Math 311-504Topics in Applied MathematicsLecture 4:Applications of systems of linear equations.Systems of linear equationsa11x1+ a12x2+ ··· + a1nxn= b1a21x1+ a22x2+ ··· + a2nxn= b2·········am1x1+ am2x2+ ··· + amnxn= bmHere x1, x2, . . . , xnare variables and aij, bjareconstants.A solution of the system is a common solution of allequations in the system. It is an n-dimensionalvector.Plenty of problems in mathematics and applicationsrequire solving systems of linear equations.ApplicationsProblem 1. Find the point of intersection of thelines x −y = −2 and 2x + 3y = 6 in R2.x −y = −22x + 3y = 6Problem 2. Find the point of intersection of theplanes x −y = 2, 2x −y − z = 3, andx + y + z = 6 in R3.x −y = 22x −y −z = 3x + y + z = 6Problem 3. Find the point of intersection of theplane t(2, 1, 1) + s(3, 2, 1) + (−2, 0, 1) with the linet(1, 1, −1) + (1, 4, 7).Let (x, y, z) be the point of intersection. Thenx = 2t + 3s − 2y = t + 2sz = t + s + 1andx = r + 1y = r + 4z = −r + 7for some t, s, r ∈ R.Problem 3. Find the point of intersection of theplane t(2, 1, 1) + s(3, 2, 1) + (−2, 0, 1) with the linet(1, 1, −1) + (1, 4, 7).Let (x, y, z) be the point of intersection. Thenx − 2t − 3s = −2y − t − 2s = 0z − t − s = 1x − r = 1y − r = 4z + r = 7for some t, s, r ∈ R.Method of undetermined coefficients often involvessolving systems of linear equations.Problem 4. Find a quadratic polynomial p(x)such that p(1) = 4, p(2) = 3, and p(3) = 4.Suppose that p(x) = ax2+ bx + c. Thenp(1) = a + b + c, p(2) = 4a + 2b + c,p(3) = 9a + 3b + c.a + b + c = 44a + 2b + c = 39a + 3b + c = 4Traffic flow450 400610 640520 600Problem. Determine the amount of trafficbetween each of the four intersections.Traffic flowx1x2x3x4450 400610 640520 600x1=?, x2=?, x3=?, x4=?Traffic flowA BCDx1x2x3x4450 400610 640520 600At each intersection, the incoming traffic has tomatch the outgoing traffic.Intersection A: x4+ 610 = x1+ 450Intersection B: x1+ 400 = x2+ 640Intersection C : x2+ 600 = x3Intersection D: x3= x4+ 520x4+ 610 = x1+ 450x1+ 400 = x2+ 640x2+ 600 = x3x3= x4+ 520⇐⇒−x1+ x4= −160x1− x2= 240x2− x3= −600x3− x4= 520Electrical network3 ohms 2 ohms4 ohms1 ohm9 volts4 voltsProblem. Determine the amount of current ineach branch of the network.Electrical network3 ohms 2 ohms4 ohms1 ohm9 volts4 voltsi1i2i3i1=?, i2=?, i3=?Electrical network3 ohms 2 ohms4 ohms1 ohm9 volts4 voltsi1i2i3Kirchhof’s law #1 (junction rule): at everynode the sum of the incoming currents equals thesum of the outgoing currents.Electrical network3 ohms 2 ohms4 ohms1 ohm9 volts4 voltsi1i2i3A BNode A: i1= i2+ i3Node B: i2+ i3= i1Electrical networkKirchhof’s law #2 (loop rule): around everyloop the algebraic sum of all voltages is zero.Ohm’s law: for every resistor the voltage drop E ,the current i, and the resistance R satisfy E = iR.Top loop: 9 − i2− 4i1= 0Bottom loop: 4 − 2i3+ i2− 3i3= 0Big loop: 4 − 2i3− 4i1+ 9 − 3i3= 0Remark. The 3rd equation is the sum of the firsttwo equations.i1= i2+ i39 − i2− 4i1= 04 − 2i3+ i2− 3i3= 0⇐⇒i1− i2− i3= 04i1+ i2= 9−i2+ 5i3= 4Stress analysis of a trussProblem. Assume t hat the leftmost and rightmostjoints are fixed. Find the forces acting on eachmember of the truss.Truss bridgeLet |fk| be the magnitude of the force in the kthmember. fk> 0 if the member is under tension.fk< 0 if the member is under compression.Static equilibrium at the joint A:horizontal projection: −1√2f1+ f4+1√2f5= 0vertical projection: −1√2f1− f3−1√2f5= 0Static equilibrium at the joint B:horizontal projection: −f4+ f8= 0vertical projection: −f7= 0Static equilibrium at the joint C:horizontal projection: −f8−1√2f9+1√2f12= 0vertical projection: −1√2f9− f11−1√2f12= 0Static equilibrium at the joint D:horizontal projection: −f2+ f6= 0vertical projection: f3− 10 = 0Static equilibrium at the joint E:horizontal projection: −1√2f5−f6+1√2f9+ f10= 0vertical projection:1√2f5+ f7+1√2f9− 15 = 0Static equilibrium at the joint F:horizontal projection: −f10+ f13= 0vertical projection: f11− 20 = 0−1√2f1+ f4+1√2f5= 0−1√2f1− f3−1√2f5= 0−f4+ f8= 0−f7= 0−f8−1√2f9+1√2f12= 0−1√2f9− f11−1√2f12= 0−f2+ f6= 0f3= 10−1√2f5− f6+1√2f9+ f10= 01√2f5+ f7+1√2f9= 15−f10+ f13= 0f11=
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