UMNW, Mathematik 1, Lösung Serie 1

Thomas Kuster

11. November 2003

1

0.001m · 2^{x} = 150 · 10^{9}m
2^{x} = \frac{150 · 10^{9}m}{0.001m}
x · ln 2 = ln \frac{150 · 10^{9}}{0.001}
x = \frac{ln 1.5 · 10^{14}}{ln 2}
x ≈ 47.09
⇒ 48 mal Falten (aufrunden)

2

N_{0} · λ^{k} = 2N_{0}
λ^{k} = 2
k ln λ = ln 2
k = \frac{ln 2}{ln λ}

Lösen mit Octave (http://www.octave.org)

octave:1> lambda = (1.01:0.01:1.1)
lambda =
1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10
octave:2> log(2)./log(lambda)
69.6607
35.0028
23.4498
17.6730
14.2067

11.8957
10.2448
 9.0065
 8.0432
 7.2725
octave:3> title('Verdoppelungszeit');
octave:4> xlabel('Lambda');
octave:5> ylabel('Jahre');
octave:6> plot(lambda,log(2)./log(lambda))

3

3.1

48503000 · 1.025^{2005−1985} = x
79477813 ≈ x

UMNW, Mathematik 1, Lösung Serie 1 3

3.2

((48503000 · 1.025^{10})1.02^{5}) 1.015^{5} = x
73847929 ≈ x

3.3

((48503000 · 1.025^{10})1.03^{5}) 1.035^{5} = x
85486025 ≈ x

3.4

N_{0} · p = N_{k}
p = \frac{N_{k}}{N_{0}}
p_{1} = \frac{79477813}{48503000} ⇒ 63.9%
p_{2} = \frac{73847929}{48503000} ⇒ 52.3%
p_{3} = \frac{85486025}{48503000} ⇒ 76.2%

4

4.1

N_{0}λ^{k} + ∑_{i=0}^{k−1} mλ^{i} = x
8 · 10^{7} · 0.99^{k} + ∑_{i=0}^{k−1} 3 · 10^{5} · 0.99^{i} = x

octave:1> 8*10^7 *0.99^10 + sum(3*10^5*0.99.^(0:(10-1)))
ans =  7.5219e+07

4.2

N_{0}λ^{k} + ∑_{i=0}^{k−1} mλ^{i} = x

10^{5} · 1.03^{k} + ∑_{i=0}^{k−1} −4 · 10^{3} · 1.03^{i} = x

octave:1> p = 100000
p = 100000
octave:2> for i=1:50
> p = [p 10^5*1.03^i - 4*10^3*sum(1.03.^(0:(i-1)))];
> end
octave:3> title('Fischentwicklung');
octave:4> xlabel('Jahre');
octave:5> ylabel('Anzahl');
octave:6> plot(p)

oder

N_{0}λ^{k} + ∑_{i=0}^{k−1} mλ^{i} = x

N_{0}λ^{k} + m\frac{λ^{k} − 1}{λ − 1} = x
10^{5} · 1.03^{k} − 4 · 10^{3} \frac{1.03^{k} − 1}{1.03 − 1} = x

octave:1> k = (1:50);
octave:2> N = 10^5*1.03.^k + (-4*10^3)*(1.03.^k-1)./(1.03-1)
octave:3> plot(k,N);

Wann hat es keine Fische mehr?

N_{0}λ^{k} + m\frac{λ^{k} − 1}{λ − 1} =^{!} 0
m\frac{λ^{k} − 1}{λ − 1} = −N_{0}λ^{k}
λ^{k} − 1 = \frac{(−N_{0}λ^{k})(λ − 1)}{m}
−1 = \frac{(−N_{0}λ^{k})(λ − 1)}{m} − λ^{k}
1 = (\frac{N_{0}(λ − 1)}{m} + 1) λ^{k}
1 = \frac{N_{0}λ − N_{0} + m}{m}λ^{k}
\frac{m}{N_{0}λ − N_{0} + m} = λ^{k}
ln \frac{m}{N_{0}λ − N_{0} + m} = k ln λ
\frac{ln \frac{m}{N_{0}λ−N_{0}+m}}{ln λ} = k
\frac{ln 4}{ln 1.03} ≈ 46.9Jahre