2017 AP Calculus AB Free Response #5
TLDRIn this educational video, Alan from Bothell STEM continues his AP Calculus 2017 free response question walkthrough, focusing on question number five. The video is aimed at the non-calculator portion of the exam, requiring manual calculations. Alan discusses the motion of two particles along the x-axis, using the given position and velocity functions to determine when particle P is moving to the left, which occurs when its velocity is negative. He then finds the times when particles P and Q move in the same direction, noting that they never move to the left at the same time. For part D, Alan calculates the acceleration of particle Q at time T equals two, explaining that the speed is decreasing despite the velocity being negative. Finally, he determines the position of particle Q when it first changes direction, using integration to find the answer. The video concludes with a review of the scoring guidelines and a summary of the correct answers, ensuring that viewers have a clear understanding of the concepts covered.
Takeaways
- 📐 The video is a continuation of AP Calculus 2017 free response questions, specifically focusing on question number five.
- ⏱ The problem is divided into four parts (A, B, C, D), each requiring different calculations and understanding of calculus concepts.
- 📍 For Part A, particle P moves to the left when its velocity is negative, which is determined by taking the derivative of its position function.
- 🔍 The derivative of particle P's position function involves using the chain rule, resulting in an expression that is always positive, indicating that particle P only moves to the left when a specific condition is met (T < 1).
- 🕒 Part B involves finding the times when both particles, P and Q, move in the same direction. This is determined by analyzing the signs of their velocities at different times.
- 🚦 The direction of particle Q is determined by the sign of its velocity function, which is a quadratic expression that changes signs at T = 3 and T = 5.
- ⏫ In Part C, the acceleration of particle Q is found by taking the derivative of its velocity function, and it is determined that at T = 2, the speed is decreasing.
- 🔢 For Part D, the position of particle Q when it first changes direction is found by integrating its velocity function from T = 0 to T = 3 and adding the initial position.
- 📉 The concept of speed and velocity is clarified, noting that speed is the absolute value of velocity, so if velocity is negative and increasing (getting more negative), the speed is also increasing.
- 🎯 The final answer for the position of particle Q when it changes direction at T = 3 is 23, which is calculated by evaluating the integral and adding the initial position.
- ✅ The video concludes with a review of the scoring guidelines and a confirmation that the answers provided align with the expected responses.
- 📚 The presenter encourages viewers to engage with the content by leaving comments, liking, or subscribing for the next video in the series.
Q & A
What is the main topic of the video?
-The video is a tutorial on AP Calculus 2017 free response question number five, focusing on the non-calculator portion of the exam.
What is the first part of the problem that Alan addresses?
-The first part of the problem, Part A, involves determining when particle P is moving to the left along the x-axis.
How does Alan determine the direction of particle P's movement?
-Alan determines the direction of particle P's movement by analyzing the velocity of particle P, which is the derivative of its position function. If the velocity is negative, the particle is moving to the left.
What mathematical concept does Alan use to find when particle P's velocity is less than zero?
-Alan uses the concept of derivatives and the chain rule to find when the velocity of particle P is less than zero.
What is the condition for particle P to be moving to the left?
-Particle P is moving to the left when the numerator of its velocity function, which is 2t - 2, is less than zero, meaning t is less than 1.
What does Part B of the problem ask for?
-Part B asks for the times during which both particles, P and Q, are traveling in the same direction between time 0 and 8.
How does Alan determine the intervals when both particles are moving in the same direction?
-Alan determines the intervals by analyzing the velocity functions of both particles and finding the times when the signs of their velocities match.
What is the condition for particle Q's velocity to be positive?
-Particle Q's velocity is positive when t is greater than 5 or less than 3, as determined by the factored form of its velocity function.
What does Alan calculate in Part C of the problem?
-In Part C, Alan calculates the acceleration of particle Q at time T equals two and determines whether the speed is increasing, decreasing, or neither.
How does Alan find the acceleration of particle Q?
-Alan finds the acceleration of particle Q by taking the derivative of the velocity function and then evaluating it at T equals two.
What does Part D of the problem involve?
-Part D involves finding the position of particle Q the first time it changes direction.
How does Alan determine the position of particle Q when it changes direction?
-Alan determines the position by integrating the velocity function of particle Q from time 0 to the time when the direction change occurs, which is T equals three, and then adding the initial position.
Outlines
📚 AP Calculus 2017 Question 5 Analysis
In this segment, Alan from Bothell Stem Coach discusses AP Calculus 2017's free-response question number five, focusing on the non-calculator portion. He addresses the motion of two particles along the x-axis, with particle P's position and particle Q's velocity given by specific functions. The key to solving the problem lies in understanding when particle P moves to the left, which is when its velocity, the derivative of its position, is negative. Alan uses the chain rule to find this derivative and then investigates the conditions under which particle P moves left. He also examines when particles P and Q move in the same direction between time 0 and 8, considering the velocity of Q and its factors. The segment concludes with an exploration of particle Q's acceleration at time T equals two and determining whether the speed is increasing, decreasing, or neither.
🔍 Direction of Particles and Acceleration Analysis
This paragraph delves into the direction of particle motion and the acceleration of particle Q. Alan explains that the direction of particle P's movement is determined by the sign of its velocity, which is the derivative of its position. He outlines the intervals during which both particles move in the same direction, highlighting that they never both move to the left at the same time. The discussion then shifts to the acceleration of particle Q, which is found by differentiating its velocity function. At time T equals two, the acceleration is calculated, and it's determined that the speed of particle Q is decreasing due to the negative rate of change of velocity. Alan cautions that one must be careful when differentiating between velocity and speed, especially when considering the direction of motion.
📌 Calculating Position and Direction Change
The final paragraph focuses on calculating the position of particle Q when it first changes direction. Alan explains that the velocity of Q switches from positive to negative at time T equals three, indicating a change in direction from right to left. To find the position at this specific time, he integrates the velocity function of Q over the interval from time 0 to 3 and adds the initial position. The integral calculation is shown step by step, leading to the conclusion that the position of particle Q at the time of direction change is 23 units. Alan also reflects on the scoring guidelines for the question, ensuring that the response aligns with the expected outcomes and acknowledges a minor oversight regarding the upper bound of the time interval.
Mindmap
Keywords
💡AP Calculus
💡Free Response Questions
💡Particle P and Particle Q
💡Velocity
💡Derivative
💡Chain Rule
💡Acceleration
💡Direction of Motion
💡Integration
💡Quadratic Formula
💡Speed
Highlights
Particle P is moving to the left when its velocity is negative, indicating a decrease in position.
The velocity of particle P is the derivative of its position function, which is found using the chain rule.
The denominator of the velocity derivative is always positive, meaning particle P only moves left when the numerator is negative.
Particle P moves to the left between times 0 and 1, and to the right between times 1 and 8.
Particles P and Q travel in the same direction between times 1 and 3 and also when T is greater than 5.
Particle Q changes direction at times T=3 and T=5, with its velocity switching from positive to negative and vice versa.
The acceleration of particle Q at T=2 is negative, indicating the speed is decreasing.
Care must be taken when interpreting acceleration and speed - a negative acceleration does not always mean the speed is decreasing.
The position of particle Q when it first changes direction at T=3 is found by integrating its velocity function from 0 to 3 and adding the initial position.
The integral of the velocity function gives the position function of particle Q.
Particle Q's position at T=3, when it changes direction, is calculated to be 23.
Scoring guidelines are provided to check the work and ensure all parts of the question are addressed.
Technically, the upper bound of the time interval should be included as 8 in the response, even though it is implied.
The video provides a detailed step-by-step solution to the AP Calculus free response question.
The presenter uses the chain rule to find derivatives and the quadratic formula to find zeros of a function.
The video concludes with a review of the solution and a reminder to leave comments, likes or subscribe.
The presenter emphasizes the importance of careful reasoning and interpretation of acceleration and speed.
Transcripts
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