Question

show the volume ( v ) of water remaining in the tank (in gallons) after ( t ) minutes.

( t ) (min)	5	10	15	20	25	30

(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 = 5, 10, 20, 25, ) and ( 30 ).
(b) estimate the slope of the tangent line at ( p ) by averaging the slopes of two secant lines.
(c) use a graph of the function 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.)

1. a cardiac monitor is used to measure the heart rate of a patient after surgery. it compiles the number of heartbeats after ( t ) min-

( t ) (min)	36	38	40	42	44

the monitor estimates this value by calculating the slope of a secant line. use the data to estimate the patient’s heart rate after 42 minutes using the secant line between the points with the given values of ( t ).
(a) ( t = 36 ) and ( t = 42 )
(b) ( t = 38 ) and ( t = 42 )
(c) ( t = 40 ) and ( t = 42 )
(d) ( t = 42 ) and ( t = 44 )
what are your conclusions?

1. the point ( p(2, -1) ) lies on the curve ( y = 1/(1 - x) ).

(a) if ( q ) is the point ( (x, 1/(1 - x)) ), use your calculator to find the slope of the secant line ( pq ) (correct to six decimal places) for the following values of ( x ):
(i) ( 1.5 )
(ii) ( 1.9 )
(iii) ( 1.99 )
(iv) ( 1.999 )
(v) ( 2.5 )
(vi) ( 2.1 )
(vii) ( 2.01 )
(viii) ( 2.001 )

Explanation:

Problem 1

Step1: Define secant slope formula

Slope: $m = \frac{V(t)-V(15)}{t-15}$

Step2: Calculate for t=5

$m = \frac{694-250}{5-15} = \frac{444}{-10} = -44.4$

Step3: Calculate for t=10

$m = \frac{444-250}{10-15} = \frac{194}{-5} = -38.8$

Step4: Calculate for t=20

$m = \frac{211-250}{20-15} = \frac{-39}{5} = -7.8$

Step5: Calculate for t=30

$m = \frac{28-250}{30-15} = \frac{-222}{15} = -14.8$

Step6: Average slopes for t=10,20

$m_{avg} = \frac{-38.8 + (-7.8)}{2} = \frac{-46.6}{2} = -23.3$

Step7: Estimate tangent slope (graphical)

Visually, tangent at t=15 is between secant slopes, ~$-23$ (matches average closely)

Step1: Define heart rate formula

Heart rate = $\frac{V(t_2)-V(t_1)}{t_2-t_1}$ beats/min

Step2: Calculate (a) t=36,42

$m = \frac{3080-2530}{42-36} = \frac{550}{6} \approx 91.67$

Step3: Calculate (b) t=38,42

$m = \frac{3080-2661}{42-38} = \frac{419}{4} = 104.75$

Step4: Calculate (c) t=40,42

$m = \frac{3080-2806}{42-40} = \frac{274}{2} = 137$

Step5: Calculate (d) t=42,44

$m = \frac{3080-2948}{44-42} = \frac{132}{2} = 66$

Step6: Summarize trend

As $t$ approaches 42, heart rate spikes then drops.

Step1: Define secant slope formula

Slope: $m = \frac{Q(x)-P(2)}{x-2} = \frac{\frac{1}{1-x} - (-1)}{x-2} = \frac{\frac{1}{1-x}+1}{x-2} = \frac{\frac{2-x}{1-x}}{x-2} = \frac{1}{x-1}$

Step2: Calculate for each x

(i) $x=1.5$: $\frac{1}{1.5-1} = 2.000000$
(ii) $x=1.9$: $\frac{1}{1.9-1} \approx 1.111111$
(iii) $x=1.99$: $\frac{1}{1.99-1} \approx 1.010101$
(iv) $x=1.999$: $\frac{1}{1.999-1} \approx 1.001001$
(v) $x=2.5$: $\frac{1}{2.5-1} \approx 0.666667$
(vi) $x=2.1$: $\frac{1}{2.1-1} \approx 0.909091$
(vii) $x=2.01$: $\frac{1}{2.01-1} \approx 0.990099$
(viii) $x=2.001$: $\frac{1}{2.001-1} \approx 0.999001$

Answer:

(a)

• t=5: $-44.4$
• t=10: $-38.8$
• t=20: $-7.8$
• t=30: $-14.8$

(b) $-23.3$
(c) Approximately $-23$ gallons per minute (rate water flows out)

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Problem 2