9 FURTHER APPLICATIONS OF INTEGRATION

f (x) = 1 2 x2 − 1 4 ln x. 486
CHAPTER 9
FURTHER APPLICATIONS OF INTEGRATION
descent always runs perpendicular to the contour lines.
Notice that this exercise involves many computations and estimates. The students must divide up the work in
order to finish in a reasonable amount of time. Make sure to have several rulers on hand.
After the exercise is completed, poll the groups to see whose path used the least cable.
Answers (will vary):
1. Lake Wobegone
Bill Turner’s
House
Cable Box
12
10
11
20
8
9
10
30
40
50
7
1 23 4 5 6
60
50
40
30
20
10
Cable Source
The path above was chosen so that the line segments are approximately orthogonal to the contour lines.
1
2.
4
inch on the map represents 100 ft. In estimating the amount of
10
cable needed, we will assume that a cross-sectional view of the path
20
between two contour lines looks like the figure at right.
Segment 1 2 3 4 5 6 7 8 9 10 11 12
Horizontal 30 25 25 25 50 125 200 75 75 90 150 85
Vertical 10 10 10 10 10 10 10 10 10 10 10 10
Length 31.6 26.9 26.9 26.9 51.0 125.4 200.2 75.7 75.7 90.6 150.3 85.6
The total length is about 966.8 ft, so we should probably order at least 1000 ft to be safe.
GROUP WORK 3: Find the Function
This is a tricky problem that reduces to solving the equation
integrating. The fact that the graph of f passes through the point
√
1 + [ f ′ (x) ] 2 = x +
1
( )
1, 1 2
4x for f ′ (x) and then
is key, as it allows the students to
determine the constant of integration.
Answer: ∫ x
1
√1 + [ f ′ (t) ] √
2 dt =
1
2
x 2 + 1 4 ln x ⇔ 1 + [ f ′ (x) ] 2 1 = x + ⇔ f ′ (x) = x − 1
4x
4x
)
⇔ f (x) = 1 2 x2 − 1 4
(1, ln x + C. Because the graph of f passes through the point 1 2
, C = 0, so