Final-Fall 2005-Solutions

Final-Fall 2005-Solutions - AMS 510.01 Fall 2005, Final...

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Unformatted text preview: AMS 510.01 Fall 2005, Final Exam Name: Student ID: Score: /200 The exam is in three parts: • The first section consists of problems involving the fundamental concepts of single and multiple variable calculus. Do all questions from this portion. (50 points) • The second section consists of comprehensive problems in Linear Algebra. Choose three out of the four problems to do. (75 points) • The third section consists of comprehensive problems in Advanced Calculus. Choose three out of the four problems to do. (75 points) Section 1. Fundamentals of Multivariate and Integral Calculus 1. Indicate whether each of the following statements is true or false. (2 points each) (a) The integral F ( x ) = R f ( x ) dx is unique for any given function f ( x ). • False. F ( x ) is only unique up to an arbitrary constant. (b) If two functions are equal everywhere in a closed interval ( f ( x ) = g ( x ) for all x in [ a, b ]), then their derivatives must be equal everywhere on the open interval ( a, b ). • True. However, the derivatives may not be equal (and may not even be defined) at the points x = a and x = b . (c) A necessary and sufficient condition for the continuity of a function f ( x, y ) at the point ( x o , y o ) is that: lim x → x o ( lim y → y o f ( x, y )) = lim y → y o ( lim x → x o f ( x, y )) • False. This is a necessary condition, but it is not sufficient. (d) R 1 R 1 xydxdy = 1. • False. R 1 R 1 xydxdy = R 1 ( R 1 xdx ) ydy = R 1 ( 1 2 ) ydy = 1 4 . (e) ∂ 2 f ∂x∂y = ∂ 2 f ∂y∂x = for any function f ( x, y ) with continuous second order derivatives. • True. This is a basic property of partial derivatives. 2. Consider the function f ( x ) = x cos x + sin xe cos x . (a) Find the indefinite integral F ( x ) = R f ( x ) dx . (5 points) • Break the integral into two terms: Z x cos x + sin xe cos x dx = Z x cos xdx + Z sin xe cos x dx The first term can be solved using integration by parts. Let u = x and v = sin x , then du = dx and dv = cos xdx . Thus: Z x cos xdx = Z udv = uv- Z vdu = x sin x- Z sin xdx = x sin x +cos x + c The second term can be solved by variable substitution. Let u = cos x , then du =- sin xdx . Thus: Z sin xe cos x dx =- Z e u du =- e u + c =- e cos x + c Combining the terms: F ( x ) = Z x cos x + sin xe cos x dx = x sin x + cos x- e cos x + c (b) Find the definite integral R π/ 2 f ( x ) dx . (5 points) • The definite integral is simply the difference of the indefinite integral, evalu- ated at each end point. Thus: R π/ 2 f ( x ) dx = F ( π 2 )- F (0) = ( π 2 ) sin π 2 ) + cos π 2- e cos π 2 )- (0 sin 0 + cos 0- e cos 0 ) = ( π 2 ) + 0- e )- (0 + 1- e 1 ) = π 2- 1- 1 + e = π 2- 2 + e 3. Consider the surface in R 4 defined by: f ( x, y, z ) = 3 x 2 + 4 xy + y 2 + cos(2 z ) (a) Evaluate the partial derivatives f x , f y and f z . (5 points) • Partial derivatives are found simply by taking the derivative with respect to one variable, holding all others fixed....
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This note was uploaded on 02/09/2010 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.

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Final-Fall 2005-Solutions - AMS 510.01 Fall 2005, Final...

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