practice4as

practice4as - SOLUTIONS TO 18.02 PRACTICE MIDTERM #4A BJORN...

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SOLUTIONS TO 18.02 PRACTICE MIDTERM #4A BJORN POONEN 1) For each of (a)-(c) below: If the statement is true, write TRUE. If the statement is false, write FALSE. (Please do not use the abbreviations T and F.) No explanations are required in this problem. (a) If C is a piecewise smooth curve in R 3 from ( a 1 , a 2 , a 3 ) to ( b 1 , b 2 , b 3 ), then R C y dx = b 1 b 2 - a 1 a 2 . Solution. FALSE. In fact, the value of R C y dx depends on more than just the endpoints of C , because y dx is not an exact differential. To see this, notice that curl h y, 0 , 0 i = ± ± ± ± ± ± i j k ∂x ∂y ∂z y 0 0 ± ± ± ± ± ± = - k 6 = 0 . ± (b) If f ( x, y, z ) is a function with continuous second partial derivatives on R 3 , then div( f ) = 0 at every point. Solution. FALSE. By definition, div( f ) = ² ∂x , ∂y , ∂z ³ · h f x , f y , f z i = f xx + f yy + f zz , but this need not be 0. For example, if f ( x, y, z ) := x 2 , then we get div( f ) = 2. (In fact, div( f ) = ∇ · ∇ f =: 2 f and 2 is called the Laplace operator .) ± 2) Let F = (cos x + 2 y 2 + 5 yz ) i + (4 xy + 5 xz ) j + (5 xy + 3 z 2 ) k on R 3 . Show that F is conservative, and find a potential function for F . For full credit, use a systematic method (not just guessing), and show your work. Solution.
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practice4as - SOLUTIONS TO 18.02 PRACTICE MIDTERM #4A BJORN...

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