The following data from Car and Driver magazine [Car and Driver, May 1998, p. 102] shows the elapsed time it took a Honda CR-V starting at rest to accelerate to 30, 60, and 90 m.p.h. Time 0 3.1 10.3 30.1 Velocity 0 30 60 90 To approximate how long it would take the CR-V to accelerate to 50 mph or to approximate the distance it would take the CR-V to accelerate to 90 mph, an explicit velocity function v(t) is needed. Such a v(t) could be found by tting a piecewise-polynomial curve to the data. The easiest approach would be to t lines between each consecutive pair of data points; the result is shown here. The drawback to this approach is that the curve is not smooth; that is, its slope changes abruptly at the data points. The expectation is that the velocity function should indeed be smoother than that produced by the line tting. In order to ensure that the velocity function v(t) is as smooth as needed, assume that v0(t) and v00(t) are continuous functions. 1 In order to make these assumptions feasible, t a third-degree (cubic) polynomial to each consecutive pair of data points. This piecewise-polynomial curve is called a cubic spline. To t a cubic spline v(t) to the velocity data, assume that on each of the three intervals [0,3.1], [3.1,10.3], and [10.3,30.1] the formula for v(t) is given by a cubic polynomial whose coecients must be determined. It is convenient to write the formulas as follows: v(t) = a1t3 + a2t2 + a3t + a4 if 0 t < 3.1 b1(t3.1)3 + b2(t3.1)2 + b3(t3.1) + b4 if 3.1 t < 10.3 c1(t10.3)3 + c2(t10.3)2 + c3(t10.3) + c4 if 10.3 t < 30.1 Since v(0) = 0, v(3.1) = 30, v(10.3) = 60, and v(30.1) = 90 to get the pieces to match up we have, a4 = 0 29.791a1 + 9.61a2 + 3.1a3 + a4 = 30 b4 = 30 373.248b1 + 51.84b2 + 7.2b3 + b4 = 60 c4 = 60 7762.392c1 + 392.04c2 + 19.8c3 + c4 = 90 Now consider the derivative v0(t): v0(t) = 3a1t2 + 2a2t + a3 if 0 t < 3.1 3b1(t3.1)2 + 2b2(t3.1) + b3 if 3.1 t < 10.3 3c1(t10.3)2 + 2c2(t10.3) + c3 if 10.3 t < 30.1 Since v0(t) is supposed to be continuous at t = 3.1 and t = 10.3, it must be true that 28.83a1 + 6.2a2 + a3 = b3 155.52b1 + 14.4b2 + b3 = c3 which may be rewritten as 28.83a1 + 6.2a2 + a3 b3 = 0 155.52b1 + 14.4b2 + b3 c3 = 0 Further consider the second derivative v00(t): v00(t) = 6a1t + 2a2 if 0 t < 3.1 6b1(t3.1) + 2b2 if 3.1 t < 10.3 6c1(t10.3) + 2c2 if 10.3 t < 30.1 2 To make v00(t) continuous at t = 3.1 and t = 10.3 requires 18.6a1 + 2a2 2b2 = 0 43.2b1 + 2b2 2c2 = 0 And so there are 10 linear equations relating the 12 variables. Two more equations are needed to hope for a unique solution, and there are several ways to do this. One way is to choose to assume that v00(0) = v00(30.1) = 0; these assumptions give the nal two equations: 2a2 = 0 118.8c1 + 2c2 = 0 The augmented matrix A for this system of equations is A = 0 0 0 1 0 0 0 0 0 0 0 0 0 29.791 9.61 3.1 1 0 0 0 0 0 0 0 0 30 0 0 0 0 0 0 0 1 0 0 0 0 30 0 0 0 0 373.248 51.84 7.2 1 0 0 0 0 60 0 0 0 0 0 0 0 0 0 0 0 1 60 0 0 0 0 0 0 0 0 7762.392 392.04 19.8 1 90 28.83 6.2 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 155.52 14.4 1 0 0 0 1 0 018 .6 2 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 43.2 2 0 0 0 2 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 118.8 2 0 0 0 To use Maple to solve this matrix we rst have to enter the command interface(rtablesize=25); Row reducing A using Maple produces the matrix 1 0 0 0 0 0 0 0 0 0 0 0 0.08470562599 0 1 0 0 0 0 0 0 0 0 0 0 0.7877623229x1010 0 0 1 0 0 0 0 0 0 0 0 0 10.49144042 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0.03451346786 0 0 0 0 0 1 0 0 0 0 0 0 0.7877623229 0 0 0 0 0 0 1 0 0 0 0 0 8.049377222 0 0 0 0 0 0 0 1 0 0 0 0 30 0 0 0 0 0 0 0 0 1 0 0 0 0.000711640096 0 0 0 0 0 0 0 0 0 1 0 0 0.0422714216 0 0 0 0 0 0 0 0 0 0 1 0 2.073134279 0 0 0 0 0 0 0 0 0 0 0 1 60 3 (Note that the exact solution for a2 is 0, there is a very small amount of roundo error that is giving the tiny value found here.) This means the unique solution of the system is found to be a1 = 0.0847056 a2 = 0 a3 = 10.491440 a4 = 0 b1 = 0.0345134 b2 = 0.787762 b3 = 8.049377 b4 = 30 c1 = 0.000711640 c2 = 0.0422714 c3 = 2.073134 c4 = 60 The function v(t) may be input into Maple using the piecewise command: v := piecewise( t> 0 and t< 3.1, -.847056e-1*t^3+10.491440*t, t> 3.1 and t<10.3, .345134e-1*(t-3.1)^3-.787762*(t-3.1)^2+8.049377*t+5.046930, t> 10.3 and t<30.1, .711640e-3*(t-10.3)^3-.422714e-1*(t-10.3)^2 +2.073134*t+38.646717 );

Question: How did they get 5.046930 and 38.646717??? I understand everything else but CANT figure it out. Please help!!!