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207 MATH Spring 2011 Total possible is 50 Dr. Smith Assignment 3 1. Find the limit if it exists, or show that the limit does not exist. x4 - y4 a) (14.2 #14) lim( x , y ) ( 0,0 ) 2 x + y2 b) (14.2 #16) lim( x , y ) ( 0, 0) x 2 sin 2 y x2 + 2 y 2 Four points c) (14.2 # 20) lim( x , y , z ) ( 0,0, 0) x 2 + 2 y 2 + 3z 2 x2 + y 2 + z 2 Four points 2. Determine the set of points at which the function is continuous. x- y a) (14.2 # 30) F ( x, y ) = 1 + x2 + y2 b) (14.2 #32) f ( x, y ) = e x y + x + y 2 2 Four points 3. (14.2 # 40) Use polar coordinates to find lim( x , y ) ( 0,0 ) ( x 2 + y 2 ) ln( x 2 + y 2 ) . 4. Find all first partial derivatives for a) (14.3 #22) f(x,y) = xy Four points (2 for each partial) b) (14.3 #26) f(x,t) = arctan(x t) c) (14.3 #38) u = sin(x1 + 2x2 + ... + nxn) Four points 5. a) (14.3 #42). Find the partial derivative fz(0,0,/4) if f ( x, y, z ) = sin 2 x + sin 2 y + sin 2 z Since f is a function of z, z is an independent variable. 2 sin z cos z sin z cos z = fx = 2 2 2 2 2 sin x + sin y + sin z sin x + sin 2 y + sin 2 z b) (14.3 #46). Use implicit differentiation to find z z and if yz = ln(x+z). x y Six points (3 each partial derivative) 6a) (14.3 #58). Verify that Clairaut's Theorem holds for u = x4y2 2xy5 2u u = 8 x3 y - 10 y 4 This is a polynomial, and so continuous, = 4 x 3 y 2 - 2 y 5 and yx x 2u u = 8 x3 y - 10 y 4 This is a polynomial, and so continuous. = 2 x 4 y - 10 xy 4 and xy y Clairaut's Theorem says that if the second cross derivatives are continuous, they are equal. Indeed, these are each continuous and we see they are equal. Six points (2 each derivative partial and 2 for noting polynomials are continuous) b) (14.3 #71). Verify that the function u = Laplace equation uxx + uyy + uzz = 0 1 x2 + y 2 + z 2 is a solution of the three-dimensional 7. Find an equation of the tangent plane to the given surface at the specified point. a) (14.4 #2) z = 3(x 1)2 + 2(y + 3)2 + 7, at (2, 2, 12) fx(2, 2) = 2*3(x 1)|x=2 = 6 and fy(2, 2) = 2*2(y+3)|y=-2 = 4 z = 6(x 2) + 4(y + 2) + 12 = 6x + 4y 12 + 8 + 12 = 6x + 4y + 8 b) (14.4 #4) z = y ln x at (1, 4, 0) fx(1, 4) = y/x|x,y=1,4 = 4 and fy(1,4) = ln x|x,y=1,4 = 0 z = 4(x 1) + 0(y 4) + 0 = 4x 4 c) (14.4 #6) z = e x - y at (1, 1, 1) 2 2 2 2 fx(1, 1) = 2 xe x - y |x , y =1, -1 = 2e0 = 2 and fy(1, 1) = - 2 ye x - y |x , y =1, -1 = -2(-1)e 0 = 2 z = 2(x 1) + 2(y + 1) + 1 = 2x + 2y 2 + 2 + 1 = 2x + 2y + 1 Six points (2 each partial derivative, and 2 for the final answer) 2 2 8. (14.4 #16) Explain why the function f(x,y) = sin(2x + 3y) is differentiable at the point (3,2). Then find the linearization L(x,y) at (3,2). Six points (1 each partial derivative, 1 for noting continuity, 3 for final answer) 9. (14.4 #28) Find the differential of the function T = dT = Tu du + Tv dv + Tw dw = v . 1 + uvw 1 - v2w uv 2 dw du + dv - (1 + uvw) 2 (1 + uvw) 2 (1 + uvw) 2 Six points (2 each partial derivative) 10. (14.4 #36) Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1 cm thick and the metal in the sides is 0.05 cm thick. (dh = 0.2 because the height includes the change in both top and bottom.) ... View Full Document