Single Variable Calculus, Chapter 2, 2.1, Section 2.1, Problem 1

The table below shows the value of the volume V of water remaining in the tank (in gallons) after $t$ minutes. Suppose that a tank holds 1000 gallons of water,
which drains from the bottom of the tank in half an hour.


$
\begin{equation}
\begin{aligned}
\begin{array}{|c|c|c|c|c|c|c|}
\hline
t (\text{min}) & 5 & 10 & 15 & 20 & 25 & 30 \\
\hline
V(\text{gal}) & 694 & 444 & 250 & 111 & 28 & 0\\
\hline
\end{array}
\end{aligned}
\end{equation}
$


(a). If $A$ is the point (15,250) on the graph of V, find the slopes of the secant lines $AB$ when $B$ is the point on the graph with $t=5,10,20,2$5 and $30$.

Slope of the secant line $AB$ at $t = 5$

slope = $\displaystyle \frac{694-250}{5-15} = -44.4$

Slope of the secant line $AB$ at $t= 10$

slope = $\displaystyle \frac{444-250}{10-15} = -38.8$

Slope of the secant line $AB$ at $t= 20$

slope = $\displaystyle \frac{111-250}{20-15} = -27.8$

Slope of the secant line $AB$ at $t= 25$

slope = $\displaystyle \frac{28-250}{25-15} = -22.2$

Slope of the secant line $AB$ at $t= 30$

slope = $\displaystyle \frac{0-250}{30-15} = -16.67$

(b). Estimate the slope of the tangent line $A$ by averaging the slopes of two secant lines.


Average = $\displaystyle \frac{-44.4 + (-16.67)}{2} = -30.54$

(c). Use a graph of the function to estimate the slope of the tangent line at $A$.







Based from the graph we obtain, the slope of the tangent line at A is approximately equal to -27.8

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