O Solucionarão CNC

Homework 2(6th edition)  

   

Problem 8.15

(a)

When the oxygen level is 5mg/l, we have:

5 = 10 - 20(e−0.15x - e−0.5x)

0 = 5 - 20(e−0.15x - e−0.5x)

Let f(x) = 5 - 20(e−0.15x - e−0.5x)

Plot f(x)

[pic 1]

According to the plot, when f(x) = 0, x is approximately equal to 1(between 0.8 to 1)

Now we can use the Bisection to find the solution:

xl = 0.8, xu = 1, xr =  , εa = ||100%[pic 2][pic 3]

iteration	xl	xu	xr	f(xl)*f(xr)	f(xu)*f(xr)	εa
0	0.8	1	0.9	>0	<0
1	0.9	1	0.95	>0	<0	5.3%
2	0.95	1	0.975	>0	<0	2.6%
3	0.975	1	0.9875	<0	>0	1.3%
4	0.975	0.9875	0.98125			0.64%

0.64% < 1% , so the distance should be x = 0.98125 with a percent relative error εa = 0.64%

Checking the result by substituting it into original equation:

c = 10 - 20(e−0.15* 0.98125 - e−0.5*0.98125)

= 4.982

(b)

c = 10 - 20(e−0.15x - e−0.5x)

let f(x) = 10 - 20(e−0.15x - e−0.5x)

f’(x) = 3e-0.15x – 10e-0.5x, when f’(x) = 0, the oxygen should be at a minimum.

0 = 3e-0.15x – 10e-0.5x

let g(x) = 3e-0.15x – 10e-0.5x

plot the g(x)

[pic 4]

According to the plot, when g(x) = 0, x should between 3 to 4.

We can use the Bisection to find the solution

xl = 3, xu = 4, xr =  , εa = ||100%[pic 5][pic 6]

iteration	xl	xu	xr	f(xl)*f(xr)	f(xu)*f(xr)	εa
0	3	4	3.5	<0	>0
1	3	3.5	3.25	>0	<0	7.7%
2	3.25	3.5	3.375	>0	<0	3.7%
3	3.375	3.5	3.4375	>0	<0	1.8%
4	3.4375	3.5	3.46875	<0	>0	0.9%
5	3.4375	3.46875	3.453125	<0	>0	0.45%
6	3.4375	3.453125	3.4453125	<0	>0	0.23%
7	3.4375	3.445313	3.4414065	<0	>0	0.11%
8	3.4375	3.4414065	3.43945325			0.06%

The distance x is about 3.43945325 when g(x) = 0 with a percent relative error εa = 0.06%. at this point , oxygen is at minimum.

Compute the minimum concentration:

c = 10 - 20(e−0.15* 3.43945325 - e−0.5*3.43945325)

= 1.643255

Problem 8.16

(a) the graphical method:

When the concentration of bacteria reduces to 15, we have:

15 = 75e−1.5t + 20e−0.075t

0 = 75e−1.5t + 20e−0.075t – 15

let f(t) = 75e−1.5t + 20e−0.075t – 15

plot f(t):

[pic 7]

According to the plot, when f(t) = 0, t ≈ 4. So when the concentration of bacteria reduces to 15, t ≈ 4

(b) Newton-Raphson method

f’(t) = -112.5e−1.5t – 1.5e−0.075t

so:

ti+1 = ti - [pic 8]

initial guess of t = 6 and a stopping criterion of 0.5%, so we start with an initial guess of t = 6 and use the above iterative equation to compute: