Question

1. a student bought a smartwatch that tracks the number of steps she walks throughout the day. the table shows the number of steps recorded t minutes after 3:00 p.m. on the first day she wore the watch.

t (min)	0	10	20	30	40
steps	3438	4559	5622	6536	7398

a. find the slopes of the secant lines corresponding to the given intervals of t. what do these slopes represent?
i. 0, 40
ii. 10, 20
iii. 20, 30

b. estimate the student’s walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines.

Explanation:

Step1: Recall slope formula for secant line

The slope of a secant line between two points $(t_1,y_1)$ and $(t_2,y_2)$ is $m=\frac{y_2 - y_1}{t_2 - t_1}$, where $y$ represents the number of steps.

Step2: Calculate slope for $[0,40]$

Let $(t_1,y_1)=(0,3438)$ and $(t_2,y_2)=(40,7398)$. Then $m_1=\frac{7398 - 3438}{40-0}=\frac{3960}{40}=99$. This slope represents the average rate of change of the number of steps from 3:00 p.m. to 3:40 p.m. in steps per minute.

Step3: Calculate slope for $[10,20]$

Let $(t_1,y_1)=(10,4559)$ and $(t_2,y_2)=(20,5622)$. Then $m_2=\frac{5622 - 4559}{20 - 10}=\frac{1063}{10}=106.3$. This slope represents the average rate of change of the number of steps from 3:10 p.m. to 3:20 p.m. in steps per minute.

Step4: Calculate slope for $[20,30]$

Let $(t_1,y_1)=(20,5622)$ and $(t_2,y_2)=(30,6536)$. Then $m_3=\frac{6536 - 5622}{30 - 20}=\frac{914}{10}=91.4$. This slope represents the average rate of change of the number of steps from 3:20 p.m. to 3:30 p.m. in steps per minute.

Step5: Estimate walking - pace at 3:20 p.m.

To estimate the walking - pace at 3:20 p.m., we average the slopes of the secant lines $[10,20]$ and $[20,30]$. The average is $\frac{m_2 + m_3}{2}=\frac{106.3+91.4}{2}=\frac{197.7}{2}=98.85$.

Answer:

a.
i. The slope for $[0,40]$ is 99.
ii. The slope for $[10,20]$ is 106.3.
iii. The slope for $[20,30]$ is 91.4.
b. The estimated walking pace at 3:20 p.m. is 98.85 steps per minute.