Chapter 1.4, Problem 48E

To determine

To graph: The volume as a function of age from the data provided between measurement of a growing plant. Also evaluate the equation of line in both point-slope and slope intercept form and volume on day $2.75$ .

$\begin{array}{cccc}\begin{array}{l}\text{Age, a}\\ \left(\text{days}\right)\end{array}& \begin{array}{l}\text{Mass, M}\\ \left(g\right)\end{array}& \begin{array}{l}\text{Volume, V}\\ \left(c{m}^3\right)\end{array}& \begin{array}{l}\text{Glucose production, G}\\ \left(mg\right)\end{array}\\ 0.5& 2.5& 5.1& 0.0\\ 1.0& 4.0& 6.2& 3.4\\ 1.5& 5.5& 7.3& 6.8\\ 2.0& 7.0& 8.4& 10.2\\ 2.5& 8.5& 9.5& 13.6\\ 3.0& 10.0& 10.6& 17.0\end{array}$

Explanation of Solution

Given information:

The table that provides relation between measurement of a growing plant.

$\begin{array}{cccc}\begin{array}{l}\text{Age, a}\\ \left(\text{days}\right)\end{array}& \begin{array}{l}\text{Mass, M}\\ \left(g\right)\end{array}& \begin{array}{l}\text{Volume, V}\\ \left(c{m}^3\right)\end{array}& \begin{array}{l}\text{Glucose production, G}\\ \left(mg\right)\end{array}\\ 0.5& 2.5& 5.1& 0.0\\ 1.0& 4.0& 6.2& 3.4\\ 1.5& 5.5& 7.3& 6.8\\ 2.0& 7.0& 8.4& 10.2\\ 2.5& 8.5& 9.5& 13.6\\ 3.0& 10.0& 10.6& 17.0\end{array}$

Graph:

Consider the table that provides relation between measurement of a growing plant.

$\begin{array}{cccc}\begin{array}{l}\text{Age, a}\\ \left(\text{days}\right)\end{array}& \begin{array}{l}\text{Mass, M}\\ \left(g\right)\end{array}& \begin{array}{l}\text{Volume, V}\\ \left(c{m}^3\right)\end{array}& \begin{array}{l}\text{Glucose production, G}\\ \left(mg\right)\end{array}\\ 0.5& 2.5& 5.1& 0.0\\ 1.0& 4.0& 6.2& 3.4\\ 1.5& 5.5& 7.3& 6.8\\ 2.0& 7.0& 8.4& 10.2\\ 2.5& 8.5& 9.5& 13.6\\ 3.0& 10.0& 10.6& 17.0\end{array}$

To graph the volume as a function of age, on the horizontal axis that is x- axis, plot the age and on vertical axis that is y -axis plot the volume.

Now plot the points $\left(0.5,5.1\right),\left(1,6.2\right),\left(1.5,7.3\right),\left(2.0,8.4\right),\left(2.5,9.5\right),\left(3.0,10.6\right)$ on the graph and connect the dots through line.

  

Interpretation:

The point slope form of a line is $y=mx+\left(y_0-m{x}_0\right)$ where m is the slope and $\left(x_0,y_0\right)$ is the point through line the line passes.

The slope-intercept form of a line is $y=mx+c$ where m is the slope and c is the intercept of the line.

The slope is $m=\frac{y-y_0}{x-x_0}$ where $\left(x,y\right)$ is an arbitrary point and $\left(x_0,y_0\right)$ is the base point.

Here independent variable is x and dependent variable is y.

According to the question, volumeV is dependent variable and age a is independent variable.

Slope of the line is evaluate when first two points are considered that is, $\left(0.5,5.1\right)\text{ and }\left(1,6.2\right)$ ,

$\begin{array}{c}m=\frac{6.2-5.1}{1-0.5}\\ =2.2\end{array}$

Slope of the line is $2.2$ .

Equation of line in point slope form is, consider $\left(0.5,5.1\right)$ as the base point.

$\begin{array}{c}V=mx+\left(y_0-m{x}_0\right)\\ =2.2a+\left(5.1-2.2\left(0.5\right)\right)\\ =2.2a+4\end{array}$

Therefore, the equation in point-slope form is $V=2.2a+4$ .

Intercept on y -axis is 4 when compared with the equation in slope-intercept form $y=mx+c$ .

The above statement is interpreted as volume of the seed when age is 0 years is $4.0{\text{ cm}}^3$ .

Biologically this could be the volume of a new seedling.

Next, volume of seed on day $2.75$ , substitute a as $2.75$ in the equation of line.

$\begin{array}{c}V=2.2\left(2.75\right)+4\\ =10.05\end{array}$

Therefore, volume is $10.05{\text{ cm}}^3$ on day $2.75$ .

Thus, equation in slope-intercept and point-slope form is $V=2.2a+4$ . All points lie on the line and volume is $10.05{\text{ cm}}^3$ on day $2.75$ .