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§8.2 Quadratic Equation Apps

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1 §8.2 Quadratic Equation Apps
Chabot Mathematics §8.2 Quadratic Equation Apps Bruce Mayer, PE Licensed Electrical & Mechanical Engineer

2 8.2 Review § Any QUESTIONS About Any QUESTIONS About HomeWork
MTH 55 Review § Any QUESTIONS About §8.2 → Quadratic Formula Any QUESTIONS About HomeWork §8.2 → HW-39

3 §8.2 Quadratic Formula The Quadratic Formula
Problem Solving with the Quadratic Formula

4 This is one of the MOST FAMOUS Formulas in all of Mathematics
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

5 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 Formula for the area, A ,of a Circle based on it’s radius, r Also the diameter, d, is half of r. Thus A in terms of d

6 Example Circular WalkWay
Familiarize: Make a DIAGRAM Translate: Use Diagram of Subtractive Geometry =

7 Example Circular WalkWay
Translate: Diagram to Equation =

8 Example Circular WalkWay
CarryOUT: Solve Eqn for Awalk Using π ≈ 3.14 find

9 Example Circular WalkWay
Check: Use Acircle = πr2 State: The Area of the Paver Walkway is about 75.4 ft2 Note that UNITS must be included in the Answer Statement

10 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 than she walked The Entire journey took 2hrs and 40min Find Devon’s Walking Speed (or Rate)

11 Example  Bike Tire BlowOut
Familiarize: Make diagram LET w ≡ Devon’s Walking Speed Recall the RATE Equation for Speed Distance = (Speed)·(Time)

12 Example  Bike Tire BlowOut
Translate: The Biking Speed, b, is 4 mph faster than the Walking Speed → From the Diagram note Distances by Rate Equation: Biking Distance = 10 miles = b·tbike Walking Distance = 6 miles = w·twalk

13 Example  Bike Tire BlowOut
Translate: Now the Total Distance of 16mi is the sum of the Biking & Walking Distances → From the Spd Eqn: Time = Dist/Spd

14 Example  Bike Tire BlowOut
Translate: Thus by Speed Eqn: Next, the sum of the Biking and Walking times is 2hrs & 40min = 2-2/3hr →

15 Example  Bike Tire BlowOut
CarryOut: Clear Fractions in the last Eqn by multiplying by the LCD of 3·(w+4) ·w:

16 Example  Bike Tire BlowOut
CarryOut: Divide the last Eqn by 8 to yield the a Quadratic Equation This Quadratic eqn does NOT factor so use Quadratic Formula with a = 1, b = −2, and c = −9

17 Example  Bike Tire BlowOut
CarryOut: find w by Quadratic Formula So Devon’s Walking Speed is

18 Example  Bike Tire BlowOut
CarryOut: Since SPEED can NOT be Negative find: Check: Test to see that the time adds up to 2.67 hrs

19 Example  Bike Tire BlowOut
State: Devon’s Walking Speed was about 4.16 mph (quite a brisk pace)

20 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

21 Example  Partition Bldg
Familiarize: by Diagram Now LET x ≡ frontage of building, in feet. Translate: The other statements into Equations involving x

22 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

23 Example  Partition Bldg
Now use the 2100 ft2 Area Constraint Area of rear = 2100 Area = x(3x−45), so

24 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

25 Example  Partition Bldg
State: The Bldg Frontage is 35ft The Bldg Depth is 3(35ft) = 105ft 60’ 105’ 35’

26 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 geometrically-similar to the original one. The GOLDEN RATION then = p/q

27 Example  Golden Rectangle
Familiarize: Make a Diagram

28 Example  Golden Rectangle
Translate: Use Diagram

29 Example  Golden Rectangle
Carry Out: LET Φ ≡ Golden Ratio = p/q

30 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

31 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 Familarize. First make a drawing and label it. Let s = length, in yards, of one leg. Then s = the length, in yards, of the other leg. 52 s s + 28

32 Pythagorean Triangle s + 28 s 52 Translate. We use the Pythagorean theorem: s2 + (s + 28)2 = 522 Carry Out. Identify the Quadratic Formula values a, b, & c s2 + (s + 28)2 = 522 s2 + s2 + 56s = 2704 2s2 + 56s − 1920 = 0 s2 + 28s − 960 = 0

33 Pythagorean Triangle Carry Out: With s2 + 28s − 960 = 0
Find: a = 1, b = 28, c = −960 Evaluate the Quadratic Formula

34 Pythagorean Triangle Carry Out: Continue Quadratic Eval s + 28 s 52

35 Pythagorean Triangle Check. Length cannot be negative, so −48 does not check. State. One leg is 20 yards and the other leg is 48 yards. 52 yds s = 20 yds s + 28yd = 48 yds

36 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 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

37 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.

38 Example  X-Games Jump h = 15, v0 = 80, and h0 = 25
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

39 Example  X-Games Jump Carry Out: Use the Quadratic Formula to find t
a = −16.1, b = 80 and c = 10

40 Example  X-Games Jump Carry 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

41 Find KW’s average speed
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

42 Example  Driving 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 r + 4 t – 1 Uses the Rate/Spd Eqn → Rate = Qty/Time

43 Example  Driving Speed
Translate: Form the Table Obtain two Equations in r & t and 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

44 Example  Driving Speed
Carry out: Next Clear Fractions Multiplying by the LCD

45 Example  Driving Speed
Carry out: The Last Eqn is Quadratic in t: Solve by Quadratic Forumula with: a = 1, b = −1, and c = −12

46 Example  Driving Speed
Carry out: By Quadratic Formula Since TIMES can NOT be NEGATIVE, then t = 4 hours Return to one of the table Eqns to find r 12 mph.

47 Example  Driving Speed
Check: To see if 12 mph checks, we increase the speed 4 mph to16 mph and see 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

48 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.

49 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:

50 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?

51 Example  Empty Tower Tank
Familiarize: Creating a Table helps to clarify the given data Pipe Time to Complete the Job Alone Work Rate Time Working Portion of Job Completed Smaller t + 20 min. 40 Larger t min. And the Job-Portions must add up to ONE complete Job:

52 Example  Empty Tower Tank
Carry Out: The Last Eqn is Quadratic

53 Example  Empty Tower Tank
Use Quadratic Formula:

54 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 or 91.2 minutes Check: Use the Work Eqn from Table

55 Example  Empty Tower Tank
State: Working alone the SMALL pipe would empty the Water Tower in about 91.2 minutes

56 WhiteBoard Work Problems From §8.2 Exercise Set
74 The Arrhenius Rate Equation

57 MotorCylcle Fatality Statistics
All Done for Today MotorCylcle Fatality Statistics

58 Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical Engineer

59 Graph y = |x| Make T-table

60


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