Problem Set 1 Solution

# Thus we have d a dt a d a r rr dt

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Unformatted text preview: from which d d !! !r # r" \$ b !r # r" dt dt ! ! r ( r rr ) * r ' r 2 + dt % r \$ C , - (2) ! !r 2ar # r \$ 2br # !! % a (1) & (3) Clearly, 2 !2 ! ! where C is the integration constant # a \$ ecbr # )." dt % ar \$ br \$ const. & ! 2ar (r v 2 tor!r! (2) 1-34. F nce 1-33. Siirst, we note that d d ( r ) ! rr ' rr ! r rr !! ! !! ! A . A !% A . A \$ A . A dt t * r + % r 2 % r ' r 2 d,But the first term on the right-hand side vanishes. Thus, we have d ! ! !! ! ( . A dt A )d & &A r ' rr ) % &%dt d (.rAdt t dt & * + 2+ *r r dt , r , so that from which !! ! & A(. A rdt)% A .rA \$ C ! ! r r ' 2 + dt % \$ C & *r r - r , MATRICES, VECTORS, AND VECTOR CALCULUS where C is a constant vector. ! ! " ! " ! " " (1) (1) (2) (2) (3) (3) 27 where C is the integration constant (a vector). 1-38. 1-34. z First, we note that z = 1 – x2 – y2 d ! !! !! A.A %A.A\$A.A dt ! " (1) But the first term on the right-hand side vanishes. Thus, y d A ! + y =! & ! A . x!! " dt % & dt xA . A1" dt (2) A A ' !# !! % a & . ! % C & ! A .\$A" "dtd% A ' A \$ds (1) (3) C 2 2 By Stoke’s theorem, we have so that S C T here C is a constant vector. whe curve C that encloses our surface S is the unit circle that lies in the xy plane. Since the element of area on the surface da is chosen to be outward from the origin, the curve is directed counterclockwise, as required by the right-hand rule. Now change to polar coordinates, so that we have ds & d( e( and A & sin ( i ) cos ( k on the curve. Since e( % i & * sin ( and e( % k & 0 , we have ' C A % ds & ' 2+ 0 ! * sin ( " d( & *+ 2 (2) 1-39. a) Let’s denote A = (1,0,0); B = (0,2,0); C = (0,0,3). Then AB & (*1, 2, 0) ; AC & (*1, 0 , 3) ; and AB \$ AC & (6 , 3, 2) . Any vector perpendicular to plane (ABC) must be parallel to AB \$ AC , so the unit vector perpendicular to plane (ABC) is n & ( 6 7 , 3 7 , 2 7 ) b) Let’s denote D = (1,1,1) and H = (x,y,z) be the point on plane (ABC) closest to H. Then DH & ( x * 1, y * 1, z * 1) is parallel to n given in a); this means x *1 6 & &2 y*1 3 and x *1 6 & &3 z*1 2 Further, AH & ( x * 1, y , z) is perpendicular to n so one has 6( x * 1) ) 3 y ) 2 z & 0 . C 0 1-39. a) Let’s denote A = (1,0,0); B = (0,2,0); C = (0,0,3). Then AB & (*1, 2, 0) ; AC & (*1, 0 , 3) ; and AB \$ AC & (6 , 3, 2) . Any vector perpendicular to plane (ABC) must be parallel to AB \$ AC , so the unit vector perpendicular to plane (ABC) is n & ( 6 7 , 3 7 , 2 7 ) b) Let’s denote D = (1,1,1) and H = (x,y,z) be the point on plane (ABC) closest to H. Then DH & ( x * 1, y * 1, z * 1) is parallel to n given in a); this means x *1 6 & &2 y*1 3 and x *1 6 & &3 z*1 2 Further, AH & ( x * 1, y , z) is perpendicular to n so one has 6( x * 1) ) 3 y ) 2 z & 0 . Solving these 3 equations one finds H & ( x , y , z) & (19 49 , 34 49 , 39 49) and DH & 5 7 1-40. a) At the top of the hill, z is maximum; 0& ,z & 2 y * 6 x * 18 ,x and 0& ,z & 2 x * 8 y ) 28 ,y...
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## This note was uploaded on 12/14/2013 for the course PHYS 301 taught by Professor Argryes during the Fall '13 term at University of Cincinnati.

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