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MA3264 Homework 1 Due Friday 12 September

MA3264 Homework 1 Due Friday 12 September. Answers. Page 50 Problem 6 a, b. Compute and plot the points.

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MA3264 Homework 1 Due Friday 12 September

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  1. MA3264 Homework 1 Due Friday 12 September Answers

  2. Page 50 Problem 6 a, b Compute and plot the points Answer (a) The model makes sense intuitively.The constant 100 is the equilibrium value of price, and 500 is the equilibrium value of supply. The model asserts that(i) prices decreases (represented by the sign of -0.1) if the supply exceeds the equilibrium supply (ii) the supply increases (represented by the sign of 0.2) if the price exceeds the equilibrium price.These facts are explained by the common sense “law of supply and demand” in economics. Consumers differ in their utility [desire] for goods and therefore differ in their the price that they are willing to pay. Therefore lower prices are required to sell a larger supply of goods. Producers differ in the price that they can charge for a good and still remain profitable because more efficient producers can produce at a lower cost. So producers supply more of a good as its price increases.

  3. Page 50 Problem 6 a, b Compute and plot the points Answer (b) The following MATLAB commands compute and plot the values of price and supply for n = 1:20 for cases A, B, C, DCase AP(1) = 100; Q(1) = 500;for n = 1:19P(n+1) = P(n) -0.1*(Q(n)-500); Q(n+1) = Q(n) + 0.2*(P(n)-100);endfigure(1); plot(P,Q,’b*’)Case B, C, D similar

  4. Page 50 Problem 6 a, b Compute and plot the points A. Initial (100,500) B. Initial (200,500) C. Initial (100,600) D. Initial (100,400)

  5. Page 50 Problem 6 a, b Compute and plot the points Answer Long-term Behavior >> M^10000 ans = 1.0e+042 * -8.2407 4.0303 -8.0606 -8.2407 Since M has eigenvalues having abosolute values > 1, cases B, C, D diverge to infinity, case A is an unstable equilibrium >> eig(M) ans = 1.0000 + 0.1414i 1.0000 - 0.1414i

  6. Page 83 Problem 3 : Solution 1. Constant At terminal velocity Terminal velocity is proportional to the square root of mass. Box with mass 800 lbs has terminal velocity1.1547 times the terminal velocity of the box with mass 600

  7. Page 83 Problem 3 : Solution 2. At terminal velocity Terminal velocity is proportional to the 6th root of mass. Box with mass 800 lbs has terminal velocity1.0491 times the terminal velocity of the box with mass 600

  8. Page 124 Problem 1 The proposed model is It is NOT a linear function of the parameter Therefore we must TRANSFORM the data and then fit the transformed data to the following model: The following MATLAB comands input data, transform the data, and compute a least squares fit of the model and plots the data and the model. x = [17 19 20 22 23 25 28 31 32 33 36 37 38 39 41];y = [19 25 32 51 57 71 113 141 123 187 192 205 252 259 294];logx = log(x); logy = log(y); m = 15; xy = logx*logy’; sx = sum(logx); sy = sum(logy); xx = logx*logx’;b = (m*xy-sx*sy)/(m*xx – sx*sx); loga = (xx*sy-xy*sx)/(m*xx-sx*sx); a = exp(loga);xs = 16:.01:42; ys = a*xs.^b; plot(xs,ys); hold on; plot(x,y,’r*’);dev = y – a*x.^b; maxdev = max(abs(dev)); S = dev*dev’;

  9. Page 124 Problem 1 a = 0.0032, b = 3.0919, sqrt(S/m) = 13.7271 <= cmax <= maxdev = 28.1377

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