# Applications Problem Solving. 6/25/2013 Applications 2 Four-step Method 1. Define variables Name the quantities to be found Write these down Example:

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Applications Problem Solving

6/25/2013 Applications 2 Four-step Method 1. Define variables Name the quantities to be found Write these down Example: Let t = time in seconds Let V = velocity at time t in mph Solution Methodology

6/25/2013 Applications 3 Four-step Method 2. Write down relationships symbolically Example: V = 2t + 20 3. Substitute and solve algebraically Combine relationships if possible E.g., write some variables in terms of others Solution Methodology

6/25/2013 Applications 4 Four-step Method 4. State and check solutions Check for reasonableness Are values within physical limitations? Are negative values reasonable? Write out response to initial question Does the result answer the question? Solution Methodology

6/25/2013 Applications 5 Four-step Method Review 1. Define variables 2. Write down relationships symbolically 3. Substitute and solve algebraically 4. State and check solutions Solution Methodology: Review

6/25/2013 Applications 6 Example: Linear Acceleration A vehicle traveling at 20 mph accelerates by 2 mph per second for 10 seconds Write an equation for velocity as a linear function of time, graph the function, find the velocity at time 6.5 seconds Applications

6/25/2013 Applications 7 Example: Linear Acceleration Identify variables Let t = time in seconds Let a = acceleration (2 mph per second) Let V = velocity in mph at time t Let V 0 = initial velocity (20 mph) Applications

6/25/2013 Applications 8 Example: Linear Acceleration Relationships Acceleration a is rate-of-change of velocity So a is the slope of the graph of V as a linear function of t Applications 20 40 60 5 10 V mph t secs  6.5 33 

6/25/2013 Applications 9 Linear Acceleration (continued) Relationships (continued) V Substitute and Solve V = a t + V 0 = 2t + 20 At t = 6.5 V = 2(6.5) + 20 = 33 Applications = a t + V 0 mph 20 40 60 5 10 V mph t secs  6.5 33 

6/25/2013 Applications 10 Linear Acceleration (continued) Applications 20 40 60 5 10 V mph t secs  6.5 33   State and Check Solution Velocity is 33 mph after 6.5 seconds Question: Why does this t (6.5) work ? Can we use any t between 0 and 10 ? Why is 10 seconds not used ?

6/25/2013 Applications 11 Rates of Completion We think of some jobs as units of work Examples: Filling a tank Painting a house Loading a truck These all allow work sharing One job done by multiple agents, such as people, pipes, machines, etc

6/25/2013 Applications 12 Rates of Completion A certain job takes T minutes to complete How much of the job will be done in 1 minute? So, per-minute rate of completion is T 1 of the job T 1 jobs/min

6/25/2013 Applications 13 Rates of Completion So, per-minute rate of completion is Question: How much work is done in 2T minutes? T 1 jobs/min T 1 jobs min ( ) 2T min ( ) 2T T 1 ( ) = min jobs min ( ) = 2 jobs * * Note: A units accounting trick

6/25/2013 Applications 14 Rates of Completion For a given job, each of n agents have individual completion times of t i where i = 1, 2, 3,..., n Individual completion rates are then titi 1 In T minutes (hours, days, …) each individual agent completes of the job titi T

6/25/2013 Applications 15 Rates of Completion In T minutes (hours, days, …) each individual agent completes of the job titi T E.g., if the i th person can complete a job in t i = 4 hours, then that person’s rate of completion is one fourth of a job per hour 1 titi 1 4 = In T = 2 hours, that person completes half the job T titi 2 4 = 1 2 =

6/25/2013 Applications 16 Rates of Completion In T hours, n persons working together each complete a fraction of the job, so total job is T t1t1 T t2t2 + + T t3t3 + + T tntn = 1 = 1 1 t1t1 1 t2t2 + + 1 t3t3 + + 1 tntn T ( ) 1 T = 1 t1t1 1 t2t2 + + 1 t3t3 + + 1 tntn job

6/25/2013 Applications 17 Rates of Completion 1 T = 1 t1t1 1 t2t2 + + 1 t3t3 + + 1 tntn Total job rate of completion is sum of individual agent rates of completion (Completion Rate)(Completion Time) ● = One Job =T 1 t1t1 1 t2t2 + + 1 t3t3 + + 1 tntn ( ) T 1 T ( ) = 1

6/25/2013 Applications 18 Example: A heavy-duty pump can empty a 60,000 gallon tank in 6 hours, while a light-duty pump can empty it in 18 hours Emptying the tank is the JOB Rates of completion First pump: Second pump: Combined Rates of Completion 6 1 of the job per hour 18 1 of the job per hour

6/25/2013 Applications 19 Example: (continued) With both pumps, overall completion time is T hours Overall completion rate is Combined Rates of Completion T 1 = 6 1 18 1 + 9 2 = T 2 9 = 4.5 = hours = 18 3 1 + = 4

6/25/2013 Applications 20 Example: An auditorium has three exits of different sizes First exit can empty the room in t 1 = 10 minutes Rate of Completion = Second exit can empty the room in t 2 = 8 minutes Rate of Completion = Combined Rates of Completion 1 t1t1 1 10 = 1 t2t2 1 8 =

6/25/2013 Applications 21  Third exit can empty the room in t 3 = 5 minutes Rate of Completion = The job : Clear the room in time T using all three exits Rate of completion Combined Rates of Completion 1 t3t3 1 5 = T 11 t1t1 = + + 1 t2t2 1 t3t3 1 10 1 8 = ++ 1 5 The Auditorium

6/25/2013 Applications 22 Combined Rates of Completion T 11 t1t1 = + + 1 t2t2 1 t3t3 1 10 1 8 = ++ 1 5 4 + 5 + 8 40 = 17 40 = 1 10 1 8 = ++ 1 5 4 4 5 5 1 8 Solving for T T 17 40 = = 2.3592411 … minutes

6/25/2013 Applications 23 Can we meet the fire code requirement for a 2-minute evacuation ? Combined Rates of Completion T 17 40 = = 2.3592411 Room can be cleared in 2.359 minutes Question: Clearly not ! What can we do to meet the requirement ? Alter the slowest door to clear the room in t minutes and force T to be under 2 minutes

6/25/2013 Applications 24 Combined Rates of Completion Alter the slowest door to clear the room in t minutes and force T to be under 2 minutes How fast would the altered door have to work to meet the code, i.e. what is t ? Combined rates of completion are then 2 1 T 1 = t 1 = 8 1 5 1 + + t 1 = 8 1 5 1 – 2 1 – The New Auditorium

6/25/2013 Applications 25 Combined Rates of Completion t 1 = 8 1 5 1 – 2 1 – t 1 = 40 7 Can you show this ? t = 40 7 ≈ 5.714 … minutes The altered door must empty the room within 5.7 minutes So alter to match fastest door (5 minutes) WHY ?

6/25/2013 Applications 26 Combined Rates of Completion Alter to match fastest door (5 minutes) 1 T 8 + 5 + 8 40 = 21 40 = ≈ 1.905 minutes … with 5.7 seconds to spare !! New completion rate is 1 5 = ++ 1 8 1 5 T 40 21 = This meets the code requirement The New Auditorium

6/25/2013 Applications 27 Finding Averages Add n values, divide by n Call Avg the average of n values of x Calculations with Data Avg = x 1 + x 2 + x 3 + ··· + x n n = n n 1  k = 1k = 1 xkxk

6/25/2013 Applications 28 Example Average of test scores 73, 85, 14, 92 Works fine for small number of values Calculations with Data 4 1 4  k = 1k = 1 xkxk = 66 = (73 + 85 + 14 + 92) 1 4 = (264) 1 4

6/25/2013 Applications 29 Example Travel d miles in t hours For d = 200 miles and t = 3 hours Works for large/infinite number of values Calculations with Data Average speed = d t = 200 miles 3 hours = 66.67 mph

6/25/2013 Applications 30 Calculations with Data Finding percentages A per-100 proportion a is to b as P is to 100 a as a percentage of number b is a b (100) = P … that is a b = 100 P

6/25/2013 Applications 31 Calculations with Data Example In a chain saw 12 ounces of oil are added per gallon of gasoline What percentage of the mixture is oil ? Note that 1 gal = 128 oz For every 100 ounces of mixture 8.571 ounces are oil % oil ≈ 8.571 % 12 128 + 12 = ( 100 ) %

6/25/2013 Applications 32 Calculations with Data Finding percent change A variable p changes by amount ∆p What percentage of p is ∆p ? Percent change in p is where p is the initial value ∆p p (100) %

6/25/2013 Applications 33 Calculations with Data Example The price of gasoline increased from \$2.56 per gallon to \$3.89 per gallon Change is: ∆p = 3.89 – 2.56 = 1.33 Percent change is ∆p p (100) = 1.33 2.56 (100) ≈ 51.95 %

6/25/2013 Applications 34 Think about it !

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