AP CALCULUS AB 2022 Exam Full Solution FRQ#6(a,b)
TLDRThe video script discusses the motion of two particles, P and Q, along the x and y axes respectively. Particle P moves horizontally with a position function given by x_p(t) = 6 - 4e^{-t}, from which its velocity function v_p(t) = 4e^{-t} is derived. Particle Q has a velocity function v_q(t) = 1/t^2, which is always positive for t > 0, indicating a constant rightward motion. The acceleration function for Q, a_q(t) = -2t^{-3}, is found to be negative for all t > 0, suggesting that the speed of Q is continuously decreasing. The key takeaway is that while particle P moves with a decreasing velocity in the horizontal direction, particle Q's speed is always decreasing in the vertical direction due to the negative acceleration acting against its velocity.
Takeaways
- 📐 The position function for particle p moving along the x-axis is given by \( x_p(t) = 6 - 4e^{-t} \) for \( t > 0 \).
- 🔄 To find the velocity of particle p, \( v_p(t) \), differentiate the position function to get \( v_p(t) = -4e^{-t} \).
- ⏱ The velocity function \( v_q(t) \) for particle q is \( v_q(t) = \frac{1}{t^2} \), which is always positive for \( t > 0 \), indicating a rightward movement.
- 🚫 There is no time when \( v_q(t) \) equals zero, as the particle is always moving in the positive direction on the y-axis.
- 📉 To determine when the speed of particle q is decreasing, we look for when acceleration \( a(t) \) is opposite to the velocity direction.
- 🔢 The acceleration function for particle q, \( a_q(t) \), is derived from \( v_q(t) \) and is \( a_q(t) = -2t^{-3} \).
- ⤵️ The acceleration \( a_q(t) \) is always negative for \( t > 0 \), indicating that particle q is always accelerating to the left.
- 🔄 Since the acceleration is always in the opposite direction to the velocity, the speed of particle q is continuously decreasing for all \( t > 0 \).
- 📌 There is no time when the direction of velocity changes to the left because \( v_q(t) \) never equals zero and is always positive.
- ⏳ The domain of the problem is \( t > 0 \), and within this domain, the speed of particle q is always decreasing due to the constant negative acceleration.
- 📈 The position function for particle p and the velocity function for particle q are both exponential functions, which are important for understanding their behavior over time.
Q & A
What is the position function of particle p along the x-axis for t > 0?
-The position function of particle p is given by x_p(t) = 6 - 4e^(-t).
How is the velocity function of particle p derived from its position function?
-The velocity function of particle p, v_p(t), is derived by differentiating the position function x_p(t) with respect to time t, resulting in v_p(t) = -4e^(-t).
What is the velocity function of particle q along the y-axis at time t=0?
-The velocity function of particle q is given by v_q(t) = 1/t^2.
What is the condition for the speed of particle q to be decreasing?
-The speed of particle q will be decreasing when the acceleration a(t) is opposite in direction to the velocity v(t).
What is the acceleration function of particle q?
-The acceleration function of particle q is found by differentiating the velocity function v_q(t), resulting in a_q(t) = -2t^(-3).
How does the direction of particle q's velocity change over time?
-The velocity function v_q(t) = 1/t^2 indicates that particle q moves to the right for all real numbers t > 0, as the velocity is always positive.
At what times is the speed of particle q decreasing?
-The speed of particle q is decreasing for all times t > 0, as the acceleration a_q(t) is negative for the entire domain, indicating a constant acceleration to the left.
What is the domain of the velocity and acceleration functions for particle q?
-The domain of both the velocity function v_q(t) and the acceleration function a_q(t) is t > 0.
What is the significance of the negative sign in the acceleration function a_q(t)?
-The negative sign in the acceleration function a_q(t) = -2t^(-3) indicates that the acceleration is acting in the opposite direction to the positive x-axis, or to the left.
Is there a time when particle q's velocity is zero?
-No, there is no time when the velocity of particle q is zero, as the velocity function v_q(t) = 1/t^2 does not equal zero for any positive t.
How can the direction of particle q's velocity be determined?
-The direction of particle q's velocity can be determined by the sign of the velocity function v_q(t). Since v_q(t) is always positive for t > 0, the particle moves to the right.
What is the implication of particle q always having a negative acceleration?
-The implication of particle q always having a negative acceleration is that the particle is continuously slowing down as it moves to the right, as the acceleration is acting in the opposite direction to its velocity.
Outlines
📊 Calculating Velocity and Acceleration
In this section, we explore the motion of two particles, P and Q, moving along the x-axis and y-axis respectively. For particle P, its position is described by the equation x_p(t) = 6 - 4e^(-t), and we derive its velocity by differentiating this position function, resulting in v_p(t) = 4e^(-t). Particle Q, on the other hand, starts with a velocity function v_q(t) = 1/t^2, which remains positive, indicating a constant rightward motion. We also look into calculating the acceleration of Q by differentiating its velocity, leading to a_q(t) = -2/t^3, a negative value suggesting acceleration to the left. This segment elaborates on the calculation methods and the physical interpretation of these functions concerning the particles' motions.
📉 Analyzing Decreasing Speed
This segment focuses on identifying when the speed of particle Q is decreasing. By analyzing the velocity and acceleration functions, we determine that particle Q, which moves rightward (positive velocity), experiences negative acceleration (acceleration to the left) for all t > 0. This opposite direction of velocity and acceleration ensures that the speed of Q is always decreasing. The summary provides a logical breakdown of how the speed's directionality and the acceleration's impact relate, leading to a consistent decrease in speed for particle Q over time.
Mindmap
Keywords
💡Particle
💡Position Function
💡Velocity
💡Differentiation
💡Acceleration
💡Speed
💡Domain
💡Chain Rule
💡Direction of Motion
💡Zeroes
💡Exponential Function
Highlights
The position function of particle p is given by x_p(t) = 6 - 4e^(-t) for t > 0, indicating motion along the x-axis.
Differentiating the position function yields the velocity function v_p(t) = -4e^(-t), which is derived using the chain rule.
Particle q has a velocity function v_q(t) = 1/t^2, which is important for understanding its motion along the y-axis.
The velocity function v_q(t) does not have a real solution for zero, indicating it is always positive for t > 0.
The domain of v_q(t) is t > 0, and it is established that the particle is moving to the right for all real numbers within this domain.
The acceleration function a_q(t) is derived from v_q(t), resulting in a_q(t) = -2t^(-3).
The acceleration a_q(t) does not equal zero for positive t, meaning the direction of motion does not change.
By testing a point, such as t = 1, it is confirmed that the acceleration is negative, indicating the particle is accelerating to the left.
The speed of particle q is always decreasing for t > 0 due to the negative acceleration.
The scenario of decreasing speed occurs when acceleration is opposite to velocity, either with a positive velocity and negative acceleration or negative velocity and positive acceleration.
The concept of speed and its relation to velocity and acceleration is central to understanding the particle's motion.
The problem-solving approach involves differentiating given functions to find velocity and acceleration, and then analyzing their behavior over time.
The use of the chain rule in calculus is demonstrated in finding the derivative of the position function.
The importance of understanding the domain of the given functions to determine the range of possible values for the variables.
The method of finding zeros of a function to determine the points where the function changes behavior.
The practical application of calculus in physics to determine the motion of particles in one-dimensional space.
The significance of the direction of acceleration in determining whether the speed of a particle is increasing or decreasing.
The comprehensive analysis of the particle's motion by considering both its velocity and acceleration over time.
Transcripts
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