Chapter 2.3, Problem 65E

Solution

a.

To graph the function $v=1.19-1.87r^2$ and to record the viewing window dimensions.

Given the function representing the velocity of blood cells:

  $v=1.19-1.87r^2$ .

Since $r$ represents the radius, it is always positive. That is, $v=1.19-1.87r^2$ is defined only for positive $r$ .

The function $v=1.19-1.87r^2$is graphed for $r\ge 0$:

  

The velocity $v$ of the blood cell is represented along the $y$ -axis and the diameter $r$ is represented along the $x$ -axis.

The dimensions of the viewing window is as follows:

  $\begin{array}{c}{y}_{\max }=1.2\\ y_{\min }=-1\\ x_{\max }=2.6\\ x_{\min }=0\\ x-\text{axis:1}\text{div = 0}\text{.2 cm}\\ y-\text{axis:1}\text{div = 0}\text{.2 cm/sec}\end{array}$ .

b.

To determine the distance from the centre of the artery that a blood cell will travel at a velocity of $0.975\text{cm/sec}$ , as predicted by the given model.

The distance of the cell, travelling at a velocity of $0.975\text{cm/sec}$ , will be approximately $0.34$ centimetres from the artery.

Given the model followed by the velocity of blood particles:

  $v=1.19-1.87r^2$ .

Calculation:

Substitute $v=0.975$ in $v=1.19-1.87r^2$ and solve for a positive $r$ :

  $\begin{array}{l}0.975=1.19-1.87r^2\\ 1.87r^2=1.19-0.975\\ 1.87r^2=0.215\\ r^2=\frac{0.215}{1.87}\\ r=\sqrt{\frac{0.215}{1.87}}\\ r\approx 0.34\end{array}$ .

Thus, $r\approx 0.34$ when $v=0.975$ .

That is, the distance of the cell, travelling at a velocity of $0.975\text{cm/sec}$ , will be approximately $0.34$ centimetres from the artery.

Conclusion:

The distance of the cell, travelling at a velocity of $0.975\text{cm/sec}$ , will be approximately $0.34$ centimetres from the artery.