Exam 2 Solution on Calculus III

Using classical curly d notation since x and z are

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using classical curly d notation, since x and z are independent variables and y is a function of x and z, we have 0 F ( x , y , z ) z F x x z F y y z F z z z 0 F y y z F z y z F / z F / y . Second, if we use subscript notation for the partial derivatives, we may set g(x,z) = F(x, h(x,z), z). Then since x/ z = 0, 0 = g z (x,z) = F y (x, h(x,z), z)h z (x, z) + F z (x, h(x,z), z), implying y/ z = h z (x,z) = -F z (x, h(x,z), z)/F y (x, h(x,z), z). (b) Is f defined by f( x , y ) sin(5( x 2 y 2 )) x 2 y 2 , ( x , y ) (0,0) 5 , ( x , y ) (0,0) continuous at (0,0)? A complete explanation is required. Details & definitions are at the heart of it.// Since f(0,0) = 5 and lim ( x , y ) (0,0) f( x , y ) lim ( x , y ) (0,0) sin(5( x 2 y 2 )) x 2 y 2 lim r 0 sin(5 r 2 ) r 2 ( L H ) lim r 0 10 r cos(5 r 2 ) 2 r 5 by using L’Hopital’s rule after doing a conversion to polar coordinates, f is continuous at (0,0). _________________________________________________________________ 10. (10 pts.) Obtain and locate the absolute extrema of the function f(x,y) = (x - y) 2 on the closed disk D with radius five centered at the origin. Observe that D is given in set builder notation as follows: D = { (x,y) : x 2 + y 2 25 }. In performing this magic, use Lagrange multipliers to deal with the behavior of f on the boundary, B = { (x,y) : x 2 + y 2 = 25 }. // Since f x (x,y) = 2(x - y) and f y (x,y) = -2(x - y), the critical points of f are the points of the form (x,x) in the interior of B. There we have f(x,x) = 0, the obvious minimum. Set g(x,y) = x 2 + y 2 - 25. Then (x,y) is on the circle defined by x 2 + y 2 = 25 precisely when (x,y) satisfies g(x,y) = 0. Since g(x,y) = <2x, 2y>, g(x,y) <0, 0> when (x,y) is on the circle given by g(x,y) = 0. Plainly f and g are smooth enough to satisfy the hypotheses of the Lagrange Multiplier Theorem. Thus, if a constrained local extremum occurs at (x,y), there is a number λ so that f(x,y) = λ∇ g(x,y). Now f(x,y) = λ∇ g(x,y) <2(x-y),-2(x-y)> = λ <2x,2y> x 2 -y 2 = 0 x = y or x = -y by performing a little routine algebraic magic. Solving each of the systems consisting of (a) x 2 + y 2 = 25 and x = y , and (b) x 2 + y 2 = 25 and x = -y , yields (5/2 1/2 ,5/2 1/2 ), (-5/2 1/2 ,-5/2 1/2 ), (5/2 1/2 ,-5/2 1/2 ), and (-5/2 1/2 ,5/2 1/2 ). It turns out f is zero, the minimum value, at the first two points, and f is 50, the maximum value, at the last two points. • Fall '06
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