Chapter 1.4, Problem 47E

To determine

To graph: The mass 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 mass on day $1.75$ .

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

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 mass 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 mass.

Now plot the points $\left(0.5,2.5\right),\left(1,4.0\right),\left(1.5,5.5\right),\left(2.0,7.0\right),\left(2.5,8.5\right),\left(3.0,10.0\right)$ 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, mass Mis dependent variable and ageais independent variable.

Slope of the line is evaluate when first two points are considered that is, $\left(0.5,2.5\right)\text{ and }\left(1,4.0\right)$ ,

$\begin{array}{c}m=\frac{4-2.5}{1-0.5}\\ =3\end{array}$

Slope of the line is 3.

Equation of line in point slope form is, consider $\left(0.5,2.5\right)$ as the base point.

$\begin{array}{c}M=mx+\left(y_0-m{x}_0\right)\\ =3a+\left(2.5-3\left(0.5\right)\right)\\ =3a+1\end{array}$

Therefore, the equation in point-slope form is $M=3a+1$ .

Intercept on y -axis is 1 when compared with the equation in slope-intercept form $y=mx+c$ .

The above statement is interpreted as mass of the seed when age is 0 years is $1.0\text{ g}$ .

Biologically this could be the mass of a new seedling.

Next, mass of seed on day $1.75$ , substitute a as $1.75$ in the equation of line.

$\begin{array}{c}M=3\left(1.75\right)+1\\ =6.25\end{array}$

Therefore, mass is $6.25\text{ g}$ on day $1.75$ .

Thus, equation in slope-intercept and point-slope formis $M=3a+1$ . All points lie on the line and mass is $6.25\text{ g}$ on day $1.75$ .