William W. answered • 06/12/20
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Q(t) = 30te-0.25t
a) Amount in bloodstream after 3 hours: Q(3) = 30(3)e-0.25(3) = 90e-0.75 = 42.513 mg
b) Rate of change after 3 hours:
Q'(t) (by the product rule) = u'v + uv' where u = 30t and v = e-0.25t so u' = 30 and v' = -0.25e-0.25t
Q'(t) = 30e-0.25t + 30t(-0.25e-0.25t) = 30e-0.25t - 7.5te-0.25t = e-0.25t(30 - 7.5t)
Q'(3) = e-0.25(3)(30 - 7.5(3)) = 3.543 mg/hr
c) Since the slope is positive, the amount is increasing.
d) Maximum is attained by setting the derivative equal to zero
Q'(t) = e-0.25t(30 - 7.5t) = 0
e-0.25t = 0 and (30 - 7.5t) = 0
e-0.25t can never be zero and 30 - 7.5t = 0 when t = 30/7.5 = 4 hours
e) The maximum quantity is Q(4) = 30(4)e-0.25(4) = 120e-1 = 44.146 mg
Ayana S.
Could you help me with this question as well please06/20/20
William W.
I’m answering on my phone and it’s a bit tricky so I hope I get you the answer without making any mistakes. To find the minimum and/or maximum, take the derivative and set it equal to zero. For Q(t) = 50te^-0.01t, the derivative is given by the product rule. If f = uv, then f ‘ = u’v + uv’. In this case u = 50t and v = e^-0.01t so u’ = 50 and v’ = -0.01e^-0.01t so Q’(t) = 50e^-0.01t + 50t(-0.01e^-0.01t) = 50e^-0.01t - 0.5te^-0.01t. Factoring gives us e^-0.01t(50 - 0.5t). Setting it equal to zero, means either e^-0.01t = 0 or 50 - 0.5t = 0. Only the latter gives an answer so 50 = 0.5t or t = 100 minutes. I’ll leave it to you to convince yourself this is a maximum. The maximum value would be the result when you plug in t = 100 into the original equation. So 50(100)e^(-0.01)(100). I don’t have a calculator handy so I’ll let you put that in a calculator to determine the maximum (mg) of the drug in the bloodstream.06/20/20
Ayana S.
thank you!06/20/20
Ayana S.
The quantity of a drug in the blood stream t minutes after the tablet is swallowed is given, in mg, by 𝑄(𝑡) = 50𝑡𝑒−.01𝑡 a.) At what time is the quantity of the drug in the bloodstream a maximum? b.) What is that maximum?06/20/20