(a.)
To sketch: a graph of elevation
(a.)
Answer to Problem 30E
The required graph is shown.
Explanation of Solution
Given:
Elevation Along Bear Creek | |
Distance Downriver | River Elevation |
0.00 | 1577 |
0.56 | 1512 |
0.92 | 1448 |
1.19 | 1384 |
1.30 | 1319 |
1.39 | 1255 |
1.57 | 1191 |
1.74 | 1126 |
1.98 | 1062 |
2.18 | 998 |
2.41 | 933 |
2.64 | 869 |
3.24 | 805 |
Calculation:
Plot the points on graph and join them
(b.)
To obtain: an approximate graph of the derivative.
(b.)
Answer to Problem 30E
The graph of the derivative is shown.
Explanation of Solution
Given:
Elevation Along Bear Creek | |
Distance Downriver | River Elevation |
0.00 | 1577 |
0.56 | 1512 |
0.92 | 1448 |
1.19 | 1384 |
1.30 | 1319 |
1.39 | 1255 |
1.57 | 1191 |
1.74 | 1126 |
1.98 | 1062 |
2.18 | 998 |
2.41 | 933 |
2.64 | 869 |
3.24 | 805 |
Concept used:
The derivative of a function is represented by slope of graph.
Calculation:
Create a table with mid point of each interval and their respective slopes.
Elevation Along Bear Creek | |
Distance Downriver | River Elevation |
Now plot the above points on the graph and join them.
(c.)
The units of measure would be appropriate for a gradient in this problem.
(c.)
Answer to Problem 30E
The unit of gradient is feet per mile.
Explanation of Solution
Given:
Elevation Along Bear Creek | |
Distance Downriver (miles) | River Elevation (feet) |
0.00 | 1577 |
0.56 | 1512 |
0.92 | 1448 |
1.19 | 1384 |
1.30 | 1319 |
1.39 | 1255 |
1.57 | 1191 |
1.74 | 1126 |
1.98 | 1062 |
2.18 | 998 |
2.41 | 933 |
2.64 | 869 |
3.24 | 805 |
Concept used:
The average change in elevation over a given distance is called a gradient.
Conclusion:
Here, the elevation is given in feet and the distance down the river is given in miles.
Thus, the units of the gradient will be feet per mile.
(d.)
The units of measure would be appropriate for derivative.
(d.)
Answer to Problem 30E
The units of the gradient will be feet per mile.
Explanation of Solution
Given:
Elevation Along Bear Creek | |
Distance Downriver (miles) | River Elevation (feet) |
0.00 | 1577 |
0.56 | 1512 |
0.92 | 1448 |
1.19 | 1384 |
1.30 | 1319 |
1.39 | 1255 |
1.57 | 1191 |
1.74 | 1126 |
1.98 | 1062 |
2.18 | 998 |
2.41 | 933 |
2.64 | 869 |
3.24 | 805 |
Concept used:
Derivative is given by the slope.
Here, the average change in elevation over a given distance is slope.
Conclusion:
Here, the elevation is given in feet and the distance down the river is given in miles.
Thus, the units of the gradient will be feet per mile.
(e)
To identify: The most dangerous section of the river by analyzing the graph in (a).
(e)
Answer to Problem 30E
The most dangerous section of the river is from 1.2 to 1.5 miles.
Explanation of Solution
Given:
Elevation Along Bear Creek | |
Distance Downriver (miles) | River Elevation (feet) |
0.00 | 1577 |
0.56 | 1512 |
0.92 | 1448 |
1.19 | 1384 |
1.30 | 1319 |
1.39 | 1255 |
1.57 | 1191 |
1.74 | 1126 |
1.98 | 1062 |
2.18 | 998 |
2.41 | 933 |
2.64 | 869 |
3.24 | 805 |
Conclusion:
In the graph, the steepest part which is from 1.2 to 1.5 miles.
Chapter 2 Solutions
AP CALCULUS TEST PREP-WORKBOOK
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