Question

A tank holds 3000 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 remaining in the tank (in gallons) after t minutes.

t (min) 5 10 15 20 25 30
V (gal) 2097 1308 780 303 87 0
(a) If P is the point (15, 780) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with the following values. (Round your answers to one decimal place.)

Q slope
(5, 2097)
(10, 1308)
(20, 303)
(25, 87)
(30, 0)

Answer 1

Answer:

The slope of the secant line PQ is the following for each point:

Q                         Slope (a)

(5, 2097)              -131.7 gal/min

(10, 1308)              -105.6 gal/min

(20, 303)               -95.4 gal/min

(25, 87)                  -69.3 gal/min

(30, 0)                   -52 gal/min

Step-by-step explanation:

The slope can be easily obtained by using a linear approximation for the 2 points P and Q. In this case, with the slope a:

$f(t)=V(t)=at+b$

$f(t_P)=V_P=at_P+b$             (1)

$f(t_Q)=V_Q=at_Q+b$           (2)

if equations 1 and 2 are subtracted:

$V_P-V_Q=at_P -at_Q=a(t_P -t_Q)\\a=\frac{V_P-V_Q}{t_P -t_Q}$

Answer 2

a) The slopes of the secant lines are listed below:

1. $m_{PQ} = -131.7\,\frac{gal}{min}$
2. $m_{PQ} = -105,6\,\frac{gal}{min}$
3. $m_{PQ} =-95.4\,\frac{gal}{min}$
4. $m_{PQ} = -69.3\,\frac{gal}{min}$
5. $m_{PQ} = -52\,\frac{gal}{min}$

b) The slope of the tangent line is approximately -100.5 gallons per minute.

Application of secant and tangent line in a water tank draining process

a) The value of the slope of line secant ($m_{PQ}$) to a curve is defined by the following expression:

$m_{PQ} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}$, where $x_{Q} \ne x_{P}$ (1)

Now we proceed to calculate for each case:

i) $Q(x,y) = (5, 2097)$

$m_{PQ} = \frac{2097-780}{5-15}$

$m_{PQ} = -131.7\,\frac{gal}{min}$ $\blacksquare$

ii) $Q(x,y) = (10, 1308)$

$m_{PQ} = \frac{1308-780}{10-15}$

$m_{PQ} = -105,6\,\frac{gal}{min}$ $\blacksquare$

iii) $Q(x,y) = (20, 303)$

$m_{PQ} = \frac{303-780}{20-15}$

$m_{PQ} =-95.4\,\frac{gal}{min}$ $\blacksquare$

iv) $Q(x,y) = (25, 87)$

$m_{PQ} = \frac{87-780}{25-15}$

$m_{PQ} = -69.3\,\frac{gal}{min}$ $\blacksquare$

v) $Q(x,y) = (30, 0)$

$m_{PQ} = \frac{0-780}{30-15}$

$m_{PQ} = -52\,\frac{gal}{min}$ $\blacksquare$

ii) The slope of the tangent line at P can be estimated by the following formula:

$m_{P} = \frac{m_{RP}+m_{PQ}}{2}$ (2)

Where $m_{RP}$ and $m_{PQ}$ are secant lines.

If we know that $R(x,y) = (10, 1308)$, $P(x,y) = (15, 780)$ and $Q(x,y) = (20, 303)$, then the slope of tangent line is:

$m_{RP} = \frac{780-1308}{15-10}$

$m_{RP} = -105.6\,\frac{gal}{min}$

$m_{PQ} = \frac{303-780}{20-15}$

$m_{PQ} = -95.4\,\frac{gal}{min}$

$m_{P} = \frac{\left(-105.6\,\frac{gal}{min} \right)+\left(-95.4\,\frac{gal}{min} \right)}{2}$

$m_{P} = -100.5\,\frac{gal}{min}$

The slope of the tangent line is approximately -100.5 gallons per minute. $\blacksquare$

Remark

The statement is incomplete and poorly formatted, the complete and correct form is described below:

A tank holds 3000 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 remaining in the tank (in gallons) after $t$ minutes.

$t(min)$             $V(gal)$

  5                   2097

  10                  1308

  15                   780

  20                  303

  25                    87

  30                     0

(a) If $P$ is the point (15, 780) on the graph of $V$, find the slopes of the secant lines $PQ$ when $Q$ is the point on the graph with the following values. (Round your answers to one decimal place)

   $m_{PQ}$

(5, 2097)

(10, 1308)

(20, 303)

(25, 87)

 (30, 0)

(b) Estimate the slope of the tangent line $P$ by averaging the slopes of two adjacent secant lines. (Round your answer to one decimal place)