Math Problem Statement

A painkiller has a concentration of 56 mg/deciliter 2.5 hours after it has been administered. At 7 hours the concentration has fallen to 36 mg/deciliter.Find the rate of decay, the initial concentration, the exponential decay model, and the half life.

Solution

To solve this problem, we will use the exponential decay model, which is commonly used to describe the decay of substances over time. The general formula for exponential decay is:

C(t)=C0ektC(t) = C_0 e^{-kt}

Where:

  • C(t)C(t) is the concentration at time tt,
  • C0C_0 is the initial concentration (at time t=0t = 0),
  • kk is the decay constant, and
  • tt is time.

Given:

  • C(2.5)=56C(2.5) = 56 mg/dL (concentration after 2.5 hours)
  • C(7)=36C(7) = 36 mg/dL (concentration after 7 hours)

1. Finding the Decay Constant (kk)

To find kk, we can use the concentrations at two different times. We have:

56=C0e2.5k56 = C_0 e^{-2.5k} 36=C0e7k36 = C_0 e^{-7k}

We can divide these equations to eliminate C0C_0:

5636=C0e2.5kC0e7k\frac{56}{36} = \frac{C_0 e^{-2.5k}}{C_0 e^{-7k}}

5636=e7k2.5k\frac{56}{36} = e^{7k - 2.5k}

5636=e4.5k\frac{56}{36} = e^{4.5k}

5636=e4.5k\frac{56}{36} = e^{4.5k}

1.5556=e4.5k1.5556 = e^{4.5k}

ln(1.5556)=4.5k\ln(1.5556) = 4.5k

k=ln(1.5556)4.5k = \frac{\ln(1.5556)}{4.5}

Now, we calculate the value of kk:

k0.44174.5k \approx \frac{0.4417}{4.5}

k0.0982per hourk \approx 0.0982 \, \text{per hour}

2. Finding the Initial Concentration (C0C_0)

Using kk and one of the equations, we can find C0C_0:

56=C0e2.5×0.098256 = C_0 e^{-2.5 \times 0.0982}

C0=56e0.2455C_0 = \frac{56}{e^{-0.2455}}

C0=560.7824C_0 = \frac{56}{0.7824}

C071.6mg/dLC_0 \approx 71.6 \, \text{mg/dL}

3. Exponential Decay Model

The model can now be written as:

C(t)=71.6e0.0982tC(t) = 71.6 e^{-0.0982t}

4. Finding the Half-Life

The half-life t1/2t_{1/2} is the time required for the concentration to reduce to half of its initial value. We use the equation:

12C0=C0ekt1/2\frac{1}{2}C_0 = C_0 e^{-kt_{1/2}}

12=ekt1/2\frac{1}{2} = e^{-kt_{1/2}}

ln(12)=kt1/2\ln\left(\frac{1}{2}\right) = -kt_{1/2}

ln(2)=kt1/2-\ln(2) = -kt_{1/2}

t1/2=ln(2)kt_{1/2} = \frac{\ln(2)}{k}

t1/2=ln(2)0.0982t_{1/2} = \frac{\ln(2)}{0.0982}

t1/27.05hourst_{1/2} \approx 7.05 \, \text{hours}

Summary

  • Decay constant kk: 0.09820.0982 per hour
  • Initial concentration C0C_0: 71.671.6 mg/dL
  • Exponential decay model: C(t)=71.6e0.0982tC(t) = 71.6 e^{-0.0982t}
  • Half-life t1/2t_{1/2}: 7.057.05 hours

Would you like further details on any of these calculations or have any additional questions?

Here are some questions you might consider:

  1. How do you derive the exponential decay formula?
  2. What are some real-world applications of exponential decay?
  3. How is the half-life concept applied in pharmacology?
  4. Can the rate of decay be influenced by external factors?
  5. How would you graph the decay of the concentration over time?

Tip: Understanding the derivation of the exponential decay formula can help in applying the concept to various real-world scenarios, such as pharmacokinetics and radioactive decay.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Exponential Decay
Decay Constant
Half-Life

Formulas

Exponential decay model: C(t) = C_0 e^{-kt}
Half-life formula: t_{1/2} = \frac{\ln(2)}{k}

Theorems

-

Suitable Grade Level

Advanced High School