2013 AP Calculus AB Free Response #1
TLDRIn this video, Alan from Bottle Stem Coaching tackles the 2013 AP Calculus free response question one, part A, which involves graphing and calculus. The problem is set in a gravel processing plant where the rate at which unprocessed gravel arrives is given by a function G(T). Alan calculates the derivative G'(T) to find the rate of change of gravel processing at T=5 hours, confirming it's decreasing. He then integrates G(T) from 0 to 8 hours to find the total amount of unprocessed gravel that arrives during the workday. Alan also explores when the gravel amount is at its maximum, using calculus to find the critical points and evaluating the function at the endpoints. After solving the problem, he checks his work against the scoring guidelines, confirming his solutions for the rate of decrease and the amounts at different times. The video concludes with an invitation for viewers to engage with the content and offers additional homework help.
Takeaways
- 📈 The rate at which unprocessed gravel arrives at a plant, denoted as G(T), is modeled by a function where T is measured in hours.
- ⏰ At the start of the workday, the plant has 500 tons of unprocessed gravel.
- 🚧 The plant processes gravel at a constant rate of 100 tons per hour.
- 🔍 To find G'(5), the derivative of the arrival rate function is computed, which represents the rate of change at 5 hours into the workday.
- 📊 G'(5) is calculated to be -24.588, indicating the rate is decreasing at that time.
- 🧮 The total amount of unprocessed gravel that arrives at the plant during operation hours is found by integrating the rate function from 0 to 8 hours.
- 🔢 The integral calculation results in 825.551 tons of gravel.
- ⬇️ At T=5 hours, G(5) is less than the processing rate, indicating the amount of gravel is decreasing.
- 🔍 To find the maximum amount of unprocessed gravel, the function G(T) is set equal to the processing rate and solved for T.
- 🏁 The maximum amount is determined by evaluating the function at the endpoints and at the critical points within the interval.
- 📝 The maximum unprocessed gravel is calculated to be 635.376 tons at approximately T=4.923 hours.
- 🎯 The scoring guidelines confirm the decreasing rate and the amounts at different times, validating the calculations.
Q & A
What is the main topic of the video?
-The main topic of the video is solving the 2013 AP Calculus free response question number one, which involves graphing and calculus concepts.
What is the function G(T) that models the rate of gravel arrival at the plant?
-The function G(T) is given as 90 + 45cos(T^2/18), where T is measured in hours.
What is the constant rate at which the plant processes gravel?
-The plant processes gravel at a constant rate of 100 tons per hour.
What is the initial amount of unprocessed gravel at the plant at the start of the workday?
-At the beginning of the workday, the plant has 500 tons of unprocessed gravel.
What does G'(T) represent?
-G'(T) represents the rate of change of the gravel arrival rate, or how fast the rate of gravel arrival is changing at a given time T.
What is the rate of change of gravel processing at 5 hours?
-The rate of change of gravel processing at 5 hours, G'(5), is -24.588 tons per hour squared, indicating a decrease in the processing rate.
How is the total amount of unprocessed gravel that arrives at the plant during the hours of operation calculated?
-The total amount of unprocessed gravel is calculated by integrating the rate function G(T) from 0 to 8 hours.
What is the amount of gravel that arrives at the plant during the hours of operation?
-The amount of gravel that arrives at the plant during the hours of operation is 825.551 tons.
Is the amount of unprocessed gravel at the plant increasing or decreasing at time T equals five hours?
-The amount of unprocessed gravel is decreasing at time T equals five hours because the arrival rate is less than the processing rate.
What is the maximum amount of unprocessed gravel at the plant during the hours of operation?
-The maximum amount of unprocessed gravel at the plant during the hours of operation is 635.376 tons, which occurs at approximately 4.923 hours.
What is the process for finding the maximum amount of unprocessed gravel?
-To find the maximum amount of unprocessed gravel, one must find when the derivative of the amount function a(T) is equal to zero, which corresponds to a local maximum. Then, evaluate a(T) at the endpoints (0 and 8 hours) and at the critical points to determine the absolute maximum.
What is the final step in the video regarding the scoring guidelines?
-The final step is to verify the work done in the video with the scoring guidelines, ensuring that the calculations and conclusions align with the expected outcomes.
Outlines
📚 AP Calculus Free Response Problem Analysis
In this segment, Alan introduces the AP Calculus free response problem, focusing on the graphing calculator part. The problem involves a gravel processing plant with an initial stock of 500 tons of unprocessed gravel. The rate at which gravel arrives at the plant, denoted by G(T), is given as a function of time T in hours. Alan computes the derivative, G'(T), to find the rate of change of gravel processing at T=5 hours, resulting in a negative rate indicating a decrease in processing rate. He then integrates G(T) from 0 to 8 hours to find the total amount of unprocessed gravel that arrives at the plant during the workday.
📉 Decreasing Gravel Arrival and Maximum Unprocessed Gravel Amount
Alan continues by calculating G(5) to determine the amount of gravel arriving at the plant per hour at T=5 hours, which is less than the processing rate of 100 tons per hour, indicating a decrease in the amount of gravel. He then explores the maximum amount of unprocessed gravel during the plant's operation. By setting up an integral to represent the amount of gravel that has arrived minus the amount processed, Alan finds the derivative of this function to locate the maximum. After plotting and analyzing the function, he identifies a critical point at T=4.923 and compares it to the initial and final conditions to determine the maximum amount of unprocessed gravel.
🧮 Final Calculations and Scoring Guidelines Review
In the final part, Alan corrects a previous miscalculation and re-evaluates the integral to find the maximum amount of unprocessed gravel. He calculates the amount of gravel at T=0 and T=8 hours, adding the initial stock of 500 tons to the amount that arrived and subtracting the amount processed. Alan then reviews the scoring guidelines, confirming his calculations for the rate of decrease and the amounts of gravel at different times. The video concludes with a summary of the findings and a prompt for viewers to engage with the content and seek further assistance through offered platforms.
Mindmap
Keywords
💡AP Calculus
💡Free Response
💡Graphing Calculator
💡Rate
💡Derivative
💡Integral
💡Processing Rate
💡Maximum Amount
💡Contextual Interpretation
💡Chain Rule
💡Sine Function
💡Fundamental Theorem of Calculus
Highlights
Alan introduces the 2013 AP Calculus free response question focusing on graphing calculator part, specifically part a.
The rate at which unprocessed gravel arrives at a plant is modeled by the function G(T), where T is measured in hours.
At the beginning of the workday, the plant has 500 tons of unprocessed gravel.
The plant processes gravel at a constant rate of 100 tons per hour.
Alan calculates the derivative G'(T) to find the rate of change of gravel processing over time.
G'(5) is computed to determine the rate of change at five hours, revealing a decrease in processing rate.
The total amount of unprocessed gravel that arrives at the plant during operation hours is found by integrating G(T) from 0 to 8.
Alan integrates the function to find that 825.55 tons of gravel arrive at the plant.
The amount of gravel is decreasing at time T equals five hours as the arrival rate is less than the processing rate.
To find the maximum amount of unprocessed gravel, Alan uses the integral of G(T) and subtracts the amount processed.
The derivative of the amount of gravel with respect to time is set to zero to find the maximum.
Alan identifies T equals 4.923 as the time when the gravel arrival rate equals the processing rate.
The maximum amount of unprocessed gravel is calculated by evaluating the integral from 0 to 4.923 and subtracting the processed gravel.
Alan corrects a mistake in setting up the equation for the maximum amount of gravel and recalculates.
The maximum amount of unprocessed gravel at the plant is determined to be 635.376 tons at approximately 4.92 hours into the workday.
Alan reviews the scoring guidelines to ensure accuracy and completeness of the solution.
The video concludes with a confirmation that the key points, such as the decreasing rate and amounts of gravel at different times, were correctly calculated.
Alan offers additional resources for free homework help on platforms like Twitch and Discord.
Transcripts
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