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The number of molecules of a certain drug in the bloodstream t hours after it has been taken is represented by the equation M(t)=−t3 +3t2 +72t, 0<t<10. Determine the number of molecules of the drug in the bloodstream 3 hours after the drug was taken.

Mathematics

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Answers

3 parts
10.1

216 molecules\text{216 molecules}

10.2

72 molecules/hour\text{72 molecules/hour}

10.3

1 hour\text{1 hour}

Graph of M(t) = −t³ + 3t² + 72t

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Key points of M(t) = −t³ + 3t² + 72t
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x-intercept00
Maximum6324

Step-by-step solution

8 steps

Part 10.1· 2 steps

  1. Substitute t=3t=3 into the given equation for M(t)M(t).

    The equation for the number of molecules is M(t)=−t3+3t2+72tM(t) = -t^3 + 3t^2 + 72t.

    M(3)=−(3)3+3(3)2+72(3)M(3) = -(3)^3 + 3(3)^2 + 72(3)

  2. Calculate the value.

    M(3)=−27+3(9)+216M(3) = -27 + 3(9) + 216 M(3)=−27+27+216M(3) = -27 + 27 + 216 M(3)=216M(3) = 216

    The number of molecules of the drug in the bloodstream 3 hours after the drug was taken is 216 molecules\boxed{\text{216 molecules}}.

Part 10.2

72 molecules/hour\text{72 molecules/hour}

  1. Find the first derivative of M(t)M(t) to determine the rate of change.

    M(t)=−t3+3t2+72tM(t) = -t^3 + 3t^2 + 72t M′(t)=ddt(−t3+3t2+72t)M'(t) = \frac{d}{dt}(-t^3 + 3t^2 + 72t) M′(t)=−3t2+6t+72M'(t) = -3t^2 + 6t + 72

  2. Substitute t=2t=2 hours into the rate equation M′(t)M'(t).

    M′(2)=−3(2)2+6(2)+72M'(2) = -3(2)^2 + 6(2) + 72

  3. Calculate the rate.

    M′(2)=−3(4)+12+72M'(2) = -3(4) + 12 + 72 M′(2)=−12+12+72M'(2) = -12 + 12 + 72 M′(2)=72M'(2) = 72

    The rate at which the number of molecules of the drug is changing at exactly 2 hours is 72 molecules/hour\boxed{\text{72 molecules/hour}}.

Part 10.3

1 hour\text{1 hour}

  1. To find when the rate of change is a maximum, we need to find the derivative of the rate function, M′(t)M'(t), and set it to zero.

    This is the second derivative of M(t)M(t).

    M′(t)=−3t2+6t+72M'(t) = -3t^2 + 6t + 72 M′′(t)=ddt(−3t2+6t+72)M''(t) = \frac{d}{dt}(-3t^2 + 6t + 72) M′′(t)=−6t+6M''(t) = -6t + 6

  2. Set M′′(t)M''(t) to zero and solve for tt.

    −6t+6=0-6t + 6 = 0 −6t=−6-6t = -6 t=1t = 1

  3. Verify that this value corresponds to a maximum.

    The third derivative M′′′(t)=−6M'''(t) = -6, which is less than 0, confirming it's a maximum. The value t=1t=1 is within the domain 0<t<100 < t < 10.
    The rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum after 1 hour\boxed{\text{1 hour}}.

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