System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 1, Problem 1.25P
To determine

(a)

The value of the parameter b and plot the function by using half-life of carbon-14.

Expert Solution
Check Mark

Answer to Problem 1.25P

The value of the parameter b is 1.2603×104 and the plot of function is shown in Figure 1.

Explanation of Solution

Given:

Half-life of carbon -14 is 5500 years

Concept Used:

Fraction of carbon -14 remaining time at 't' is C(t)C(0)

The exponential decay, C(t)C(0)=ebt.

Calculation:

Obtain b using half-life.

C(t)C(0)=12=0.5

Thus, C(t)C(0)=ebt=0.5

Substitute carbon half-life, 5500 years for t.

0.5=eb(5500)ln(0.5)=b(5500)0.693=5500bb=1.2603×104

Thus the value of the parameter b is 1.2603×104

Now the exponential decay function C(t)C(0)=e1.2603×104i

Enter the following code in MATLAB:

>>t=0:10:5500;>>c=exp(-1.2603E-4.*t);>>plot(t,c)>>xlabel t>>ylabel fractiondecay

The following is the plot of the exponential decay function:

System Dynamics, Chapter 1, Problem 1.25P

Figure 1.

Conclusion:

Thus, the value of the parameter b is 1.2603×104 and the plot of function is shown in Figure 1.

To determine

(b)

The time when the organism died.

Expert Solution
Check Mark

Answer to Problem 1.25P

Organism died 836yearsago.

Explanation of Solution

Given:

Half-life of carbon -14 is 5500 years.

Concept Used:

Fraction of carbon -14 remaining time at 't' is C(t)C(0).

The exponential decay, C(t)C(0)=ebt.

Calculation:

We have,

C(t)C(0)=0.9[because90%]Also, C(t)C(0)=e1.2603×104t

Solve for t.

e1.2603×104t=0.9ln (0.9)=1.2603×104tt=ln(0.9)1.2603×104836years.

Conclusion:

Thus,organism died 836years ago.

To determine

(c)

The effect of b on the age obtained in part (b).

Expert Solution
Check Mark

Answer to Problem 1.25P

The organism died age is reduced to 760years ago.

Explanation of Solution

Given:

Half-life of carbon-14 is 5500 years and b is off by ±1%.

Concept Used:

Fraction of carbon - 14 remaining time at 't' is C(t)C(0)

The exponential decay, C(t)C(0)=ebt.

Calculation:

Determine b for an off by ±1%

b=1.1(1.2603×104)

Determine the time estimate

t=ln(0.9)b=ln(0.9)(1.1)(1.2603×104)=760years.

Conclusion:

Thus, the organism died age is reduced to 760years ago.

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Chapter 1 Solutions

System Dynamics

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