Answer:
a) The expression for the height, 'H', of the plant after 't' day is;
[tex]H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}[/tex]
b) The height of the plant after 30 days is approximately 19.426 inches
Step-by-step explanation:
The given maximum theoretical height of the plant = 30 in.
The height of the plant at the beginning of the experiment = 5 in.
a) The logistic differential equation can be written as follows;
[tex]\dfrac{dH}{dt} = K \cdot H \cdot \left( M - {P} \right)[/tex]
Using the solution for the logistic differential equation, we get;
[tex]H = \dfrac{M}{1 + A\cdot e^{-(M\cdot k) \cdot t}}[/tex]
Where;
A = The condition of height at the beginning of the experiment
M = The maximum height = 30 in.
Therefore, we get;
[tex]5 = \dfrac{30}{1 + A\cdot e^{-(30\cdot k) \cdot 0}}[/tex]
[tex]1 + A = \dfrac{30}{5} = 6[/tex]
A = 5
When t = 20, H = 12
We get;
[tex]12 = \dfrac{30}{1 + 5\cdot e^{-(30\cdot k) \cdot 20}}[/tex]
[tex]1 + 5\cdot e^{-(30\cdot k) \cdot 20} = \dfrac{30}{12} = 2.5[/tex]
[tex]5\cdot e^{-(30\cdot k) \cdot 20} = 2.5 - 1 = 1.5[/tex]
∴ -(30·k)·20 = ㏑(1.5)
k = ㏑(1.5)/(30 × 20) ≈ 6·7577518 × 10⁻⁴
k ≈ 6·7577518 × 10⁻⁴
Therefore, the expression for the height, 'H', of the plant after 't' day is given as follows
[tex]H = \dfrac{30}{1 + 5\cdot e^{-(30\times 6.7577518 \times 10^{-4}) \cdot t}} = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}[/tex]
b) The height of the plant after 30 days is given as follows
[tex]H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}[/tex]
At t = 30, we have;
[tex]H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \times 30}} \approx 19.4258866473[/tex]
The height of the plant after 30 days, H ≈ 19.426 in.