testFw06solns

testFw06solns - Math 214 Calculus III Final Exam Solutions...

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Unformatted text preview: Math 214, Calculus III Final Exam Solutions 1. (25 points) Suppose we know that for a certain function f ( x,y ), f (3 , 4) = 25 , f x (3 , 4) = 6 , f y (3 , 4) = 8 , and f (4 , 5) = 41 . (a) Find a linear function L ( x,y ) that approximates f as well as possible near (3 , 4). Solution: The linear function L that best approximates f near (3 , 4) is the linearization , L ( x,y ) = f (3 , 4)+[ f x (3 , 4)]( x- 3)+[ f y (3 , 4)]( y- 4) = [25]+[6]( x- 3)+[8]( y- 4) = 6 x +8 y- 25 . That is, z = 6 x +8 y- 25 is the equation of the tangent plane to z = f ( x,y ) at the point (3 , 4 , 25). (b) Use L to estimate f (2 . 9 , 3 . 9), f (3 . 1 , 4 . 1) and f (4 , 5). Solution: Using the linear function L ( x,y ) = 6 x + 8 y- 25 above as an approximation of f ( x,y ), we estimate f (2 . 9 , 3 . 9) ≈ L (2 . 9 , 3 . 9) = 25 + 6([2 . 9]- 3) + 8([3 . 9]- 4) = 25- . 6- . 8 = 23 . 6 , f (3 . 1 , 4 . 1) ≈ L (3 . 1 , 4 . 1) = 25 + 6([3 . 1]- 3) + 8([4 . 1]- 4) = 25 + 0 . 6 + 0 . 8 = 26 . 4 , f (4 , 5) ≈ L (4 , 5) = 25 + 6([4]- 3) + 8([5]- 4) = 25 + 6 + 8 = 39 . (c) Could f itself be a linear function? Why or why not? Solution: If f is a linear function, then we must have f = L , since L is the linear function that is the best approximation of f . However, f (4 , 5) = 41 while L (4 , 5) = 39 implies f 6 = L . Thus, we conclude that f is not a linear function . 2. (25 points) Consider the function f ( x,y ) = x 2 + y 2 + Cxy where C is some constant. (a) Show that (0 , 0) is a critical point of f . Solution: The point (0 , 0) will be a critical point of f provided f x (0 , 0) = f y (0 , 0) = 0. So consider f x ( x,y ) = 2 x + Cy and f y ( x,y ) = 2 y + Cx. Then, no matter what the value of C happens to be, we see that f x (0 , 0) = 2[0] + C [0] = 0 and f y (0 , 0) = 2[0] + C [0] = 0 . Therefore, (0 , 0) is a critical point of f . (b) For what values of C , if any, does f have a local minimum at (0 , 0)? Solution: To determine whether f has a local max, min or saddle point at any of its critical points, we employ the Second Derivative Test . So, to get things started, we compute the second partial derivatives of f : f xx ( x,y ) = 2 , f xy ( x,y ) = C = f yx ( x,y ) , and f yy ( x,y ) = 2 . Thus the discriminant of f is D ( x,y ) = f xx ( x,y ) f yy ( x,y )- f xy ( x,y ) f yx ( x,y ) = [2][2]- [ C ][ C ] = 4- C 2 Math 214, Calculus III Final Exam Solutions for all points ( x,y ). Recall that if D ( x,y ) > 0, then all of the concavities of f at ( x,y ) agree (i.e., the graph of f is either concave up in all directions at ( x,y ) or it is concave down in all directions at ( x,y )); if D ( x,y ) < 0, then there are at least two directions in one of which the graph of f is concave up and in the other it is concave down (i.e., the concavities don’t all agree at ( x,y )); if D ( x,y ) = 0, then the Second Derivative Test doesn’t know what to do....
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testFw06solns - Math 214 Calculus III Final Exam Solutions...

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