Let `P(t)` be a function such that `(dP)/(dt)=3000 e^{2t}`.
When `t=0, P=4000`.
Find an expression for `P(t)`. (2 marks)
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Let `P(t)` be a function such that `(dP)/(dt)=3000 e^{2t}`.
When `t=0, P=4000`.
Find an expression for `P(t)`. (2 marks)
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`P(t)=1500e^{2t}+2500`
`P(t)` | `=int (dP)/(dt)\ dt` | |
`=int 3000e^{2t}\ dt` | ||
`=1500e^{2t}+c` |
`text{When}\ t=0, P=4000`
`4000` | `=1500e^0+c` | |
`c` | `=2500` |
`:.P(t)=1500e^{2t}+2500`
The population, `D`, of Tasmanian Devils in a sanctuary is given by `D(t)`, where `t` is the time in years after the sanctuary was established.
The devil population changes at a rate modelled by the function `(dD)/(dt) = 28 e^(0.35t)`.
Calculate the increase in the number of Tasmanian Devils at the end of the first 8 years. Give your answer correct to three significant figures. (3 marks)
`1240 \ text((to 3 sig. fig.))`
`int_0^8 28e^(0.35t)` | `= [28 xx (1)/(0.35) e^(0.35t)]_0^8` |
`= 80(e^(0.35 xx 8) – e°)` | |
`= 80(16.44 … – 1)` | |
`= 1235.57 …` | |
`= 1240 \ text((to 3 sig. fig.))` |