📐 Remember: Average Rate of Change = (f(b) - f(a)) / (b - a)
🎯 Problem-Solving Strategy
- Read carefully: Identify what quantity is changing
- Find the function: Determine f(x) or the given data points
- Identify the interval: Find the starting and ending values
- Calculate: Apply the average rate of change formula
- Interpret: Explain what the result means in context
⚡ Physics
💰 Economics
🧬 Biology
🏠 Everyday Life
⚡ Physics Problems
Problem 1: Falling Object
A ball is dropped from a 100-foot building. Its height after t seconds is given by h(t) = 100 - 16t². Find the average velocity of the ball between t = 1 second and t = 2 seconds.
Step 1: Identify the function and interval
Function: h(t) = 100 - 16t²
Interval: [1, 2]
Function: h(t) = 100 - 16t²
Interval: [1, 2]
Step 2: Calculate h(2)
h(2) = 100 - 16(2)² = 100 - 16(4) = 100 - 64 = 36 feet
Step 3: Calculate h(1)
h(1) = 100 - 16(1)² = 100 - 16(1) = 100 - 16 = 84 feet
Step 4: Apply the formula
Average velocity = (h(2) - h(1)) / (2 - 1) = (36 - 84) / 1 = -48 ft/s
Answer: -48 feet per second
💡 The negative sign indicates the ball is falling (moving downward) at an average rate of 48 feet per second during this time interval.
Problem 2: Car Acceleration
A car's position along a highway is given by s(t) = 2t³ - 12t² + 18t + 5 miles after t hours. Find the average velocity between t = 1 hour and t = 3 hours.
Step 1: Identify the function and interval
Function: s(t) = 2t³ - 12t² + 18t + 5
Interval: [1, 3]
Function: s(t) = 2t³ - 12t² + 18t + 5
Interval: [1, 3]
Step 2: Calculate s(3)
s(3) = 2(27) - 12(9) + 18(3) + 5 = 54 - 108 + 54 + 5 = 5 miles
Step 3: Calculate s(1)
s(1) = 2(1) - 12(1) + 18(1) + 5 = 2 - 12 + 18 + 5 = 13 miles
Step 4: Apply the formula
Average velocity = (5 - 13) / (3 - 1) = -8 / 2 = -4 mph
Answer: -4 miles per hour
💡 The car moved backward on average at 4 mph, ending up 8 miles behind its starting position after 2 hours.
Problem 3: Temperature Change
The temperature of a cooling object follows T(t) = 80e^(-0.1t) + 20 degrees Fahrenheit after t minutes. Find the average rate of temperature change from t = 0 to t = 10 minutes.
Step 1: Identify the function and interval
Function: T(t) = 80e^(-0.1t) + 20
Interval: [0, 10]
Function: T(t) = 80e^(-0.1t) + 20
Interval: [0, 10]
Step 2: Calculate T(10)
T(10) = 80e^(-1) + 20 = 80(0.368) + 20 ≈ 29.4 + 20 = 49.4°F
Step 3: Calculate T(0)
T(0) = 80e^0 + 20 = 80(1) + 20 = 100°F
Step 4: Apply the formula
Average rate = (49.4 - 100) / (10 - 0) = -50.6 / 10 = -5.06°F/min
Answer: -5.06°F per minute
💡 The object cools at an average rate of about 5°F per minute during the first 10 minutes.
💰 Economics Problems
Problem 4: Company Revenue
A company's monthly revenue is modeled by R(x) = -2x² + 24x + 100 thousand dollars, where x is the number of months since January. Find the average rate of revenue change from month 3 to month 8.
Step 1: Identify the function and interval
Function: R(x) = -2x² + 24x + 100
Interval: [3, 8]
Function: R(x) = -2x² + 24x + 100
Interval: [3, 8]
Step 2: Calculate R(8)
R(8) = -2(64) + 24(8) + 100 = -128 + 192 + 100 = 164 thousand
Step 3: Calculate R(3)
R(3) = -2(9) + 24(3) + 100 = -18 + 72 + 100 = 154 thousand
Step 4: Apply the formula
Average rate = (164 - 154) / (8 - 3) = 10 / 5 = 2 thousand/month
Answer: $2,000 per month
💡 The company's revenue increased by an average of $2,000 per month during this 5-month period.
Problem 5: Stock Price Analysis
A stock's price follows P(t) = 50 + 10sin(πt/6) dollars after t weeks. Find the average rate of price change from week 0 to week 6.
Step 1: Identify the function and interval
Function: P(t) = 50 + 10sin(πt/6)
Interval: [0, 6]
Function: P(t) = 50 + 10sin(πt/6)
Interval: [0, 6]
Step 2: Calculate P(6)
P(6) = 50 + 10sin(π) = 50 + 10(0) = 50 dollars
Step 3: Calculate P(0)
P(0) = 50 + 10sin(0) = 50 + 10(0) = 50 dollars
Step 4: Apply the formula
Average rate = (50 - 50) / (6 - 0) = 0 / 6 = 0 dollars/week
Answer: $0 per week
💡 Despite fluctuations, the stock price returned to its starting value, resulting in zero average change over 6 weeks.
Problem 6: Production Cost
The total cost to produce x units is C(x) = 0.01x³ - 0.6x² + 15x + 1000 dollars. Find the average rate of cost change when production increases from 20 to 40 units.
Step 1: Identify the function and interval
Function: C(x) = 0.01x³ - 0.6x² + 15x + 1000
Interval: [20, 40]
Function: C(x) = 0.01x³ - 0.6x² + 15x + 1000
Interval: [20, 40]
Step 2: Calculate C(40)
C(40) = 0.01(64000) - 0.6(1600) + 15(40) + 1000
= 640 - 960 + 600 + 1000 = $1,280
= 640 - 960 + 600 + 1000 = $1,280
Step 3: Calculate C(20)
C(20) = 0.01(8000) - 0.6(400) + 15(20) + 1000
= 80 - 240 + 300 + 1000 = $1,140
= 80 - 240 + 300 + 1000 = $1,140
Step 4: Apply the formula
Average rate = (1280 - 1140) / (40 - 20) = 140 / 20 = $7/unit
Answer: $7 per unit
💡 On average, each additional unit costs $7 to produce when increasing from 20 to 40 units.
🧬 Biology Problems
Problem 7: Bacterial Growth
A bacterial culture grows according to N(t) = 1000e^(0.3t) bacteria after t hours. Find the average growth rate from hour 2 to hour 5.
Step 1: Identify the function and interval
Function: N(t) = 1000e^(0.3t)
Interval: [2, 5]
Function: N(t) = 1000e^(0.3t)
Interval: [2, 5]
Step 2: Calculate N(5)
N(5) = 1000e^(1.5) = 1000(4.482) ≈ 4,482 bacteria
Step 3: Calculate N(2)
N(2) = 1000e^(0.6) = 1000(1.822) ≈ 1,822 bacteria
Step 4: Apply the formula
Average rate = (4482 - 1822) / (5 - 2) = 2660 / 3 ≈ 887 bacteria/hour
Answer: ≈887 bacteria per hour
💡 The bacterial population increases by an average of 887 bacteria per hour during hours 2-5.
Problem 8: Plant Height
A plant's height is modeled by H(t) = 20√(t + 1) centimeters after t weeks. Find the average growth rate from week 3 to week 8.
Step 1: Identify the function and interval
Function: H(t) = 20√(t + 1)
Interval: [3, 8]
Function: H(t) = 20√(t + 1)
Interval: [3, 8]
Step 2: Calculate H(8)
H(8) = 20√(8 + 1) = 20√9 = 20(3) = 60 cm
Step 3: Calculate H(3)
H(3) = 20√(3 + 1) = 20√4 = 20(2) = 40 cm
Step 4: Apply the formula
Average rate = (60 - 40) / (8 - 3) = 20 / 5 = 4 cm/week
Answer: 4 cm per week
💡 The plant grows at an average rate of 4 centimeters per week during weeks 3-8.
Problem 9: Drug Concentration
The concentration of a drug in the bloodstream is C(t) = 15te^(-0.2t) mg/L after t hours. Find the average rate of concentration change from t = 1 to t = 6 hours.
Step 1: Identify the function and interval
Function: C(t) = 15te^(-0.2t)
Interval: [1, 6]
Function: C(t) = 15te^(-0.2t)
Interval: [1, 6]
Step 2: Calculate C(6)
C(6) = 15(6)e^(-1.2) = 90(0.301) ≈ 27.1 mg/L
Step 3: Calculate C(1)
C(1) = 15(1)e^(-0.2) = 15(0.819) ≈ 12.3 mg/L
Step 4: Apply the formula
Average rate = (27.1 - 12.3) / (6 - 1) = 14.8 / 5 = 2.96 mg/L per hour
Answer: ≈2.96 mg/L per hour
💡 The drug concentration increases by an average of about 3 mg/L per hour during hours 1-6.
🏠 Everyday Life Problems
Problem 10: Water Tank
A water tank is being filled. The volume of water after t minutes is V(t) = 50t - 0.5t² gallons. Find the average rate at which water is being added from minute 10 to minute 30.
Step 1: Identify the function and interval
Function: V(t) = 50t - 0.5t²
Interval: [10, 30]
Function: V(t) = 50t - 0.5t²
Interval: [10, 30]
Step 2: Calculate V(30)
V(30) = 50(30) - 0.5(900) = 1500 - 450 = 1050 gallons
Step 3: Calculate V(10)
V(10) = 50(10) - 0.5(100) = 500 - 50 = 450 gallons
Step 4: Apply the formula
Average rate = (1050 - 450) / (30 - 10) = 600 / 20 = 30 gallons/minute
Answer: 30 gallons per minute
💡 Water is being added to the tank at an average rate of 30 gallons per minute during minutes 10-30.
Problem 11: Coffee Temperature
Hot coffee cools according to T(t) = 70e^(-0.05t) + 20 degrees Fahrenheit after t minutes. Find the average cooling rate from minute 5 to minute 20.
Step 1: Identify the function and interval
Function: T(t) = 70e^(-0.05t) + 20
Interval: [5, 20]
Function: T(t) = 70e^(-0.05t) + 20
Interval: [5, 20]
Step 2: Calculate T(20)
T(20) = 70e^(-1) + 20 = 70(0.368) + 20 ≈ 25.8 + 20 = 45.8°F
Step 3: Calculate T(5)
T(5) = 70e^(-0.25) + 20 = 70(0.779) + 20 ≈ 54.5 + 20 = 74.5°F
Step 4: Apply the formula
Average rate = (45.8 - 74.5) / (20 - 5) = -28.7 / 15 ≈ -1.91°F/minute
Answer: ≈-1.91°F per minute
💡 The coffee cools at an average rate of about 1.9°F per minute during minutes 5-20.
Problem 12: Savings Account
Money in a savings account grows according to A(t) = 5000(1.03)^t dollars after t years. Find the average rate of growth from year 2 to year 8.
Step 1: Identify the function and interval
Function: A(t) = 5000(1.03)^t
Interval: [2, 8]
Function: A(t) = 5000(1.03)^t
Interval: [2, 8]
Step 2: Calculate A(8)
A(8) = 5000(1.03)^8 = 5000(1.267) ≈ $6,335
Step 3: Calculate A(2)
A(2) = 5000(1.03)^2 = 5000(1.061) ≈ $5,305
Step 4: Apply the formula
Average rate = (6335 - 5305) / (8 - 2) = 1030 / 6 ≈ $171.67/year
Answer: ≈$171.67 per year
💡 The savings account grows by an average of about $172 per year during years 2-8.