Question

A tank hold 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaning in the tank (in gallons) after t minutes.

t(min)	5	10	15	20	25	30
V(gal)	694	444	250	111	28	0

(a) If P is the point (15, 250) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with \(t = {\bf{5}},\;{\bf{10}}{\rm{,}}\,{\bf{20}}{\rm{,}}\,{\bf{25}}{\rm{,}}\,{\bf{and}}\,\,{\bf{30}}\).

(b) Estimate the slope of the tangent line at P by averaging the slopes of two secant ines.

(c) Use a graph of V to estimate the slope of the tangent line at P. (This slope represents the rate at which the water is flowing from the tank after 15 minutes.)

Short answer

(a)

\(t\)	\(Q\)	\({m_{PQ}}\)
5	(5, 694)	-44.4
10	(10, 444)	-38.8
20	(20, 111)	-27.8
25	(25, 28)	-22.2
30	(30, 0)	-16.6

(b) The slope of the line is \( - 33.3\).

(c) slope of the line is \( - 33.3\).

Step-by-step solution

Step 1:Find slopes of the secant line

Find the slope of secant line at \(t = 5\) (\(\left( {5,694} \right)\)).

\(\begin{aligned}{c}{m_{PQ}} = \frac{{694 - 250}}{{5 - 15}}\\ = - \frac{{444}}{{10}}\\ = - 44.4\end{aligned}\)

The table below represents the slope of secant line for remaning points.

\(t\)	\(Q\)	\({m_{PQ}}\)
5	(5, 694)	-44.4
10	(10, 444)	-38.8
20	(20, 111)	-27.8
25	(25, 28)	-22.2
30	(30, 0)	-16.6

Find slopes of the tangent line at P

Find the slope of the tangent line at P , by using average of slope at \(t = 10\)and\(t = 20\).

\(\begin{aligned}{c}m &= \frac{{ - 38.8 + \left( { - 27.8} \right)}}{2}\\ &= - 33.3\end{aligned}\)

Thus, the average slope of tangent line at Pis \( - 33.3\).

Find slopes of the tangent line at P

The figure below represents the graph of V with respect to t.

So, the slope of tangent line at P can be calculated as,

\(\frac{{ - 300}}{9} = - 33.3\)

Thus, the slope of the line is \( - 33.3\).