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mat203finalsolutions_1

# mat203finalsolutions_1 - Please write out the honor pledge...

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Please write out the honor pledge and sign it: NAME (print): Instructor / class section: MAT 203 – Final January 13th, 2010 1:30-5:00PM Information Please read and sign the exam conditions first before turning the page: No books / notes / calculators / collaborations are allowed. The final has to be completed in a single time stretch of 210 minutes . There are 10 problems. Each problem is worth 20 points. The total score is 200 points. The first page of the exam lists useful facts . Be sure to read it. I have read these conditions and will follow them (initials): Score: Problem 1: points Problem 2: points Problem 3: points Problem 4: points Problem 5: points Problem 6: points Problem 7: points Problem 8: points Problem 9: points Problem 10: points Total score: points 1

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For Reference The following formulas may (or may not, depending on your solutions) be useful: 1. sin 3 ( x ) = 3 sin( x ) - sin(3 x ) 4 2. sin 5 ( x ) = 10 sin( x ) - 5 sin(3 x ) + sin(5 x ) 16 2
Problem 1 (20 points) Let f : R R be a differentiable function such that 0 f 0 ( x ) 1 for all x . Show that the flux of the vector field F ( x, y, z ) = ( f ( x )+ f ( y )+ f ( z ) , f ( x )+ f ( y )+ f ( z ) , f ( x )+ f ( y )+ f ( z )) over the unit sphere is at most 4 π . Solution Compute: ZZ S F · dS = ZZZ B ∇ · FdV = ZZZ B f 0 ( x ) + f 0 ( y ) + f 0 ( z ) dV 3 V ol ( B ) =4 π 3

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Problem 2 (20 points) Which of the following vector fields F are given by F = f for some function f ? Here f is assumed to have the same domain as F . If such an f exists, find it. Otherwise, show that no such f can exist. 1. ( 7pts ) F ( x, y, z ) = ( ye y , xe y + yxe y , e z + ze z ) 2. ( 7pts ) F ( x, y, z ) = ( xy + zx, x + z, xz + yz ) 3. ( 6pts ) F ( x, y, z ) = x - y x 2 + y 2 , x + y x 2 + y 2 , z 2 Solution 1. One can check that F is curl free, and hence is the gradient of some function. R F 1 dx = xye y , so call this function f 0 . Then ∂f ∂x = F 1 and ∂f ∂y = F 2 , but ∂f ∂z 6 = F 3 . Noticing that R e z + ze z dz = ze z , we find consider the new function f = f 0 + ze z . Then F = f . 2. ∂F 2 ∂x - ∂F 1 ∂y = 1 - x 2 xy + zx 6 = 0, so F has nonzero curl and hence is not a conservative vector field. 3. Away from the z -axis, ∇× F = 0. However, F is not the gradient of a function because it has nonzero circulation around, say, the unit circle C in the xy -plane: Z C F · ds = Z C x - y x 2 + y 2 dx + x + y x 2 + y 2 dy = Z C ( x - y ) dx + ( x + y ) dy = Z C x dx + y dy = Z 2 π 0 (cos 2 ( t ) + sin 2 ( t )) dt =2 π 4
Problem 3 (20 points) Let m ( u, v ) = f ( g ( u, v ) , h ( j ( u, v ) , k ( u, v ))). Find m uv = 2 m ∂v∂u .

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mat203finalsolutions_1 - Please write out the honor pledge...

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