Answer :
To find the cardiac output using the dye dilution method, we are provided with a dye concentration function [tex]\( c(t) = 27t e^{-0.9t} \)[/tex], which models the dye concentration over time [tex]\( t \)[/tex] in seconds. The cardiac output in liters per second is given by [tex]\( 0.2001 \times \int_{0}^{10} c(t) \, dt \)[/tex].
Let's calculate this step by step:
1. Set up the integral:
We need to integrate the given function over the interval from 0 to 10:
[tex]\[ \int_{0}^{10} 27t e^{-0.9t} \, dt. \][/tex]
2. Use integration by parts:
To integrate [tex]\( 27t e^{-0.9t} \)[/tex], we apply integration by parts. The formula for integration by parts is:
[tex]\[ \int u \, dv = uv - \int v \, du, \][/tex]
where [tex]\( u = t \)[/tex] and [tex]\( dv = 27 e^{-0.9t} \, dt \)[/tex].
Differentiate and integrate:
[tex]\[ du = dt, \][/tex]
[tex]\[ v = \int 27 e^{-0.9t} \, dt = 27 \left( -\frac{1}{0.9} e^{-0.9t} \right) = -30 e^{-0.9t}. \][/tex]
3. Substitute into the integration by parts formula:
Now, substitute [tex]\( u = t \)[/tex], [tex]\( du = dt \)[/tex], [tex]\( v = -30 e^{-0.9t} \)[/tex], and [tex]\( dv = 27 e^{-0.9t} \, dt \)[/tex]:
[tex]\[ \int 27t e^{-0.9t} \, dt = t \cdot (-30 e^{-0.9t}) - \int (-30 e^{-0.9t}) \, dt. \][/tex]
Simplify both terms:
[tex]\[ = -30t e^{-0.9t} + 30 \int e^{-0.9t} \, dt. \][/tex]
Integrate [tex]\( \int e^{-0.9t} \, dt \)[/tex]:
[tex]\[ \int e^{-0.9t} \, dt = -\frac{1}{0.9} e^{-0.9t} = -\frac{10}{9} e^{-0.9t}. \][/tex]
4. Complete the integration:
Plug this back into our expression:
[tex]\[ \int 27t e^{-0.9t} \, dt = -30t e^{-0.9t} + 30 \left(-\frac{10}{9} e^{-0.9t}\right). \][/tex]
Simplify:
[tex]\[ = -30t e^{-0.9t} - \frac{300}{9} e^{-0.9t}. \][/tex]
Further simplify:
[tex]\[ = -30t e^{-0.9t} - \frac{100}{3} e^{-0.9t}. \][/tex]
Evaluate this integral from 0 to 10:
[tex]\[ \left[ -30t e^{-0.9t} - \frac{100}{3} e^{-0.9t} \right]_{0}^{10}. \][/tex]
5. Evaluate the definite integral:
Calculate at the upper limit [tex]\( t = 10 \)[/tex]:
[tex]\[ -30(10) e^{-0.9(10)} - \frac{100}{3} e^{-0.9(10)} = -300 e^{-9} - \frac{100}{3} e^{-9}. \][/tex]
Simplify:
[tex]\[ = -\left( 300 + \frac{100}{3} \right)e^{-9}. \][/tex]
Calculate at the lower limit [tex]\( t = 0 \)[/tex]:
[tex]\[ -30(0) e^{-0.9(0)} - \frac{100}{3} e^{-0.9(0)} = 0 - \frac{100}{3}. \][/tex]
6. Combine the results:
[tex]\[ \left[ -\left( 300 + \frac{100}{3} \right)e^{-9} \right] - \left[ 0 - \frac{100}{3} \right] = -\left( \frac{900}{3} + \frac{100}{3} \right)e^{-9} + \frac{100}{3}. \][/tex]
Simplify further:
[tex]\[ -\frac{1000}{3} e^{-9} + \frac{100}{3}. \][/tex]
7. Approximate solution:
Evaluate [tex]\( e^{-9} \approx 1.2341 \times 10^{-4} \)[/tex]
Thus,
[tex]\[ -\frac{1000}{3} \times 1.2341 \times 10^{-4} + \frac{100}{3} = -0.4113 + 33.3333 \approx 32.9220. \][/tex]
8. Cardiac output:
Multiply by the given coefficient [tex]\( 0.2001 \)[/tex]:
[tex]\[ 0.2001 \times 32.9220 \approx 6.5869. \][/tex]
Thus, the cardiac output is approximately [tex]\( 6.5869 \)[/tex] liters per second (rounded to four decimal places).
Let's calculate this step by step:
1. Set up the integral:
We need to integrate the given function over the interval from 0 to 10:
[tex]\[ \int_{0}^{10} 27t e^{-0.9t} \, dt. \][/tex]
2. Use integration by parts:
To integrate [tex]\( 27t e^{-0.9t} \)[/tex], we apply integration by parts. The formula for integration by parts is:
[tex]\[ \int u \, dv = uv - \int v \, du, \][/tex]
where [tex]\( u = t \)[/tex] and [tex]\( dv = 27 e^{-0.9t} \, dt \)[/tex].
Differentiate and integrate:
[tex]\[ du = dt, \][/tex]
[tex]\[ v = \int 27 e^{-0.9t} \, dt = 27 \left( -\frac{1}{0.9} e^{-0.9t} \right) = -30 e^{-0.9t}. \][/tex]
3. Substitute into the integration by parts formula:
Now, substitute [tex]\( u = t \)[/tex], [tex]\( du = dt \)[/tex], [tex]\( v = -30 e^{-0.9t} \)[/tex], and [tex]\( dv = 27 e^{-0.9t} \, dt \)[/tex]:
[tex]\[ \int 27t e^{-0.9t} \, dt = t \cdot (-30 e^{-0.9t}) - \int (-30 e^{-0.9t}) \, dt. \][/tex]
Simplify both terms:
[tex]\[ = -30t e^{-0.9t} + 30 \int e^{-0.9t} \, dt. \][/tex]
Integrate [tex]\( \int e^{-0.9t} \, dt \)[/tex]:
[tex]\[ \int e^{-0.9t} \, dt = -\frac{1}{0.9} e^{-0.9t} = -\frac{10}{9} e^{-0.9t}. \][/tex]
4. Complete the integration:
Plug this back into our expression:
[tex]\[ \int 27t e^{-0.9t} \, dt = -30t e^{-0.9t} + 30 \left(-\frac{10}{9} e^{-0.9t}\right). \][/tex]
Simplify:
[tex]\[ = -30t e^{-0.9t} - \frac{300}{9} e^{-0.9t}. \][/tex]
Further simplify:
[tex]\[ = -30t e^{-0.9t} - \frac{100}{3} e^{-0.9t}. \][/tex]
Evaluate this integral from 0 to 10:
[tex]\[ \left[ -30t e^{-0.9t} - \frac{100}{3} e^{-0.9t} \right]_{0}^{10}. \][/tex]
5. Evaluate the definite integral:
Calculate at the upper limit [tex]\( t = 10 \)[/tex]:
[tex]\[ -30(10) e^{-0.9(10)} - \frac{100}{3} e^{-0.9(10)} = -300 e^{-9} - \frac{100}{3} e^{-9}. \][/tex]
Simplify:
[tex]\[ = -\left( 300 + \frac{100}{3} \right)e^{-9}. \][/tex]
Calculate at the lower limit [tex]\( t = 0 \)[/tex]:
[tex]\[ -30(0) e^{-0.9(0)} - \frac{100}{3} e^{-0.9(0)} = 0 - \frac{100}{3}. \][/tex]
6. Combine the results:
[tex]\[ \left[ -\left( 300 + \frac{100}{3} \right)e^{-9} \right] - \left[ 0 - \frac{100}{3} \right] = -\left( \frac{900}{3} + \frac{100}{3} \right)e^{-9} + \frac{100}{3}. \][/tex]
Simplify further:
[tex]\[ -\frac{1000}{3} e^{-9} + \frac{100}{3}. \][/tex]
7. Approximate solution:
Evaluate [tex]\( e^{-9} \approx 1.2341 \times 10^{-4} \)[/tex]
Thus,
[tex]\[ -\frac{1000}{3} \times 1.2341 \times 10^{-4} + \frac{100}{3} = -0.4113 + 33.3333 \approx 32.9220. \][/tex]
8. Cardiac output:
Multiply by the given coefficient [tex]\( 0.2001 \)[/tex]:
[tex]\[ 0.2001 \times 32.9220 \approx 6.5869. \][/tex]
Thus, the cardiac output is approximately [tex]\( 6.5869 \)[/tex] liters per second (rounded to four decimal places).
