Emmy is a civil engineer who has to approximate the slope of the water table along a certain line in the Hanford Nuclear Reservation. She can dig only three test holes to evaluate the depth of groundwater at certain points.
She finds that the water table lies at $(0,35)$, $(300,38)$, and $(600,42)$, with all distances given in feet and positive vertical distances representing distances below the surface.

(a) Using the data from the first two wells, estimate the slope of the water table. Use this slope to write an equation of a line that describes the water table.

(b) Considering the equation of the line from part (a) as the first two terms of a Maclaurin series for the function $w\left(x\right)$ describing the water table, write an
expression for the error in this linear approximation.

(c) Verify that the graph of the quadratic

$w_2\left(x\right)=\frac{1}{180000}{x}^2+\frac{1}{120}x+35$

passes through the three data points describing the
water table.

(c) Use the quadratic from part (c) to estimate the error
in the linear approximation to the water table.

Consider the points $(0,35)$, $(300,38)$, and $(600,42)$.

The objective is to find the equation of the line that defines the slope of the water table.

The slope of the line is $m=\frac{y_2-y_1}{x_2-x_1}$.
Thus,
$\begin{array}{rcl}m& =& \frac{38-35}{300-0}\\ & =& \frac{3}{300}\\ & =& \frac{1}{100}\end{array}$

The formula for the equation of line is $y-y_1=m\left(x-x_1\right)$.
Thus, $w\left(x\right)-35=\frac{1}{100}(x-0)$.
Therefore, the equation of the line that defines the slope of the water table is
$w\left(x\right)=\frac{1}{100}x+35\text{.}$

If $f$is a function that can be differentiated $n+1$times in some open interval $I$containing the points $x_0$, and let $R_n\left(x\right)$be the $n$th remainder for $f$at $x=x_0$, then there exists at least one $c$ between $x_0$and $x$such that $R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}$, here, $R_n\left(x\right)$ can also be termed as error in the linear approximation.
The function $w\left(x\right)$is differentiable two times, therefore, $n+1=2$.
Therefore, the error in approximation is $R_1\left(x\right)=\frac{w^{\text{'}\text{'}}\left(c\right)}{2!}{x}^2$.
Hence, the expression for the error in this linear approximation is $R_1\left(x\right)=\frac{w^{\text{'}\text{'}}\left(c\right)}{2}{x}^2$, here, $c\in [0,300]$

Consider the function $w_2\left(x\right)=\frac{1}{180000}{x}^2+\frac{1}{120}x+35$.
The objective is to verify that the above function passes through three points $(0,35),(300,38)$and $(600,42)$.
Substitute $x=0$in the function $w_2\left(x\right)=\frac{1}{180000}{x}^2+\frac{1}{120}x+35$
$\begin{array}{rcl}{w}_2\left(x\right)& =& \frac{1}{180000}{x}^2+\frac{1}{120}x+35\\ & =& 35\end{array}$

Thus, when $x=0$the value of $w_2\left(x\right)$is $35$. Therefore, the point $(0,35)$is satisfied. Substitute $x=300$in the function $w_2\left(x\right)=\frac{1}{180000}{x}^2+\frac{1}{120}x+35$.
$\begin{array}{rcl}{w}_2\left(x\right)& =& \frac{1}{180000}{x}^2+\frac{1}{120}x+35\\ & =& \frac{1}{180000}\times 90000+\frac{1}{120}\times 300+35\\ & =& \frac{1}{2}+\frac{5}{2}+35\\ & =& 38\end{array}$

Thus, when $x=300$the value of $w_2\left(x\right)$is $38$. Therefore, the point $(300,38)$is satisfied. Substitute $x=600$in the function $\begin{array}{rcl}{w}_2\left(x\right)& =& \frac{1}{180000}{x}^2+\frac{1}{120}x+35\\ & =& \frac{1}{180000}\times 360000+\frac{1}{120}\times 600+35\\ & =& 2+5+35\\ & =& 42\end{array}$.

Thus, when $x=600$ the value of$w_2\left(x\right)$is $42$ . Therefore, the point $(600,42)$ is satisfied.

Assume that the quadratic is somewhat close to $w\left(t\right)$, the error term is approximated by $R_i\left(x\right)=\frac{w^{\text{'}\text{'}}\left(c\right)}{2!}{x}^2$, here, $c\in [0,300]$. This is bounded above by

Thus, the estimation is that the line is at most half a foot off the table over the entire $300$-foot interval.