Chabot Mathematics
§8.2 Quadratic
Equation Apps
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Review § 8.2
MTH 55
 Any QUESTIONS About
• §8.2 → Complete-the-Square
 Any QUESTIONS About HomeWork
• §8.2 → HW-37
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
§8.2 Quadratic Formula
 The Quadratic
Formula
 Problem
Solving with
the Quadratic
Formula
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
The Quadratic Formula
 The solutions of
ax2 + bx + c = 0
are given by
This is one of the
MOST FAMOUS
Formulas in all
of Mathematics
 b  b  4ac
x
2a
2
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example Circular WalkWay
 A Circular Pond 10 feet in diameter is to
be surrounded by a “Paver” walkway
that will be 2 feet wide.
Find the AREA of the WalkWay
 Familiarize: Recall the
2
Formula for the area, A ,of
A r
a Circle based on it’s radius, r
2
2
 Also the diameter, d, is half A    d   d
4
2
of r. Thus A in terms of d

Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example Circular WalkWay
 Familiarize: Make
a DIAGRAM
 Translate: Use Diagram
of Subtractive Geometry
Chabot College Mathematics
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Acirle 
d 2
4
10
14
=
10
2
−
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example Circular WalkWay
 Translate: Diagram to Equation
10
14
−
=
 14
2
Awalk 
Chabot College Mathematics
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4
 10
2

4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example Circular WalkWay
 CarryOUT: Solve Eqn for Awalk
 14
2
Awalk
Awalk


4

4
 10
2

4
196  100 


4

14  10 
4

2
96  24
 Using π ≈ 3.14 find
Awalk
Chabot College Mathematics
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 243.14  75.36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
2
Example Circular WalkWay
 Check: Use Acircle = πr2
Achk
  7    5  3.14  49  3.14  25
Achk
 153.86  78.5  75.36
2
2

 State: The Area of the Paver Walkway
is about 75.4 ft2
• Note that UNITS must
be included in the
Answer Statement
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Partition Bldg
 A rectangular building whose depth
(from the front of the building) is three
times its frontage is divided into two
parts by a partition that is 45 feet from,
and parallel to, the front wall. If the rear
portion of the building contains 2100
square feet, find the dimensions of the
building.
 Familiarize: REALLY needs a Diagram
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Partition Bldg
 Familiarize:
by Diagram
 Now LET x ≡
frontage of
building, in feet.
 Translate: The
other statements
into Equations
involving x
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Partition Bldg
 The Bldg depth is three
times its frontage, x
→ 3x = depth of building,
in feet
 The Bldg Depth is divided
into two parts by a partition
that is 45 feet from, and
parallel to, the front wall
→ 3x – 45 = depth of rear portion, in ft
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Partition Bldg
 Now use the 2100 ft2
Area Constraint
• Area of rear = 2100
• Area = x(3x−45), so
x 3x  45   2100
3x  45x  2100  0
2
x  15x  700  0
2
x  35 x  20 
Chabot College Mathematics
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x  35
x  20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Partition Bldg
 So x is either 35ft or −20ft
 But again Distances can
NOT be negative
 Thus x = 35 ft
 Check: Use 2100 ft2 Area
x3 x  45  2100
35335  45  2100
?
35105  45  2100
?
Chabot College Mathematics
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3560  2100 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Partition Bldg
 State:
 The Bldg
Frontage is 35ft
 The Bldg Depth
is 3(35ft) = 105ft
60’
105’
35’
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 Devon set out on a 16 mile Bike Ride.
Unfortunately after 10 miles of Biking
BOTH tires Blew Out. Devon Had to
complete the trip on Foot.
 Devon biked four miles per hour
(4 mph) faster than she walked
 The Entire journey took 2hrs and 40min
Find Devon’s Walking Speed (or Rate)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 Familiarize:
Make
diagram
16
10
 LET
w ≡ Devon’s
Bike
Walk
Walking Speed
 Recall the RATE Equation for Speed
Distance = (Speed)·(Time)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 Translate: The Biking
Speed, b, is 4 mph
faster than the
Walking Speed →
b  w 4
16
10
Bike
Walk
 From the Diagram note Distances by
Rate Equation:
• Biking Distance = 10 miles = b·tbike
• Walking Distance = 6 miles = w·twalk
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 Translate: Now the
Total Distance of
16mi is the sum of the
Biking & Walking
Distances →
16
10
Bike
16  btbike  wtwalk
Walk
and b  w  4
16  w  4tbike  wtwalk
 From the Spd Eqn: Time = Dist/Spd
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 Translate: Thus by
Speed Eqn:
10mi
tbike 
b
6mi
t walk 
w
16
10
10mi

w4
Bike
Walk
 Next, the sum of the Biking and Walking
times is 2hrs & 40min = 2-2/3hr →
2
10
6
2  tbike  t walk 

3
w4 w
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 CarryOut: Clear Fractions in the last
Eqn by multiplying by the LCD of
3·(w+4) ·w:
2
10
6
8
10
6
2 
  

3 w4 w
3 w4 w
10
6
8
 
 3  ww  4
3 w4 w
 8  ww  4  10  3w  6  3w  4
 8w  32 w  30 w  18w  72
2
 8w
2
Chabot College Mathematics
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 16 w  72  0  8w
2

 2 w  9 by GCF
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 CarryOut: Divide the last Eqn by 8 to
yield a Quadratic Equation


0  8 w  2w  9
2
 0  w  2w  9
8
 This Quadratic eqn does NOT factor so
use Quadratic Formula with a = 1,
b = −2, and c = −9
2
 b  b  4ac   2 
w

2a
2
Chabot College Mathematics
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 2
21
2
 41 9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 CarryOut: find w by Quadratic Formula
2  4  36 2  40 2  4 10 2  4 10
w



2
2
2
2
2  2 10 2 1  10
w

 1  10
2
2


 So Devon’s Walking Speed is


2  2 10 2 1  10
w

 1  10
2
2
w  1  10  4.16 or w  1  10  2.16
Chabot College Mathematics
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 CarryOut: Since SPEED can NOT be
Negative find: w  1  10  4.16 mph
 Check: Test to see that the time adds
up to 2.67 hrs
2
10
6
2  tbike  t walk 

3
w4 w
?
10
6
2.67 

 1.226  1.443  2.668
4.16  4 4.16

Chabot College Mathematics
24
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Bike Tire BlowOut
 State: Devon’s Walking Speed was
about 4.16 mph (quite a brisk pace)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Golden Rectangle
 Let’s Revisit the Derivation of the
GOLDEN RATIO
 A rectangle of length p and width q, with
p > q, is called a golden rectangle if
you can divide the rectangle into a
square with side of length q and a
smaller rectangle that is geometricallysimilar to the original one.
 The GOLDEN RATION then = p/q
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Golden Rectangle
 Familiarize: Make a Diagram
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Golden Rectangle
 Translate: Use Diagram
Longer side of A Longer side of B

Shorter side of A Shorter side of B
p
q

q pq
p p  q  q  q
p 2  pq  q 2
2
 p
p
 q   q  1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Golden Rectangle
 Carry
Out:
LET
Φ≡
Golden
Ratio
= p/q
2
 p
p
 q   q  1
2    1  0

 1 
1  4 11
2 1
2
1 5
 0.618 or

2
Chabot College Mathematics
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 1.618
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Golden Rectangle
 Carry Out: Since both p & q are
distances they are then both POSITIVE
 Thus Φ = p/q must be POSITIVE
 State: GOLDEN RATIO as defined by
the Golden Rectangle
1 5

 1.618
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Pythagorus
 The hypotenuse of a right triangle is 52
yards long. One leg is 28 yards longer
than the other. Find the lengths of the legs
1. Familarize. First make a drawing and
label it. Let s = length, in yards, of one
leg. Then s + 28
52
= the length,
s
in yards, of
the other leg.
s + 28
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Pythagorean Triangle
2. Translate. We use the
52
s
Pythagorean theorem:
s2 + (s + 28)2 = 522
s + 28
3. Carry Out. Identify the Quadratic
Formula values a, b, & c
s2 + (s + 28)2 = 522
s2 + s2 + 56s + 784 = 2704
2s2 + 56s − 1920 = 0
s2 + 28s − 960 = 0
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Pythagorean Triangle
3. Carry Out: With s2 + 28s − 960 = 0
Find: a = 1, b = 28, c = −960
Evaluate the Quadratic Formula
28  (28)2  4(1)(960)
s
2(1)
28  784  3840
s
2
28  4624
s
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Pythagorean Triangle
3. Carry Out: Continue
Quadratic Eval
28  4624
s
2
28  68
s
2
s
28  68
2
40
s
 20
2
Chabot College Mathematics
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52
s + 28
or s 
or
s
28  68
2
96
s
 48
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Pythagorean Triangle
4. Check. Length cannot be negative,
so −48 does not check.
5. State. One leg is 20 yards and the
other leg is 48 yards.
52 yds
s = 20 yds
s + 28yd = 48 yds
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Vertical Ballistics
 In Physics 4A, you will learn the general
formula for describing the height of an
object after it has been thrown upwards:
• Where
1 2
h  gt  v0t  h0 ,
2
– g ≡ the acceleration due to gravity
 A CONSTANT = 32.2 ft/s2 = 9.81 m/s2
– t ≡ the time in flight, in s
– v0 ≡ the initial velocity in ft/s or m/s
– h0 ≡ the initial height in ft or m
Chabot College Mathematics
36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  X-Games Jump
 In an extreme games competition, a
motorcyclist jumps with an initial velocity
of 80 feet per second from a ramp
height of 25 feet, landing on a ramp with
a height of 15 feet.
 Find the time
the motorcyclist
is in the air.
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  X-Games Jump
 Familiarize: We are given the initial
velocity, initial height, and final height
and we are to find the time the bike is in
the air. Can use the Ballistics Formula
 Translate: Use the formula with
h = 15, v0 = 80, and h0 = 25.
 Use
h = 15, v0 = 80, and h0 = 25
Chabot College Mathematics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  X-Games Jump
1 2
h  gt  v0t  h0
2
1
15  (32.2)t 2  80t  25
2
 Carry
Out:
15  16.1t 2  80t  25
0  16.1t 2  80t  10
 Use the Quadratic Formula to find t
• a = −16.1, b = 80 and c = 10
 b  b  4ac  80 
t

2a
2
Chabot College Mathematics
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80  4 16.110
2 16.1
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  X-Games Jump
 80  7044
 Carry
t
 t  5.09 or t  0.122
 32.2
Out:
 Since Times can NOT be Negative
t ≈ 5.09 seconds
 Check: Check by substuting 5.09 for t in
the ballistics Eqn.
• The Details are left for later
 State: The MotorCycle Flight-Time is
very nearly 5.09 seconds
Chabot College Mathematics
40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Biking Speed
 Kang Woo (KW to his friends) traveled
by Bicycle 48 miles at a certain speed.
If he had gone 4 mph faster, the trip
would have taken 1 hr less.
Find KW’s average speed
Chabot College Mathematics
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Biking Speed
 Familiarize: In this can tabulating the
information can help. Let r represent the
rate, in miles per hour, and t the time, in
hours for Kang Woo’s trip
Distance
Speed
Time
48
r
t
48
r+4
t–1
r  48/ t
48
r4
t 1
• Uses the Rate/Spd Eqn → Rate = Qty/Time
Chabot College Mathematics
42
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Biking Speed
 Translate: From the Table Obtain two
Equations
48
48
and r  4 
r
.
in r & t
t
t 1
 Carry out: A system of equations has
been formed. We substitute for r from
the first equation into the second and
solve the resulting equation
48
48
4
t
t 1
Chabot College Mathematics
43
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Driving Speed
 Carry out: Next Clear Fractions
48
 48 
Multiplying
t (t  1)    4  t (t  1) 
t  1 by the LCD
 t

48
48
t (t  1)  t (t  1)  4  t (t  1) 
t
t 1
48(t  1)  4t (t  1)  48t
2
48t  48  4t  4t  48t
4t 2  4t  48  0
2
Chabot College Mathematics
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t  t  12  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Driving Speed
 Carry out: The Last
Eqn is Quadratic in t:
2
t  t  12  0
 Solve by Quadratic Forumula with:
• a = 1, b = −1, and c = −12
t
b
2




b  4ac   4   1  41 12

2a
2 1
2
1  1  48 1  49 1  7
8
t



or
2
2
2
2
Chabot College Mathematics
45
6
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Driving Speed
 Carry out: By
Quadratic Formula
t  4 or  3
 Since TIMES can NOT be NEGATIVE,
then t = 4 hours
 Return to one of the table Eqns to find r
48 48
r

 12 mph.
t
4
Chabot College Mathematics
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Driving Speed
 Check: To see if 12 mph checks, we
increase the speed 4
48 mi
 3 hr
mph to16 mph and see t16 
16 mi hr
how long the trip would
have taken at that speed:
 The Answer checks a 3hrs is indeed 1hr
less than 4 hrs that the trip actually took
 State: KW rode his bike at an average
speed of 12 mph
Chabot College Mathematics
47
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
ReCall The WORK Principle
 Suppose that A requires a units of time
to complete a task and B requires b
units of time to complete the same task.
 Then A works at a rate of
1/a tasks per unit of time.
 B works at a rate of
1/b tasks per unit of time,
 Then A and B together work at a
rate of [1/a + 1/b] per unit of time.
Chabot College Mathematics
48
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
The WORK Principle
 If A and B,
working together,
require t units of
time to complete a
task, then their
rate is 1/t and the
following
equations hold:
1
1
t  t 1
a
b
1 1
t    1
a b
t t
 1
a b
1 1 1
 
a b t
Chabot College Mathematics
49
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Empty Tower Tank
 A water tower has two drainpipes
attached to it. Working alone, the
smaller pipe would take 20 minutes
longer than the larger pipe to empty the
tower. If both drainpipes work together,
the tower can be drained in 40 minutes.
 How long would it take the small
pipe, working alone, to drain the
tower?
Chabot College Mathematics
50
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Empty Tower Tank
 Familiarize: Creating a Table helps to
clarify the given data
Pipe
Time to
Work
Complete the Rate
Job Alone
Smaller
t + 20 min.
Larger
t min.
1
t  20
1
t
Combined
Working
Time
Portion of
Job
Completed
40
40
t  20
40
40
t
 And the Job-Portions 40
40

1
must add up to
t
ONE complete Job: t  20
Chabot College Mathematics
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Empty Tower Tank
40 
 40
t (t  20) 
   t (t  20) 1
 t  20 t 
40
40
t (t  20)
 t (t  20)
 t (t  20) 1
t  20
t
 Carry
Out:
40t  40(t  20)  t (t  20)
40t  40t  800  t 2  20t
80t  800  t 2  20t
0  t 2  60t  800
 The Last Eqn is Quadratic
Chabot College Mathematics
52
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Empty Tower Tank
 Use Quadratic Formula:
0  t 2  60t  800
(60)  (60) 2  4(1)(800)
t
2
60  3600  3200
t
2
60  6800
t
2
60  6800
60  6800
t
 71.2
t
 11.2
2
2
Chabot College Mathematics
53
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Empty Tower Tank
 Omit the negative solution as times
cannot be negative
 The amount of time required by the
small pipe is represented by t + 20, it
would take approximately 20 + 71.2 or
91.2 minutes
 Check: Use the Work Eqn from Table
?

40
40
40
40

1 

 0.4386  0.5618  1.004
71.2  20 71.2
92.1 71.2
Chabot College Mathematics
54
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Example  Empty Tower Tank
 State: Working alone the SMALL pipe
would empty the Water Tower in about
91.2 minutes
Chabot College Mathematics
55
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
WhiteBoard Work
 Problems From §8.2 Exercise Set
• 74

The Arrhenius Rate Equation
Chabot College Mathematics
56
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
All Done for Today
MotorCyle
Fatality
Statistics
Chabot College Mathematics
57
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
58
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
59
5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
4
5
6
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
-6
-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
60
-4
M55_§JBerland_Graphs_0806.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt
8
10