Polynomial Rate Dynamics

Next: Transforming these computations into AP Exam justifications... Slide 4 of 8

AP Rubric Masterclass Free-Response Justification Protocol

The 3-Part AP Justification Sentence Frame

1

Identify Interval

Explicitly state the exact closed domain interval: "On the interval \( [a, b] \)..."

2

State Value & Sign

State whether the rate is positive or negative: "the function increases/decreases at an average rate of..."

3

Compound Units

Include dependent over independent units: "...of \( 9\text{ meters per second} \)."

AP Exemplar Response:

"On the interval \( [1, 4] \) seconds, the drone's height increases at an average rate of \( 9 \) meters per second because \( \frac{h(4)-h(1)}{4-1} = 9 > 0 \)."

AP Practice 2.B: Justifying and Communicating Mathematical Reasoning Slide 5 of 8

Phase 2: "We Do" Collaboration Investigating Rate Behavior

Analyzing Successive Intervals (Equal Width \(\Delta x = 2\))

A cubic polynomial \( P(x) \) values are given in the table below. Calculate the average rates of change across consecutive intervals:

Interval \([0, 2]\) \( \frac{10-2}{2-0} = \mathbf{+4} \)

Interval \([2, 4]\) \( \frac{30-10}{4-2} = \mathbf{+10} \)

Interval \([4, 6]\) \( \frac{68-30}{6-4} = \mathbf{+19} \)

Partner Discussion: The rates are \( +4 \to +10 \to +19 \). Since the rates are positive and increasing, what does this tell us about the graph's curvature (concavity)?

Increasing Rates of Change = Graph is Concave Up (Bending Upwards) Slide 6 of 8

Phase 3: "You Do" Practice Independent / Paired Mastery (25 Min)

AP Free-Response Challenge (On Student Handout)

The volume of chemical solution in a reactor tank is modeled by \( V(t) = 0.5t^3 - 6t^2 + 20t + 50 \), where \( t \) is measured in hours (\( 0 \le t \le 8 \)) and \( V(t) \) is in liters.

Task 1: Calculator Strategy

Calculate the average rate of change on \([1, 4]\) and \([4, 7]\). Show numerical quotient setups.

Setup: (V(4) - V(1)) / (4 - 1)

Task 2: Full AP Justification

Write two complete justification sentences adhering to the 3-part AP rubric standards including units of L/hr.

Include: Interval • Direction • Value • Units

Turn to Page 2 of your Student Handout to execute work. Slide 7 of 8

Phase 4: Synthesis & Closure 10 Minutes Remaining

AP Precalculus Rate Matrix Synthesis

Rate Behavior Summary

• Positive Rates: Function values are net increasing.
• Negative Rates: Function values are net decreasing.
• Increasing Rates: Function curve bends upwards (Concave Up).
• Decreasing Rates: Function curve bends downwards (Concave Down).

Exit Ticket Protocol

Please clear your desks and independently complete the Rate Dynamics Exit Ticket.

✓ 1 Calculation with explicit difference quotient

✓ 1 AP Justification with contextual compound units

Turn in Exit Ticket before the bell rings! Slide 8 of 8

(b) Using correct units, interpret the meaning of your answer to part (a) in the context of the problem.

(c) Find the average rate of change on \( [4, 7] \). Compare this to part (a) to justify whether the rate is increasing or decreasing.

Polynomial Rate Dynamics • Student Packet Page 2 of 2

Phase 3: "You Do" FRQ Reactor Chamber Model

Model: \( V(t) = 0.5t^3 - 6t^2 + 20t + 50 \), \( 0 \le t \le 8 \)

Part (a) Average Rate of Change on \([1, 4]\):

\( V(1) = 0.5(1)^3 - 6(1)^2 + 20(1) + 50 = 64.5\text{ liters} \)
\( V(4) = 0.5(64) - 6(16) + 20(4) + 50 = 32 - 96 + 80 + 50 = 66\text{ liters} \)
\( \text{AROC}_{[1, 4]} = \frac{V(4) - V(1)}{4 - 1} = \frac{66 - 64.5}{3} = \frac{1.5}{3} = \mathbf{0.5\text{ liters per hour}} \)

[1 pt: Difference quotient setup, 1 pt: Value of 0.5 with L/hr]

Part (b) Contextual Interpretation:

"On the time interval from \( t = 1 \) hour to \( t = 4 \) hours, the volume of solution in the reaction chamber increases at an average rate of \( 0.5 \) liters per hour."

[1 pt: Cites interval \([1, 4]\), direction (increases), value 0.5, and correct compound units L/hr]

Part (c) Rate on \([4, 7]\) & Behavior Comparison:

\( V(7) = 0.5(343) - 6(49) + 20(7) + 50 = 171.5 - 294 + 140 + 50 = 67.5\text{ liters} \)
\( \text{AROC}_{[4, 7]} = \frac{67.5 - 66}{7 - 4} = \frac{1.5}{3} = \mathbf{0.5\text{ liters per hour}} \)

Analysis: Across these symmetrical 3-hour windows around \(t=4\), the net average rate is identical (\(0.5\text{ L/hr}\)). However, evaluating smaller sub-intervals shows the rate decreases toward \(t=4\) then increases after \(t=4\), showing an inflection point in rate behavior.

Formative Exit Ticket Official Solution & Scoring Rubric Total: 3 Points

Question 1 Solution (\(2\) pts):

\( \text{AROC}_{[2, 5]} = \frac{T(5) - T(2)}{5 - 2} = \frac{25.0 - 85.6}{3} = \frac{-60.6}{3} = \mathbf{-20.2\,^\circ\text{C/min}} \)

• Point 1: Correct difference quotient expression: \( \frac{25.0 - 85.6}{5 - 2} \) or equivalent.
• Point 2: Correct value of \(-20.2\) with compound units (\(^\circ\text{C/min}\) or degrees Celsius per minute).

Question 2 Solution (\(1\) pt):

"On the time interval from \( t = 2 \) minutes to \( t = 5 \) minutes, the temperature of the alloy decreases at an average rate of \( 20.2 \) degrees Celsius per minute." (Alternatively: "...changes at an average rate of \(-20.2\,^\circ\text{C/min}\).")

• Point 3: Includes all 4 AP components: interval \([2, 5]\), direction ("decreases"), magnitude (\(20.2\)), and compound units (\(^\circ\text{C/min}\)). (Do not accept "decreases at a rate of \(-20.2\)" due to double negative error).

Polynomial Rate Dynamics • Complete Answer Key Page 2 of 2