a.) The amount of drug in the patient's bloodstream reaches a maximum when the derivative of the function V(t) is equal to zero. Taking the derivative of V(t) gives:
V'(t) = 60e^-t - 60te^-t
Setting V'(t) equal to zero and solving for t gives:
60e^-t - 60te^-t = 0
t = 1 hour
Therefore, the amount of drug in the patient's bloodstream reaches a maximum after 1 hour. Plugging t=1 into V(t) gives:
V(1) = 60e^-1 ≈ 22.05 mg
b.) The function that describes the rate of change in the amount of the drug in the bloodstream is the derivative of V(t), which is V'(t) = 60e^-t - 60te^-t. If V'(t) is positive, it means that the drug is being absorbed into the bloodstream. If V'(t) is negative, it means that the drug is being removed from the bloodstream.
c.) To find the time at which the rate at which the drug is being absorbed into the bloodstream is greatest, we need to find the maximum of V'(t). Taking the derivative of V'(t) gives:
V''(t) = 60te^-t - 120e^-t
Setting V''(t) equal to zero and solving for t gives:
t = 2
Therefore, the rate at which the drug is being absorbed into the bloodstream is greatest after 2 hours.
d.) To find the time at which the rate at which the drug is being removed from the bloodstream is greatest, we need to find the minimum of V'(t). Taking the derivative of V'(t) gives:
V''(t) = 60te^-t - 120e^-t
Setting V''(t) equal to zero and solving for t gives:
t = 0
Therefore, the rate at which the drug is being removed from the bloodstream is greatest immediately after the drug is administered.
e.) To find the long-term behavior of V(t), we take the limit as t approaches infinity:
lim V(t) = lim 60te^-t
t->∞ = 0
Therefore, the amount of drug in the patient's system in the long run is zero.