The concentration of a drug in a body is `F(t)`, where `t` is the time in hours after the drug is taken.
Initially the concentration of the drug is zero. The rate of change of concentration of the drug is given by
`F′(t) = 50e^(−0.5t) - 0.4F(t)`.
- By differentiating the product `F(t)e^(0.4t)` show that
`qquadd/(dt)(F(t)e^(0.4t)) = 50e^(−0.1t)`. (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
- Hence, or otherwise, show that `F(t) = 500(e^(−0.4t) - e^(−0.5t))`. (2 marks)
--- 8 WORK AREA LINES (style=lined) ---
- The concentration of the drug increases to a maximum.
For what value of `t` does this maximum occur? (2 marks)
--- 6 WORK AREA LINES (style=lined) ---