Answer:
A) f(1) =38.43mg
B) f'(1) = -15.37 mg/hour
C) f(6) = 11.47mg
D) f'(6) = -4.97mg/hour
F) 23.06mg
H) 0
Step-by-step explanation:
Given the function:
Q = 70te^-0.6t........(1)
Where Q is measured in mg and t is in hours
(A) f(1) is the quantity Q in mg after 1 hour
f(1) = 70(1)e^-0.6(1) = 70e^-0.6
= 70×0.5488
= 38.43 mg
(B) f'(1) = dQ/dt, at t=1
= 70t × -0.6e^-0.6t +70e^-0.6t
= -42te^-0.6t + 70e^-0.6t
= -42(1) × 0.5488 + 70 × 0.5488
=28 × 0.5488
= -15.37mg/hour
(C) f(6) = 70te^-0.6t
= 70(6)e^-0.6(6)
= 420e^-3.6
= 420 × 0.0273
= 11.47mg
(D) f'(6) = dQ/dt, at t= 6
f' = -42te^-0.6t + 70e^-0.6t
f'(6) = -42(6)e^-3.6 + 70e^-3.6
= -252 × 0.0273 + 70 × 0.0273
= -6.8796 + 1.911
= -4.9686
= -4.97mg/hour
(E) f(1) is the quantity of the drug present in the body 1 hour after injection.
f'(1) is the rate of decrease of the drug 1 hour after injection or it is the rate at which the drug is decreasing 1 hour after injection.
(F) The quantity Q is given by
Q= 38.43 - 15.37
= 23.06mg
(G) f(6) is the quantity of the drug present in the body 6 hours after injection
f'(6) is the rate at which the drug is removed or decreased from the body 6 hours after injection.
(H) The quantity is given by
Q = 11.47 - 4.97×6
Q= 11.47-29.82
Q= -18.35mg
This is not possible, hence after 6 hours of injection the quantity of drug remaining is zero.
