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During an experiment, the number of organisms present at time $t$ days is denoted by $N$, where $N$ is treated as a continuous variable. It is given that

$\frac{{dN}}{{dt}} = 1.2{e^{0.02t}}{N^{0.5}}$.

When $t = 0$, the number of organisms present is 100.

a) Find an expression for $N$ in terms of $t$.

b) State what happens to the number of organisms present after a long time.

پاسخ تشریحی :
نمایش پاسخ

a) Separate variables and attempt integration on both sides

Obtain $2{N^{0.5}}$ on left-hand side or equivalent

Obtain $ - 60{e^{ - 0.02t}}$ on right-hand side or equivalent

Use $0$ and $100$ to evaluate a constant or as limits in a solution containing terms $\alpha {N^{0.5}}$ and $b{e^{ - 0.02t}}$

Obtain $2{N^{0.5}} =  - 60{e^{ - 0.02t}} + 80$ or equivalent

Conclude with $N = {\left( {40 - 30{e^{ - 0.02t}}} \right)^2}$ or equivalent

b) State number approaches $1600$ or equivalent, following expression of form ${\left( {c + d{e^{ - 0.02t}}} \right)^n}$

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