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

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

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=(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\,(gal)/(min)`
2. 
   `m_(PQ) = -105,6\,(gal)/(min)`
3. 
   `m_(PQ) =-95.4\,(gal)/(min)`
4. 
   `m_(PQ) = -69.3\,(gal)/(min)`
5. 
   `m_(PQ) = -52\,(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) = (y_(Q)-y_(P))/(x_(Q)-x_(P))`, where
`x_(Q) \\e x_(P)` (1)

Now we proceed to calculate for each case:

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

`m_(PQ) = (2097-780)/(5-15)`

`m_(PQ) = -131.7\,(gal)/(min)`
`\blacksquare`

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

`m_(PQ) = (1308-780)/(10-15)`

`m_(PQ) = -105,6\,(gal)/(min)`
`\blacksquare`

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

`m_(PQ) = (303-780)/(20-15)`

`m_(PQ) =-95.4\,(gal)/(min)`
`\blacksquare`

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

`m_(PQ) = (87-780)/(25-15)`

`m_(PQ) = -69.3\,(gal)/(min)`
`\blacksquare`

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

`m_(PQ) = (0-780)/(30-15)`

`m_(PQ) = -52\,(gal)/(min)`
`\blacksquare`

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

`m_(P) = (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) = (780-1308)/(15-10)`

`m_(RP) = -105.6\,(gal)/(min)`

`m_(PQ) = (303-780)/(20-15)`

`m_(PQ) = -95.4\,(gal)/(min)`

`m_(P) = (\left(-105.6\,(gal)/(min) \right)+\left(-95.4\,(gal)/(min) \right))/(2)`

`m_(P) = -100.5\,(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)