two contours realizeci as curves in the Jay—plane. 3 of4 11/30/2009 4:57 PM Gradients http://www.ma.utexas.edtflusers/gllbert/asciimatla/Gradéents.html But if we differentiate the equation f (0(1)) = k, then (1 . £11600) “ (VfXCUD’c'U) “1 0- N'ow e‘{t) is the tangent vector to the contour, so the fact that (V f)(c(t) - 0'0) e 0 means that the V f is eerpendieular at o (r) to the contour. That's what the right hand graph shows. But why are some vectors longer than others? Look on the graph where those short vectors occur, it’s close to the y—axis where the surface isn‘t vezy steep, whereas the vectors get longer and longer the more we move along the contour away from the y—axis. But the surface gets steeper anti steeper as one moves away from the y—axjs. In other words the slope increases! 4 of4 11/30/2009 4:57 PM ...
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