The dye dilution method is used to measure cardiac output with 4 mg of dye. The dye concentrations, in mg/L, are modeled by 

*c*(*t*) = 27*te*−0.8*t*,
 
0 ≤ *t* ≤ 10,

 where *t* is measured in seconds. Find the cardiac output. (Hint: Integration by parts is required. Round your answer to four decimal places.)

To determine the cardiac output using the dye dilution method, we need to calculate the total amount of dye that passes through the system over time. The cardiac output (CO) can be found by integrating the concentration of the dye over the given time period and then dividing the total amount of dye injected by this integral.

Given:
\[ c(t) = 27t e^{-0.8t}, \quad 0 \leq t \leq 10 \]

We need to integrate \( c(t) \) over the interval from \( 0 \) to \( 10 \) seconds.

### Step-by-Step Solution:

1. **Set up the integral:**
   \[
   \int_{0}^{10} 27t e^{-0.8t} \, dt
   \]

2. **Integration by parts:**
   We use integration by parts, where we let:
   \[
   u = 27t \quad \text{and} \quad dv = e^{-0.8t} \, dt
   \]

   Then:
   \[
   du = 27 \, dt \quad \text{and} \quad v = \int e^{-0.8t} \, dt = -\frac{1}{0.8} e^{-0.8t} = -1.25 e^{-0.8t}
   \]

3. **Apply the integration by parts formula:**
   \[
   \int u \, dv = uv - \int v \, du
   \]

   Substituting \( u \), \( v \), \( du \), and \( dv \):
   \[
   \int_{0}^{10} 27t e^{-0.8t} \, dt = 27 \left[ -1.25t e^{-0.8t} \right]_{0}^{10} - 27 \int_{0}^{10} -1.25 e^{-0.8t} \, dt
   \]
   Simplify the expression:
   \[
   -33.75 \left[ t e^{-0.8t} \right]_{0}^{10} + 33.75 \int_{0}^{10} e^{-0.8t} \, dt
   \]

4. **Evaluate each part:**

   First term:
   \[
   -33.75 \left[ t e^{-0.8t} \right]_{0}^{10} = -33.75 \left( 10 e^{-8} - 0 \right) = -337.5 e^{-8} \approx -337.5 \times 0.0003355 \approx -0.1132
   \]

   Second term:
   \[
   33.75 \int_{0}^{10} e^{-0.8t} \, dt = 33.75 \left[ -1.25 e^{-0.8t} \right]_{0}^{10} = 33.75 \left[ -1.25 e^{-8} + 1.25 \right]
   \]

   Simplify:
   \[
   33.75 \left[ -1.25 (0.0003355) + 1.25 \right] = 33.75 \left[ -0.000419375 + 1.25 \right] = 33.75 \times 1.249580625 \approx 42.481
   \]

5. **Combine results:**
   \[
   \int_{0}^{10} 27t e^{-0.8t} \, dt \approx -0.1132 + 42.481 \approx 42.3678
   \]

6. **Cardiac output:**
   Cardiac output (CO) is calculated by dividing the total amount of dye injected by the area under the concentration-time curve.
   \[
   \text{CO} = \frac{\text{Total amount of dye}}{\text{Area under concentration-time curve}} = \frac{4 \text{ mg}}{42.3678 \text{ mg}\cdot\text{s/L}} \approx 0.0944 \text{ L/s}
   \]

   Converting \( \text{L/s} \) to \( \text{L/min} \):
   \[
   \text{CO} \approx 0.0944 \times 60 \approx 5.664 \text{ L/min}
   \]

So, the cardiac output is approximately \( \boxed{5.664 \text{ L/min}} \).

Would you like more details or have any questions?

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### Additional Questions:
1. How would the cardiac output change if the initial dye concentration was doubled?
2. What would be the effect on the cardiac output if the integration period was extended to 15 seconds?
3. How does the constant \( -0.8 \) in the exponent affect the concentration model?
4. What is the significance of using integration by parts in this problem?
5. How would you solve this problem if the concentration function was a different form, such as a polynomial?
6. What assumptions are inherent in this model of dye dilution?
7. Can this method be applied to measure cardiac output for different types of patients or conditions?
8. How does the accuracy of the model change with different time intervals?

### Tip:
For complex integrals, consider substitution or breaking the problem into smaller parts to simplify the calculations. Integration by parts is particularly useful when the integrand is a product of two functions, one of which can be easily integrated after differentiation.

Learn how to calculate cardiac output using the dye dilution method with a concentration modelled by c(t) = 27t e^-0.8t over 0 ≤ t ≤ 10 seconds. Detailed step-by-step solution with integration by parts. Suitable for advanced mathematics students and professionals.

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