18 02A Topic 26 Tangent plane approximation directional derivatives Read TB 19 1 19 2 SN TA From last time we have the tangent plane approximation x x x0 y y y0 w f x y f x0 y0 w w x w y x 0 y z 0 The supplemental notes TA give an analytic argument for this and its genaralization to 3 dimensions Suppose w f x y z and x x x0 y y y0 z z z0 then x w y w z w f x y f x0 y0 w x 0 y z 0 0 Gradient w w rate of change in bi direction x 0 y w w gradient of w w x 0 y 0 K K K K K K K K K K K K K K KK K K K KK K K K KKK x y rate of change in bj direction 0 To evaluate at P0 we write w P0 Will use this in a moment w Directional derivative b and a point P0 in the plane Fix a direction u b is The directional derivative of w at P0 in the direction u defined as dw ds SSSS SSSS w w s 0 s dw ds P0 b u s b w P0 u x Proof The tangent plane approximation and the bottom picture at right show w s w x x 0 s w y 0 y s w x 0 cos S S S s Here w is the change in w caused by a step of length s in b all in the xy plane the direction of u It is a fact that S S S P0 lim P0 b u w y sin 0 b hcos sin i since it is a unit vector Thus the last But u b In the limit the approximations become formula is just w u exact and we get the boxed equation QED continued 1 y s y P0 x y RRRR RRR b u b hcos sin i u x 18 02A topic 26 2 Example Algebraic example Let w x3 3y 2 dw Compute at P0 1 2 in the direction of v 3i 4k ds i w h3x2 3y 2 i w 1 2 h15 12i 15 i 12 j v 3 4 b ii u i j v 5 5 3 4 93 dw b 15 i 12 j i j w 1 2 u iii ds P0 bu 4 5 5 b be the direction of h1 1i Example Geometric example Let u w w dw Using the picture at right estimate and x P y p ds P bu y By measuring from P to the next in level curve in the x direction we see that x 5 w 10 w 20 x P x 5 1 w 20 Similarly we get y P w 25w 15 w 5 P o y O x b u Measuring in the u direction we get s 3 dw ds P b u x 1 w 10 33 3 s 3 Direction of maximum change b u w Proof angle 0 w The Gradient is perpendicular to level curves w level curve surface w c Example Consider the graph of y ex Find a vector perpindicular to the tangent to at the point 1 e u b P0 w Old method Find the slope take the negative reciprocal and make the vector New method This graph is the level curve of w y ex with w 0 w h ex 1i normal w 1 e h e 1i b is tangent to the level curve at P0 then proof If u since w is constant along the level curve i e continued dw ds dw ds P0 b u 0 P0 b u b 0 w P0 u QED 18 02A topic 26 3 Example Find the tangent plane to the surface x2 2y 2 3z 2 6 at the point P 1 1 1 Introduce a new variable w x2 2y 2 3z 2 6 Our surface is the level surface w 6 normal to surface is w h2x 4y 6zi At the point P we have w P h2 4 6i Using point normal form the equation of the tangent plane is 2 x 1 4 y 1 6 z 1 0 2x 4y 6z 12
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