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where k = absorption rate coetlicient. The principal debits 1 associated with such an operation might be losses of valuable material in the exit gas and costs of processing the absorbent: 1 = vlFy i- VZL where v1 = product value v2 = processing cost The debit equation may be rewritten on the basis of independent variabIes alone : 1 = vlkx; + VZL The minimum point on a curve of debits vs. L would occur where the slope db/dL is zero. dl -= -I&~ + 212 = 0 dL L2 This defines the locus of optimum L. L, ,, = VlkZ F2 v,

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486 Worked-Out Solutions to Exercises: 1-9

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7 To find the polar negative of the vector we derived in the solution to Problem 6, we reverse the direction but leave the magnitude the same In this situation, 0 qc < p, so we should add p to the angle to reverse the direction That gives us the exact answer as -(a + b) = {[p + p + Arctan ( 4/3)],5} = {[2p + Arctan ( 4/3)],5} If we say that 2p 62832, then we can approximate the angle to four decimal places and define the vector as -(a + b) (53559,5) 8 The original two vectors are a = (p /2,4) and b = (p,3) To find the polar negatives, we reverse the directions but leave the magnitudes the same We want to keep the angles less than 2p without letting either of them become negative In this case, that means we should add p to qa, but we should subtract p from qb When we make these changes, we get -a = (3p /2,4) and -b = (0,3) We must be careful to avoid confusion about what the coordinates of b actually mean The first entry in the ordered pair is an angle, while the second entry is a radius 9 This time, we want to find the polar sum of the vectors -a = (3p /2,4) and -b = (0,3) Converting them to Cartesian form, we get -a = {[4 cos (3p /2)],[4 sin (3p /2)]} = {(4 0),[ 4 ( 1)]} = (0, 4)

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4

The operations support TT member learns the system and how to install it, does this work during the pilot test, and trains the rest of operations support in how to maintain the system If the work of the transition team goes well, then the project team does not need to have any role in providing routine support as of the go-live date The bene t of a smooth transition period is that the system is supported by the production operations sta and help desk as soon as it is implemented The users do not identify the project manager or project team as the source of system support Instead, they call the same help desk that they call for all other systems I have spoken to many project managers in organizations where this is not done.

and -b = [(3 cos 0),(3 sin 0)] = [(3 1),(3 0)] = (3,0) Adding, we get the Cartesian vector sum -a + (-b) = [(0 + 3),( 4 + 0)] = (3, 4) Let s call this Cartesian sum vector d = (xd,yd ) We have xd = 3 and yd = 4 This is in the fourth quadrant of the Cartesian plane We seek the polar sum vector d = (qd,rd ), where qd is the direction angle of d and rd is the magnitude of d Using the applicable angle-conversion formula, we get qd = 2p + Arctan ( 4/3) The formula for the polar magnitude tells us that rd = (xd 2 + yd 2)1/2 = [32 + ( 4)2]1/2 = (9 + 16)1/2 = 251/2 = 5 This gives us the ordered pair d = -a + (-b) = (qd,rd ) = {[2p + Arctan ( 4/3)],5} This is precisely the same vector that we got when we solved Problem 7 Now we know that in the specific polar-vector case where a = (p /2,4) and b = (p,3) the following formula holds: -(a + b) = -a + (-b) Of course, demonstrating this single example doesn t prove the general case We know it works in general for Cartesian vectors If you re ambitious and would like some extra credit, go ahead and rigorously prove that polar vector negation always distributes through polar vector addition You re on your own!

488 Worked-Out Solutions to Exercises: 1-9

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