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2. Partial Differentiation Solutions

# 2. Partial Differentiation Solutions - 2 Partial...

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- - 2. Partial Differentiation 2A. Functions and Partial Derivatives 2A-1 In the pictures below, not d l of the level curves are labeled. In (c) and (d), the picture is the same, but the labelings are different. In more detail: b) the origin is the level curve 0; the other two unlabeled level curves are .5 and 1.5; c) on the left, two level curves are labeled; the unlabeled ones are 2 and 3; the origin is the level curve 0; d) on the right, two level curves are labeled; the unlabeled ones are -1 and -2; the origin is the level curve 1; The crude sketches of the graph in the first octant are at the right. 2A-3 a) both sides are mnxm-' yn-' x-Y . -x b) f, = (x + yI2 ' fx, = (fx), = --- f, = - f y x = -(Y - x) (x + yI3 ' (x + yI2 ' (x + Y ) ~ c) f, = -2xsin(x2+y), f,, = (f,), = -2xcos(x2 +y); f, = -sin(x2+y), f,, = -cos(x2+y).2x. d) both sides are fl(x)g'(y). 2A-4 (f,), = ax + 6y, (f,), = 22 + 6y; therefore f,, = f,, H a = 2. By inspection, one sees that if a = 2, f (x, y) = x2y + 3xy2 is a function with the given f, and f,. 2A-5 a) w, = aeax sin ay, w,, = a2eax sin ay; W, = eaxa cos ay, w,, = eaxa2 (- sin ay); therefore wyy = -w,,. 22 2(y2 - x2) b) We have w, = - WXX = (x2 + y2)2 ' If we interchange x and y, the function x2 + y2 ' w = ln(x2 + y2) remains the same, while w,, gets turned into w,,; since the interchange just changes the sign of the right hand side, it follows that w,, = -w,,. 2B. Tangent Plane; Linear Approximation 2B-1 a) z, = y2, z, = 2xy; therefore at (1,1,1), we get z, = 1, z, = 2, so that the tangent plane is z = 1 + (x - 1) + 2(y - I), or z = x + 2y - 2.

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- - 1 2. PARTIAL DIFFERENTIATION b) w, = -y2/x2, W, = 2ylx; therefore at (1,2,4), we get w, = -4, w, = 4, so that the tangent plane is w = 4 - 4(x - 1) + 4(y - 2), or w = -4x + 4y. x x 2B-2 a) z, = - by symmetry (interchanging x and Y), z, = -; Y then the I / - z z xo Yo xo Yo tangentplaneisz=zo+-(x-xo)+-(y-yo), or z = --x+--Y ,sincexi+~:=z:. zo 20 zo zo b) The line is x = xot, y = yot, z = zot; substituting into the equations of the cone and the tangent plane, both are satisfied for all values of t; this shows the line lies on both the cone and tangent plane (this can also be seen geometrically). 2B-3 Letting x, y, z be respectively the lengths of the two legs and the hypotenuse, we have z = I / - thus the calculation of partial derivatives is the same as in 2B-2, and 3 4 7 we get Az M -Ax + -Ay. Taking Ax = Ay = .01, we get Az M -(.01) = .014. 5 5 5 2B-4 From the formula, we get R = R1R2 . From this we calculate R1+ R2 dR 2 , and by symmetry, 4 4 1 Substituting R1 = 1, R2 = 2 the approximation formula then gives AR = -ARl+ -AR2. 9 9 4 1 5 By hypothesis, lARil 5 .l, for i = 1,2, so that 1 ARl 5 -(.I) + -(.I) = -(.I) M .06; thus 9 9 9 n 2B-5 a) Wehave f ( x , y ) = ( ~ + y + 2 ) ~ , fX=2(x+y+2), fy=2(x+y+2). Therefore at (0,0), f,(O, 0) = f, (0,O) = 4, f (0,O) = 4; linearization is 4 + 42 + 4y; at (1,2), fx(l,2) = fy(l,2) = 10, f (l,2) = 25; linearization is 10(x - 1) + 10(y - 2) + 25, or lox + 10y - 5. linearization at (0,O): 1 + x; linearization at ( 0 , ~ / 2 ) :-y dV dV dV 2B-6 We have V = rr2h, - = 2 ~ r h , - " AT+ dr a h = r r 2 ; A V M ( ~ ) , ( = ) , ~ h . Evaluating the partials at T = 2, h = 3, we get Assuming the same accuracy lArl 5 6, [Ah1 < 6 for both measurements, we get 1 + 47r 1 r ~ = 1 6 ~ which is < .1 if 6 < 1- < .002 .
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2. Partial Differentiation Solutions - 2 Partial...

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