Dye dilution is a method of measuring cardiac output. If $A \mathrm { mg }$ of dye is used and $c ( t )$ is the concentration of the dye at time $t$, then the cardiac output over the time interval $[ 0 , T ]$ is given by $F = \frac { A } { \int _ { 0 } ^ { T } c ( t ) d t }$ Find the cardiac output over the time interval $( 0,15 )$ if the dye dilution method is used with $10 \mathrm { mg }$ of dye and the dye concentration, in mg/L, is modeled by
$c ( t ) = \frac { 1 } { 2 } t ( 15 - t ) , 0 \leq t \leq 15$
where $t$ is measured in seconds.
A) $0.029$
B) $0.039$
C) $0.026$
D) $0.044$
E) $0.036$
F) $0.020$

Question

Dye dilution is a method of measuring cardiac output. If $A \mathrm { mg }$ of dye is used and $c ( t )$ is the concentration of the dye at time $t$, then the cardiac output over the time interval $[ 0 , T ]$ is given by $F = \frac { A } { \int _ { 0 } ^ { T } c ( t ) d t }$ Find the cardiac output over the time interval $( 0,15 )$ if the dye dilution method is used with $10 \mathrm { mg }$ of dye and the dye concentration, in mg/L, is modeled by 
$c ( t ) = \frac { 1 } { 2 } t ( 15 - t ) , 0 \leq t \leq 15$
where $t$ is measured in seconds.
A) $0.029$
B) $0.039$
C) $0.026$
D) $0.044$
E) $0.036$
F) $0.020$

Quizplus · Accepted Answer

To find the cardiac output, we first need to calculate the integral of $c(t)$ from $0$ to $15$. The integral of $c(t) = \frac{1}{2}t(15-t)$ is $\int_{0}^{15} \frac{1}{2}t(15-t) dt$. This simplifies to $\frac{1}{2}\int_{0}^{15} (15t - t^2) dt$, which further simplifies to $\frac{1}{2}[\frac{15t^2}{2} - \frac{t^3}{3}]_{0}^{15}$. Evaluating this from $0$ to $15$ gives $\frac{1}{2}[\frac{15*225}{2} - \frac{3375}{3}]$ = $\frac{1}{2}[1687.5 - 1125] = \frac{1}{2}[562.5] = 281.25$.The cardiac output $F$ is then $F = \frac{A}{\int_{0}^{15} c(t) dt} = \frac{10}{281.25} = 0.0355$, which rounds to $0.036$ when considering the options provided.

Dye Dilution Is a Method of Measuring Cardiac Output AmgA \mathrm { mg }Amg

Dye Dilution Is a Method of Measuring Cardiac Output $A \mathrm { mg }$