A hospital patient is receiving a certain drug through a drop at a constant rate of \(50\) mg per hour. The rate of loss of the drug from the patient’s body is proportional to \(x\), where \(x\) (in mg) is the amount of drug in the patient at time \(t\) (in hour). Form a differential equation connecting \(x\) and \(t\) and show that
\(x=\frac{1}{k}\left( 50-A{{e}^{-kt}} \right)\)[3]
Where \(A\) and \(k\) are constants.
Given that \(k=\frac{1}{20}\).
Find the time needed for the drug in the patient to reach \(200\) mg, if initially there is no trace of this drug in the patient’s body.
[2]When there is \(80\) mg of the drug in the patient’s body, the drip is disconnected. Assuming that the rate of loss remains the same, find the time taken or the amount of drug in the patient to fall from \(80\) mg to \(20\) mg.
[3]Video solution locked
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