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Earlier, I said that reading a book could be a project. Well, it is, if you set a deadline and set speci c learning goals. If you have no extra time beyond what it takes to read this book, then reading this book is your project. When you nish 3 make a list of speci c things you want to learn from reading the book. Make sure you pick tools and methods, not just ideas. In your project plan, write down what you will do di erently after reading the book. Then your project is to read the book, learn the tools, and start using them.

Another example Now let s find the midpoint between the origin and Q = (3,5, 2) This time, we let P = (0,0,0), so the midpoint M has the coordinates (xm,ym,zm) = [(xp + xq)/2,(yp + yq)/2,(zp + zq)/2]

= {(0 + 3)/2,(0 + 5)/2,[(0 + ( 2)]/2} = (3/2,5/2, 2/2) = (3/2,5/2, 1)

In the steady state, the velocity of flow in a pipe varies with pressure drop :

Here are some good ideas for projects for learning that some of my students have used. Passing the Project Management Professional exam and becoming a certi ed PMP. Building a small web site for a non-pro t organization. Cleaning out a closet, including putting in new hanging rods and shelves.

Still another example Now let s work out a tougher problem Suppose we want to find the coordinates of the midpoint M between P = ( 5, 4,3) and Q = (3,5, 2) Pairing off the coordinates for convenience, we have

xp = 5 and xq = 3 yp = 4 and yq = 5 zp = 3 and zq = 2 Plugging the values into the formula and working through the arithmetic, we obtain the coordinates of M as (xm,ym,zm) = [(xp + xq)/2,(yp + yq)/2,(zp + zq)/2] = {( 5 + 3)/2,( 4 + 5)/2,[(3 + ( 2)]/2} = ( 2/2,1/2,1/2) = ( 1,1/2,1/2)

Does it seem obvious that the midpoint between two points, say P and Q, doesn t depend on whether we go from P to Q or from Q to P That s indeed the case; but if we demand proof, we must show that for real numbers xp, yp, zp, xq, yq, and zq, it s always true that [(xp + xq)/2,(yp + yq)/2,(zp + zq)/2] = [(xq + xp)/2,(yq + yp)/2,(zq + zp)/2] This proof is almost trivial, but it s good mental exercise to put it down in rigorous form The commutative law for addition of real numbers tells us that xp + xq = xq + xp Dividing each side by 2 gives us (xp + xq)/2 = (xq + xp)/2 Using the same logic with the y and z coordinates, we get (yp + yq)/2 = (yq + yp)/2 and (zp + zq)/2 = (zq + zp)/2 Based on these facts, we know that the coordinates on both sides of the original equation are identical It follows that the midpoint along a straight-line segment connecting any two points in Cartesian three-space is the same, regardless of which way we go

In Part Two we will come back to your project for learning in each chapter. We will also have a case study, a project I did last year, building a home theater system, to highlight key points of each project stage. So, what will your project for learning be

Here s a challenge!

where u = velocity, ft/sec C = flow coefficient g = gravity, ft/sec2 Ap = pressure drop, lh/ftz p = density, lb/ft But velocity is proportional to flow: F ZL = A where F = flow, ft3/sec A = inside area, ftz Therefore the pressure drop due to flow in the steady stat e is AP=~= 2gA2C2 If the applied force A Ap exweds resistance to flow, acceleration takes place. An equation can he written for the unsteady state: net force equals mass times accelcrat,ion. AAp AF2p

Imagine two points in Cartesian three-space where corresponding coordinates are negatives of each other Show that the midpoint is exactly at the origin