Math Problem Statement

The quantity, Q , of a drug in the blood stream begins with 300 mg and decays to one-fifth its value over every 120 minute period. Give the constants a , b , and T so that Q is represented by a function of the form Q=ab−t/T , where t is the time in minutes after the drug is injected.

Solution

To represent the quantity QQ of a drug in the bloodstream as a function of time tt using the form Q=abt/TQ = ab^{-t/T}, we need to determine the constants aa, bb, and TT given the initial conditions and the decay behavior of the drug.

Given Information:

  • Initial quantity of the drug: Q(0)=300Q(0) = 300 mg.
  • The drug decays to one-fifth of its value every 120 minutes.

Function Form:

The function form given is:

Q(t)=abt/TQ(t) = ab^{-t/T}

where:

  • aa is the initial quantity of the drug.
  • bb is the base of the exponential decay.
  • TT is the time constant associated with the decay.

Step-by-Step Solution:

  1. Determine aa:

    The initial quantity of the drug in the bloodstream is 300 mg. Therefore, when t=0t = 0:

    Q(0)=ab0=a×1=aQ(0) = ab^{0} = a \times 1 = a

    So, a=300a = 300.

  2. Determine bb and TT:

    The problem states that the drug decays to one-fifth of its value every 120 minutes. This means:

    Q(120)=15Q(0)Q(120) = \frac{1}{5} Q(0)

    Substituting Q(0)=300Q(0) = 300 and the functional form:

    Q(120)=300b120/T=15×300Q(120) = 300b^{-120/T} = \frac{1}{5} \times 300

    Simplifying this:

    300b120/T=60300b^{-120/T} = 60

    Dividing both sides by 300:

    b120/T=60300=15b^{-120/T} = \frac{60}{300} = \frac{1}{5}

    Taking the reciprocal to simplify:

    b120/T=5b^{120/T} = 5

  3. Solve for bb:

    We know that b120/T=5b^{120/T} = 5, and since we want a base for our exponential decay:

    b=5T/120b = 5^{T/120}

    To maintain a consistent decay form, we can choose T=120T = 120 because every 120 minutes, the drug reduces to one-fifth:

    b=5120/120=5b = 5^{120/120} = 5

  4. Final Values for aa, bb, and TT:

    • a=300a = 300 (initial quantity of the drug)
    • b=5b = 5 (base of the exponential decay)
    • T=120T = 120 (time constant for decay to one-fifth)

Final Function:

The function that represents the quantity QQ of the drug in the bloodstream over time tt is:

Q(t)=300×5t/120Q(t) = 300 \times 5^{-t/120}

Or, equivalently:

Q(t)=300×(15)t/120Q(t) = 300 \times \left(\frac{1}{5}\right)^{t/120}

Would you like further details, or do you have any questions?

Related Questions:

  1. How does changing the decay time constant TT affect the rate at which the drug decays?
  2. What would the function look like if the drug decayed to one-tenth of its value every 120 minutes?
  3. How can we convert this exponential decay model into a continuous decay model using ee?
  4. What is the half-life of the drug according to this model?
  5. How much of the drug remains in the bloodstream after 240 minutes?

Tip: Understanding exponential decay is crucial in pharmacokinetics, where it models how drugs dissipate in the body.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Initial Value Problems

Formulas

Exponential decay formula Q(t) = ab^(-t/T)

Theorems

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Suitable Grade Level

Grades 11-12