palm_ch8

# palm_ch8 - Solutions to Problems in Chapter Eight Solutions...

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Solutions to Problems in Chapter Eight Solutions to Test Your Understanding Problems T8.1-1 Z π 0 sin xdx = - cos x | π 0 = - cos π +cos0=1+1=2 T8.1-2 The equation of motion is m ˙ v = f ,or ˙ v = f/m =5 g . (a) v (10) = Z 10 0 5 gdt g (10) = 50(9 . 81) = 490 . 5m / sec (b) v ( t )= Z t 0 5 gt h ( t Z 10 0 v ( t ) dt = Z 10 0 5 gt dt g t 2 2 ± ± ± ± ± 10 0 = 250 g = 2452 . 5 meters T8.1-3 (a) Let y =3 x .Then d sin 3 x dx = d sin y dy dy dx = (cos y )3 = 3 cos 3 x (b) Let y =cos x d cos 2 x dx = dy 2 dy dy dx =2 y ( - sin x - 2cos x sin x (c) dx 3 ln x dx = dx 3 dx ln x + x 3 d ln x dx x 2 ln x + x 3 x = x 2 (3 ln x +1) (d) Let f =sin x and g = x 2 .No t etha t d ( f/g ) dx = g df dx - f dg dx g 2 Then d (sin x ) /x 2 dx = x 2 cos x - 2 x sin x x 4 = x cos x - 2s in x x 3 T8.2-1 Use the velocity calculations given in the table in Example 8.2-1. The script ﬁle is similar to the third script ﬁle in Example 8.2-1, with v replacing a and x replacing v . 8-1

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t = [0:10]; v = [0,1,4,9.5,18.5,32.5,53,81,117,162,211.5]; x(1) = 0; for k = [1:10] x(k+1) = trapz(t(k:k+1),v(k:k+1))+x(k); end disp([t 0 ,x 0 ]) The results are (time is in the ﬁrst column; displacement is in the second column): 00 1.0000 0.5000 2.0000 3.0000 3.0000 9.7500 4.0000 23.7500 5.0000 49.2500 6.0000 92.0000 7.0000 159.0000 8.0000 258.0000 9.0000 397.5000 10.0000 584.2500 T8.2-2 The closed form solution is Z 5 2 1 x dx =ln x | 5 2 =ln5 - ln 2 = ln 5 2 =0 . 9163 The Matlab session is as follows. ± quad( 0 1./x 0 ,2,5) ans = 0.9163 ± quadl( 0 fn 0 ,2,5) ans = 0.9163 Here quad and quadl give the same result to four decimal places. T8.2-3 The session is ± quad( 0 sqrt 0 ,0,1,.001) ans = 0.6635 8-2
± quadl( 0 sqrt 0 ,0,1,.001) ans = 0.6666 The value obtained with quad and quadl using the default tolerance is 0.6667 to four decimal places. Thus a tolerance of 0.001 produces a less accurate answer. T8.3-1 The forward diﬀerence estimate is simply shifted one index relative to the backward diﬀerence estimate. Thus, simply replace the ﬁrst plot command in the program given in the text above the problem with plot(x(2:n),td(2:n),x(1:n-1),d1,’o’),xlabel(’x’),. .. , and change the gtext labeling function. The results are shown in the ﬁgure. As you would expect, the backward estimate performs similarly to the forward estimate. 0 0.5 1 1.5 2 2.5 3 -2 -1 0 1 2 x Derivative Forward Difference Estimate 0 0.5 1 1.5 2 2.5 3 -2 -1 0 1 2 x Central Difference Estimate Figure : for Problem T8.3-1. T8.4-1 (a) The roots are s = - 4and - 7. The form of the free response is y ( t )= C 1 e - 4 t + C 2 e - 7 t (b) The roots are s = - 3 ² 5 i . The form of the free response is y ( t Be - 3 t sin(5 t + φ ) 8-3

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(c) The roots are s = - 3 and 5. The form of the free response is y ( t )= C 1 e - 3 t + C 2 e 5 t (d) The roots are s =4and - 10. The form of the free response is y ( t C 1 e 4 t + C 2 e - 10 t T8.5-1 Solve for the derivative: dy dt = te - 2 t - y 10 The model’s time constant is 10, but the time constant of the input is 0.5. So we select Δ to be a small fraction of 0.5, say Δ = 0 . 5 / 20. The script ﬁle for the Euler method is delta = .5/20; y(1)=2; k=0; for time = [delta:delta:2] k = k+1; y(k+1) = y(k) + 0.1*(time*exp(-2*time)-y(k))*delta; end t = [0:delta:2]; y_true = (732*exp(-0.1*t)-19*t.*exp(-2*t)-10*exp(-2*t))/361; plot(t,y,’o’,t,y_true),xlabel(’t’),ylabel(’y’) The plot is shown in the ﬁgure. The numerical solution is shown by the circles. It matches the true solution.
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## This note was uploaded on 10/21/2009 for the course ENGR 290 taught by Professor Priritera during the Spring '05 term at S. Alabama.

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palm_ch8 - Solutions to Problems in Chapter Eight Solutions...

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