Math Problem Statement

The concentration of a drug in an organ at any time t (in seconds) is given by the function: x(t) = 0.06 + 0.16(1 − e^(-0.02t)), where x(t) is measured in milligrams per cubic centimeter (mg/cm³). (a) What is the initial concentration of the drug in the organ? (b) What is the concentration of the drug in the organ after 20 sec?

Solution

Let's analyze the given problem.

The concentration of the drug in the organ is given by the function:

x(t)=0.06+0.16(1−e−0.02t)x(t) = 0.06 + 0.16 \left( 1 - e^{-0.02t} \right)

Where:

  • x(t)x(t) is the concentration in mg/cm3^3.
  • tt is time in seconds.

Part (a): Initial Concentration

The initial concentration is the value of x(t)x(t) when t=0t = 0. Substituting t=0t = 0 into the equation:

= 0.06 + 0.16 \left( 1 - e^0 \right) = 0.06 + 0.16 \times (1 - 1) = 0.06$$ So, the initial concentration is **0.06 mg/cm³**, which is correctly marked. ### Part (b): Concentration at $$t = 20$$ seconds Now, let's compute the concentration when $$t = 20$$: $$x(20) = 0.06 + 0.16 \left( 1 - e^{-0.02 \times 20} \right) = 0.06 + 0.16 \left( 1 - e^{-0.4} \right)$$ First, calculate $$e^{-0.4}$$: $$e^{-0.4} \approx 0.6703$$ Substitute this value back into the equation: $$x(20) = 0.06 + 0.16 \left( 1 - 0.6703 \right) = 0.06 + 0.16 \times 0.3297 = 0.06 + 0.05275 = 0.11275$$ So, the concentration at $$t = 20$$ seconds is approximately **0.1128 mg/cm³**, not 0.2200 mg/cm³ as shown in the image. Would you like further details on any step? Here are 5 questions to deepen your understanding: 1. How does the exponential decay factor $$e^{-0.02t}$$ affect the concentration over time? 2. What would the concentration be after 40 seconds? 3. How would you modify the function if the drug's decay rate were faster? 4. How does the initial concentration influence the long-term behavior of the function? 5. What role does the constant 0.16 play in the concentration equation? **Tip:** When solving exponential decay problems, pay close attention to how the decay factor $$e^{-kt}$$ behaves as $$t$$ increases. The concentration typically approaches a limiting value over time.

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Math Problem Analysis

Mathematical Concepts

Exponential decay
Drug concentration
Exponential functions

Formulas

x(t) = 0.06 + 0.16(1 − e^(-0.02t))

Theorems

Exponential decay: e^(-kt) represents decay over time

Suitable Grade Level

Grades 10-12

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