Chapter 5.6, Problem 44E

To determine

To find: what is happening to cardiac output.

Answer to Problem 44E

The value of $\frac{dy}{dt}$ is $\text{0}\text{.277 L}\text{./min}{\text{.}}^2$

Explanation of Solution

Given information :

Cardiac output formula-

  $y=\frac{Q}{D}$

D is decreasing.

  $\frac{d\text{D}}{dt}=-2\text{units/min}\text{.}$

Q remains constant.

  $Q=233$

  $D=41$

Calculation:

We have to calculate what is happening to cardiac output.

Since the cardiac output formula is given,

  $\begin{array}{l}y=\frac{Q}{D}\\ y=Q{D}^{-1}\end{array}$

Since Q is constant, we differentiate only y and D with respect to t .

  $\begin{array}{l}\frac{dy}{dt}=Q\left[\left(-1\right)D^{-1-1}\frac{d\text{D}}{dt}\right]\\ \frac{dy}{dt}=Q\left[-D^{-2}\frac{d\text{D}}{dt}\right]\\ \frac{dy}{dt}=-\frac{Q}{D^2}\frac{d\text{D}}{dt}\end{array}$

Putting the value of $Q=233$ , $D=41$ and $\frac{d\text{D}}{dt}=-2$ in above equation,

  $\begin{array}{l}\frac{dy}{dt}=-\frac{233}{{\left(41\right)}^2}\left(-2\right)\\ \frac{dy}{dt}=\frac{233\times 2}{1681}\\ \frac{dy}{dt}=\frac{466}{1681}\\ \frac{dy}{dt}=0.277\text{ L/min}{\text{.}}^2\end{array}$

Hence, the value of $\frac{dy}{dt}$ is $\text{0}\text{.277 L}\text{./min}{\text{.}}^2$