Question

researchers measured the average blood alcohol concentration c(t) of eight men starting one hour after consumption of 30 ml of ethanol (corresponding to two alcoholic drinks). selected values of c(t) are given in the table. (round your answers to two decimal places.)

t (hours)	1.0	1.5	2.0	2.5	3.0

(a) find the average change of c with respect to t over each time interval.

(i) 1.0, 2.0

(ii) 1.5, 2.0

(iii) 2.0, 2.5

(iv) 2.0, 3.0

(b) estimate the instantaneous rate of change at t = 2 and interpret your result.

what are the units?

○ h
○ \\(\frac{\text{dl/g}}{\text{h}}\\)
○ \\(\frac{\text{g/dl}}{\text{h}}\\)
○ g/dl
○ \\(\frac{\text{h}}{\text{g/dl}}\\)
○ \\(\frac{\text{h}}{\text{dl/g}}\\)

Explanation:

Part (a)

The average rate of change of a function \( C(t) \) over the interval \([a, b]\) is given by the formula:
\[
\text{Average rate of change} = \frac{C(b) - C(a)}{b - a}
\]

(i) Interval \([1.0, 2.0]\)

Step1: Identify \( C(1.0) \) and \( C(2.0) \)

From the table: \( C(1.0) = 0.34 \, \text{g/dL} \), \( C(2.0) = 0.18 \, \text{g/dL} \), \( b - a = 2.0 - 1.0 = 1.0 \, \text{hours} \).

Step2: Apply the average rate of change formula

\[
\frac{C(2.0) - C(1.0)}{2.0 - 1.0} = \frac{0.18 - 0.34}{1.0} = \frac{-0.16}{1.0} = -0.16
\]

(ii) Interval \([1.5, 2.0]\)

Step1: Identify \( C(1.5) \) and \( C(2.0) \)

From the table: \( C(1.5) = 0.24 \, \text{g/dL} \), \( C(2.0) = 0.18 \, \text{g/dL} \), \( b - a = 2.0 - 1.5 = 0.5 \, \text{hours} \).

Step2: Apply the average rate of change formula

\[
\frac{C(2.0) - C(1.5)}{2.0 - 1.5} = \frac{0.18 - 0.24}{0.5} = \frac{-0.06}{0.5} = -0.12
\]

(iii) Interval \([2.0, 2.5]\)

Step1: Identify \( C(2.0) \) and \( C(2.5) \)

From the table: \( C(2.0) = 0.18 \, \text{g/dL} \), \( C(2.5) = 0.12 \, \text{g/dL} \), \( b - a = 2.5 - 2.0 = 0.5 \, \text{hours} \).

Step2: Apply the average rate of change formula

\[
\frac{C(2.5) - C(2.0)}{2.5 - 2.0} = \frac{0.12 - 0.18}{0.5} = \frac{-0.06}{0.5} = -0.12
\]

(iv) Interval \([2.0, 3.0]\)

Step1: Identify \( C(2.0) \) and \( C(3.0) \)

From the table: \( C(2.0) = 0.18 \, \text{g/dL} \), \( C(3.0) = 0.08 \, \text{g/dL} \), \( b - a = 3.0 - 2.0 = 1.0 \, \text{hours} \).

Step2: Apply the average rate of change formula

\[
\frac{C(3.0) - C(2.0)}{3.0 - 2.0} = \frac{0.08 - 0.18}{1.0} = \frac{-0.10}{1.0} = -0.10
\]

Part (b)

To estimate the instantaneous rate of change at \( t = 2 \), we use the average rates of change of the intervals surrounding \( t = 2 \) (e.g., \([1.5, 2.0]\) and \([2.0, 2.5]\)).

Step1: Average the rates from \([1.5, 2.0]\) and \([2.0, 2.5]\)

From part (a):

• Rate over \([1.5, 2.0]\): \( -0.12 \)
• Rate over \([2.0, 2.5]\): \( -0.12 \)

\[
\text{Instantaneous rate} \approx \frac{-0.12 + (-0.12)}{2} = \frac{-0.24}{2} = -0.12
\]

Units

The rate of change of \( C(t) \) (in \(\text{g/dL}\)) with respect to \( t \) (in hours) has units of \(\frac{\text{g/dL}}{\text{hour}}\) (or \(\frac{\text{g/dL}}{\text{h}}\)).

Final Answers

(a)

(i) \(\boldsymbol{-0.16}\)
(ii) \(\boldsymbol{-0.12}\)
(iii) \(\boldsymbol{-0.12}\)
(iv) \(\boldsymbol{-0.10}\)

(b)

Instantaneous rate: \(\boldsymbol{-0.12}\) (units: \(\boldsymbol{\frac{\text{g/dL}}{\text{h}}}\))

Answer:

Part (a)

The average rate of change of a function \( C(t) \) over the interval \([a, b]\) is given by the formula:
\[
\text{Average rate of change} = \frac{C(b) - C(a)}{b - a}
\]

(i) Interval \([1.0, 2.0]\)

Step1: Identify \( C(1.0) \) and \( C(2.0) \)

From the table: \( C(1.0) = 0.34 \, \text{g/dL} \), \( C(2.0) = 0.18 \, \text{g/dL} \), \( b - a = 2.0 - 1.0 = 1.0 \, \text{hours} \).

Step2: Apply the average rate of change formula

\[
\frac{C(2.0) - C(1.0)}{2.0 - 1.0} = \frac{0.18 - 0.34}{1.0} = \frac{-0.16}{1.0} = -0.16
\]

(ii) Interval \([1.5, 2.0]\)

Step1: Identify \( C(1.5) \) and \( C(2.0) \)

From the table: \( C(1.5) = 0.24 \, \text{g/dL} \), \( C(2.0) = 0.18 \, \text{g/dL} \), \( b - a = 2.0 - 1.5 = 0.5 \, \text{hours} \).

Step2: Apply the average rate of change formula

\[
\frac{C(2.0) - C(1.5)}{2.0 - 1.5} = \frac{0.18 - 0.24}{0.5} = \frac{-0.06}{0.5} = -0.12
\]

(iii) Interval \([2.0, 2.5]\)

Step1: Identify \( C(2.0) \) and \( C(2.5) \)

From the table: \( C(2.0) = 0.18 \, \text{g/dL} \), \( C(2.5) = 0.12 \, \text{g/dL} \), \( b - a = 2.5 - 2.0 = 0.5 \, \text{hours} \).

Step2: Apply the average rate of change formula

\[
\frac{C(2.5) - C(2.0)}{2.5 - 2.0} = \frac{0.12 - 0.18}{0.5} = \frac{-0.06}{0.5} = -0.12
\]

(iv) Interval \([2.0, 3.0]\)

Step1: Identify \( C(2.0) \) and \( C(3.0) \)

From the table: \( C(2.0) = 0.18 \, \text{g/dL} \), \( C(3.0) = 0.08 \, \text{g/dL} \), \( b - a = 3.0 - 2.0 = 1.0 \, \text{hours} \).

Step2: Apply the average rate of change formula

\[
\frac{C(3.0) - C(2.0)}{3.0 - 2.0} = \frac{0.08 - 0.18}{1.0} = \frac{-0.10}{1.0} = -0.10
\]

Part (b)

To estimate the instantaneous rate of change at \( t = 2 \), we use the average rates of change of the intervals surrounding \( t = 2 \) (e.g., \([1.5, 2.0]\) and \([2.0, 2.5]\)).

Step1: Average the rates from \([1.5, 2.0]\) and \([2.0, 2.5]\)

From part (a):

• Rate over \([1.5, 2.0]\): \( -0.12 \)
• Rate over \([2.0, 2.5]\): \( -0.12 \)

\[
\text{Instantaneous rate} \approx \frac{-0.12 + (-0.12)}{2} = \frac{-0.24}{2} = -0.12
\]

Units

The rate of change of \( C(t) \) (in \(\text{g/dL}\)) with respect to \( t \) (in hours) has units of \(\frac{\text{g/dL}}{\text{hour}}\) (or \(\frac{\text{g/dL}}{\text{h}}\)).

Final Answers

(a)

(i) \(\boldsymbol{-0.16}\)
(ii) \(\boldsymbol{-0.12}\)
(iii) \(\boldsymbol{-0.12}\)
(iv) \(\boldsymbol{-0.10}\)

(b)

Instantaneous rate: \(\boldsymbol{-0.12}\) (units: \(\boldsymbol{\frac{\text{g/dL}}{\text{h}}}\))