Chapter 1.4, Problem 49E

To determine

To graph: The glucose production as a function of mass 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 glucose productionwhen mass reaches

$20.0\text{ g}$ .

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

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 glucose production as a function of mass, on the horizontal axis that is x- axis, plot the mass and on vertical axis that is y -axis plot the glucose production.

Now plot the points $\left(2.5,0.0\right),\left(4.0,3.4\right),\left(5.5,6.8\right),\left(7.0,10.2\right),\left(8.5,13.6\right),\left(10.0,17.0\right)$ 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, glucose productionG is dependent variable and massMis independent variable.

Slope of the line is evaluate when first two points are considered that is, $\left(2.5,0.0\right)\text{ and }\left(4.0,3.4\right)$ ,

$\begin{array}{c}m=\frac{3.4-0.0}{4-2.5}\\ =2.27\end{array}$

Slope of the line is $2.27$ .

Equation of line in point slope form is, consider $\left(2.5,0.0\right)$ as the base point.

$\begin{array}{c}G=mx+\left(y_0-m{x}_0\right)\\ =2.27M+\left(0.0-2.27\left(2.5\right)\right)\\ =2.27M-5.7\end{array}$

Therefore, the equation in point-slope form is $G=2.27M-5.7$ .

Intercept on y -axis is $-5.7$ when compared with the equation in slope-intercept form $y=mx+c$ .

The above statement is interpreted as plant won’t produce glucose until the mass go beyond a particular value.

Next, glucose productionwhen mass reaches $20.0\text{ g}$ is substitute Mas $20.0$ in the equation of line.

$\begin{array}{c}G=2.27\left(20.0\right)-5.7\\ =39.7\end{array}$

Therefore, glucose production is $39.7mg$ when mass reaches $20.0\text{ g}$ .

Thus, equation in slope-intercept and point-slope form is $G=2.27M-5.7$ . All points lie on the line and glucose production is $39.7mg$ when mass reaches $20.0\text{ g}$ .