[email protected] MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

[email protected] • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§8.2 Quadratic§8.2 QuadraticEquation AppsEquation Apps



Review §Review §

Any QUESTIONS About• §8.2 → Complete-the-Square

Any QUESTIONS About HomeWork• §8.2 → HW-37

8.2 MTH 55



§8.2 Quadratic Formula§8.2 Quadratic Formula

The Quadratic Formula

Problem Solving with the Quadratic Formula



The Quadratic FormulaThe Quadratic Formula

The solutions of ax2 + bx + c = 0 are given by

a

acbbx

2

42

This is one of theMOST FAMOUSFormulas in allof Mathematics



Example Example Circular WalkWayCircular 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

2rA

Also the diameter, d, is half of r. Thus A in terms of d 42

22dd

A




Familiarize: Makea DIAGRAM

Translate: Use Diagram of Subtractive Geometry

4

2dAcirle

2 10

14 10

= −




Translate: Diagram to Equation

4

10

4

14 22 walkA

14 10

= −




CarryOUT: Solve Eqn for Awalk

24964

1001964

101444

10

4

14 2222

walk

walk

A

A

Using π ≈ 3.14 find

367514324 .. walkA




Check: Use Acircle = πr2

367557886153

251434914357 22

...

..

chk

chk

A

A

State: The Area of the Paver Walkway is about 75.4 ft2

• Note that UNITS must be included in the Answer Statement



Example Example Partition Bldg 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




Familiarize: by Diagram

Now LET x ≡ frontage of building, in feet.

Translate: The other statements into Equations involving x




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




Now use the 2100 ft2 Area Constraint• Area of rear = 2100

• Area = x(3x−45), so

x 3x 45 2100

3x2 45x 2100 0

x2 15x 700 0

x 35 x 20 x 35 x 20




So x is either 35ft or −20ft But again Distances can

NOT be negative Thus x = 35 ft Check: Use 2100 ft2 Area

21004510535

21004535335

2100453

?

?

xx

21006035




State: The Bldg

Frontage is 35ft The Bldg Depth

is 3(35ft) = 105ft105’

35’

60’



Example Example Bike Tire BlowOut 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)




Familiarize: Make diagram

LET w ≡ Devon’s Walking Speed

Recall the RATE Equation for Speed

Distance = (Speed)·(Time)




Translate: The BikingSpeed, b, is 4 mphfaster than theWalking Speed →

From the Diagram note Distances by Rate Equation:• Biking Distance = 10 miles = b·tbike

• Walking Distance = 6 miles = w·twalk

4wb




Translate: Now the Total Distance of16mi is the sum of theBiking & Walking Distances →

walkbike

walkbike

wttw

wbandwtbt

416

416

From the Spd Eqn: Time = Dist/Spd




Translate: Thus bySpeed Eqn:

Next, the sum of the Biking and Walking times is 2hrs & 40min = 2-2/3hr →

wwtt walkbike

6

4

10

3

22

wt

wbt

walk

bike

mi64

mi10mi10




CarryOut: Clear Fractions in the last Eqn by multiplying by the LCD of 3·(w+4) ·w:

GCFby 928072168

721830328

43631048

436

4

10

3

8

6

4

10

3

86

4

10

3

22

22

2

wwww

wwww

wwww

wwww

wwww




CarryOut: Divide the last Eqn by 8 to yield a Quadratic Equation

920

8

9280 22

wwww

This Quadratic eqn does NOT factor so use Quadratic Formula with a = 1, b = −2, and c = −9

12

91422

2

422

a

acbbw




CarryOut: find w by Quadratic Formula

2

1042

2

1042

2

402

2

3642

w

101

2

1012

2

1022

w

So Devon’s Walking Speed is

162101or164101

1012

1012

2

1022

..

ww

w




CarryOut: Since SPEED can NOT be Negative find:

Check: Test to see that the time adds up to 2.67 hrs

mph 164101 .w

668244312261164

6

4164

10672

6

4

10

3

22

.....

.?

ww

tt walkbike




State: Devon’s Walking Speed was about 4.16 mph (quite a brisk pace)



Example Example Golden Rectangle 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




Familiarize: Make a Diagram



Example Example Golden Rectangle Golden Rectangle Translate: Use Diagram

Longer side of A

Shorter side of A

Longer side of B

Shorter side of Bp

q

q

p qp p q qq

p2 pq q2

p

q

2

p

q1




CarryOut: LETΦ ≡ GoldenRatio= p/q

p

q

2

p

q1

2 1 0

1 1 2 4 1 1

2 1

1 5

26181or6180 ..




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

21.618



Example Example Pythagorus 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 = the length, in yards, of the other leg. s + 28

s 52



Pythagorean TrianglePythagorean Triangle

2. Translate. We use the Pythagorean theorem:s2 + (s + 28)2 = 522

3. Carry Out. Identify the Quadratic Formula values a, b, & c

s + 28

s 52

s2 + (s + 28)2 = 522

s2 + s2 + 56s + 784 = 2704

2s2 + 56s − 1920 = 0

s2 + 28s − 960 = 0




3. Carry Out: With s2 + 28s − 960 = 0

Find: a = 1, b = 28, c = −960

Evaluate the Quadratic Formula2( ) 4( )(28 )

2

28 1 960

1( )s

28 784 3840

2s

28 4624

2s




3. Carry Out: Continue Quadratic Eval

28 4624

2s

28 68

2s

28 68 28 68 or

2 2s s

40 9620 or 48

2 2s s

s + 28

s 52




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.

s + 28yd = 48 yds

s = 20 yds

52 yds



Vertical BallisticsVertical Ballistics

In Physics 4A, you will learn the general formula for describing the height of an object after it has been thrown upwards:

20 0

1,

2h gt v t h

• 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



Example Example X-Games Jump 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.




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




CarryOut:

20 0

1

2h gt v t h

2115 ( 32.2) 80 25

2t t

215 16.1 80 25t t 20 16.1 80 10t t

Use the Quadratic Formula to find t• a = −16.1, b = 80 and c = 10

1162

1011648080

2

422

.

.

a

acbbt




CarryOut:

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

1220or095232

704480..

.

ttt



Example Example Biking Speed 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




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

48/r t48

41

rt

• Uses the Rate/Spd Eqn → Rate = Qty/Time




Translate: From the Table Obtain two Equations in r & t

48r

t and

484 .

1r

t

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

48 484

1t t



Example Example Driving Speed Driving Speed

Carry out: Next Clear Fractions48 48

( 1) 4 ( 1)1

t t t tt t

Multiplying by the LCD

48 48( 1) ( 1) 4 ( 1)

1t t t t t t

t t

48( 1) 4 ( 1) 48t t t t

248 48 4 4 48t t t t 24 4 48 0t t

2 12 0t t




Carry out: The Last Eqn is Quadratic in t:

2 12 0t t

Solve by Quadratic Forumula with:• a = 1, b = −1, and c = −12

12

121414

2

422

a

acbbt

2

6or

2

8

2

71

2

491

2

4811

t




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

3or4 t

48 48

4r

t 12 mph.




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

hr 3hrmi 16

mi 4816 t



ReCall The WORK PrincipleReCall 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.



The WORK PrincipleThe 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:

111

tb

ta

111

bat

1b

t

a

t

tba

111



Example Example Empty Tower Tank 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?




Familiarize: Creating a Table helps to clarify the given data

Pipe Time to Complete the Job Alone

Work Rate

CombinedWorkingTime

Portion of Job Completed

Smaller t + 20 min. 40

Larger t min. 40

1

20t

1

t

40

t

40

20t

And the Job-Portions must add up to ONE complete Job:

40 401

20t t




CarryOut:

40 40( 20) ( 20) 1

20t t t t

t t

40 40( 20) ( 20) ( 20) 1

20t t t t t t

t t

40 40( 20) ( 20)t t t t 240 40 800 20t t t t

280 800 20t t t 20 60 800t t

The Last Eqn is Quadratic




Use Quadratic Formula:20 60 800t t

2( 60) ( 60) 4(1)( 800)

2t

60 3600 3200

2t

60 6800

2t

60 680071.2

2t

60 680011.2

2t




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

00415618043860271

40

192

401

271

40

20271

40...

....

?




State: Working alone the SMALL pipe would empty the Water Tower in about 91.2 minutes



WhiteBoard WorkWhiteBoard Work

Problems From §8.2 Exercise Set• 74

The Arrhenius Rate Equation



All Done for TodayAll Done for Today

MotorCyleFatality

Statistics



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

x

y

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls



-3

-2

-1

0

1

2

3

4

5

-3 -2 -1 0 1 2 3 4 5

M55_§JBerland_Graphs_0806.xls -5

-4

-3

-2

-1

0

1

2

3

4

5

-10 -8 -6 -4 -2 0 2 4 6 8 10

M55_§JBerland_Graphs_0806.xls

x

y

[email protected] MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

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