2023 AP Calculus AB FRQ #2
TLDRThe video script discusses a calculus problem from the 2023 Calc AB exam involving a velocity function. The problem is about Steven swimming in a 50-meter pool for 90 seconds, with his velocity modeled by a function V(T). Part A asks when Steven changes direction, which is determined by the sign change of V(T) at T=56 seconds. Part B involves calculating Steven's acceleration at T=60 seconds, revealing a negative acceleration, indicating he is slowing down. Part C requires finding the distance between Steven's positions at T=20 and T=80 seconds, which is the absolute value of the displacement, calculated as the integral of V(T) from 20 to 80. Part D asks for the total distance Steven swims from T=0 to T=90 seconds, which is the integral of the absolute value of V(T). The problem is solved using calculus concepts, including derivatives for acceleration and integrals for distance and displacement, with attention to units and the interpretation of signs.
Takeaways
- ๐โโ๏ธ The problem involves calculating the behavior of Steven's swimming velocity in a 50-meter pool over 90 seconds.
- ๐ Steven's velocity is modeled by a function V(T) = 2.38e^(-0.02T)sin(ฯ/56T), where T is time in seconds and V(T) is in meters per second.
- ๐ For part A, the question asks to find the times when Steven changes direction, which is indicated by a sign change in the velocity function.
- ๐ Steven changes direction at T = 56 seconds, as this is the only time the velocity function V(T) changes sign.
- ๐งฎ In part B, acceleration is calculated as the derivative of velocity, denoted as a(60) = V'(60), with units of meters per second squared (m/sยฒ).
- โก๏ธ At T = 60 seconds, Steven is slowing down because the acceleration is negative and the velocity is also negative, indicating they have the same sign.
- ๐ For part C, the task is to find the distance between Steven's positions at times 20 and 80 seconds, which is the absolute value of displacement.
- ๐ The displacement is calculated by integrating the velocity function from T = 20 to T = 80, resulting in approximately 23.384 meters.
- ๐ข In part D, the total distance Steven swims from T = 0 to T = 90 seconds is found by integrating the absolute value of the velocity function, yielding approximately 62.164 meters.
- ๐ค The script emphasizes the importance of understanding the difference between distance (total path length) and displacement (straight-line distance from start to end).
- ๐ The problem-solving approach involves using calculus to analyze the velocity function, its derivative for acceleration, and integral calculus for distance and displacement calculations.
Q & A
What is the problem presented in the transcript?
-The problem involves calculating various aspects of Steven's swimming in a 50-meter pool, using a velocity function given by V(T) = 2.38e^(-0.02T)sin(ฯ/56T), where T is in seconds and V(T) is in meters per second.
What is the significance of the velocity function's sign change?
-The sign change in the velocity function indicates a change in Steven's direction while swimming.
At what time does Steven change direction according to the transcript?
-Steven changes direction at T equals 56 seconds.
What is the acceleration of Steven at T equals 60 seconds?
-The acceleration of Steven at T equals 60 seconds is approximately -0.036 meters per second squared.
Is Steven speeding up or slowing down at T equals 60 seconds?
-Steven is slowing down at T equals 60 seconds because the velocity and acceleration have the same sign, which is negative.
What is the distance between Steven's position at time 20 and his position at time 80?
-The distance between Steven's position at time 20 and time 80 is approximately 23.384 meters.
How is the total distance Steven swims from 0 to 90 seconds calculated?
-The total distance Steven swims is calculated by integrating the absolute value of the velocity function from 0 to 90 seconds.
What is the total distance Steven swims over the time interval from 0 to 90 seconds?
-The total distance Steven swims over the time interval from 0 to 90 seconds is approximately 62.164 meters.
Why is the absolute value used in the integral for the total distance swum?
-The absolute value is used because distance traveled is always positive, regardless of the direction of Steven's swim.
What is the formula for Steven's velocity as given in the transcript?
-Steven's velocity is modeled by the formula V(T) = 2.38e^(-0.02T)sin(ฯ/56T), where T is time in seconds.
What is the unit for the acceleration calculated at T equals 60 seconds?
-The unit for the acceleration is meters per second squared (m/sยฒ).
How does one determine if Steven is speeding up or slowing down?
-One determines if Steven is speeding up or slowing down by looking at the signs of velocity and acceleration; if they are the same, he is speeding up, and if they are different, he is slowing down.
What is the concept of displacement in the context of the problem?
-Displacement in this context is the straight-line distance between two points of Steven's swim, which is the absolute value of the integral of velocity over the time interval.
Outlines
๐โโ๏ธ Steven's Velocity and Direction Change
The video discusses a problem from the 2023 calculus exam involving Steven swimming in a 50-meter pool for 90 seconds. Steven's velocity is given by a function V(T) = 2.38e^(-0.02T)sin(ฯ/56T), where T is in seconds and V is in meters per second. The video explains how to find the time T at which Steven changes direction by looking for a sign change in the velocity function, which occurs at T=56 seconds. The explanation includes using a graphing calculator to visualize and find the intersection points of the velocity function and the x-axis.
๐ Calculating Steven's Acceleration at T=60
The video moves on to part B, which involves finding Steven's acceleration at T=60 seconds. Acceleration is the derivative of velocity, denoted as a(T) = V'(T). The video demonstrates how to calculate the derivative of the velocity function at T=60 seconds, resulting in an approximate value of -0.036 m/sยฒ. The units for acceleration are thus m/sยฒ. The video also discusses how to determine if Steven is speeding up or slowing down at T=60 by checking the sign of the velocity at that time, which is found to be negative, indicating that Steven is slowing down.
๐ Displacement Between Two Points in Time
Part C of the problem asks for the distance between Steven's positions at times 20 and 80 seconds. The video explains that the problem is about finding the displacement, not the distance traveled. The correct approach is to integrate the velocity function from T=20 to T=80 and take the absolute value of the result. The video clarifies that the integral does not result in a negative value, so the absolute value is not necessary in the calculation, leading to an approximate displacement of 23.384 meters.
๐๏ธโโ๏ธ Total Distance Swum by Steven Over 90 Seconds
The final part, D, requires finding the total distance Steven swims from T=0 to T=90 seconds. This is different from the previous part as it involves the distance traveled, not displacement. The video shows how to set up the integral of the absolute value of the velocity function over the given interval, resulting in a total distance of approximately 62.164 meters. The units for this distance are meters, as Steven is swimming.
Mindmap
Keywords
๐กVelocity
๐กAcceleration
๐กDisplacement
๐กDistance Traveled
๐กIntegration
๐กDirection Change
๐กSign Change
๐กDerivative
๐กCalculator
๐กGraphing
๐กUnits of Measure
Highlights
Steven swims back and forth along a straight path in a 50 meter long pool.
Steven's velocity is modeled by a function V(T) = 2.38 * e^(-0.02T) * sin(ฯ/56 * T), where T is in seconds and V(T) is in meters per second.
To find when Steven changes direction, the sign change of V(T) is considered, which occurs at T = 56 seconds.
For Part B, Steven's acceleration at T = 60 seconds is calculated using the derivative of velocity.
The acceleration at T = 60 seconds is approximately -0.036 m/sยฒ, indicating a negative value.
The velocity at T = 60 seconds is negative, which means Steven is slowing down at that time.
The reason for slowing down is that velocity and acceleration have the same sign.
In Part C, the distance between Steven's position at time 20 and time 80 is calculated using the integral of V(T) from 20 to 80.
The calculated displacement between the two positions is approximately 23.384 meters.
For Part D, the total distance Steven swims from 0 to 90 seconds is found using the integral of the absolute value of V(T).
The total distance traveled by Steven is approximately 62.164 meters.
The problem involves calculus concepts such as velocity, acceleration, and integration to solve a real-world scenario.
Graphing the velocity function helps to visualize when Steven changes direction in the pool.
Units of measure are carefully considered when calculating acceleration and distance.
Understanding the concept of displacement and distance traveled is crucial for solving the problem accurately.
The use of a calculator is essential for finding derivatives and integrals in the problem-solving process.
The problem tests the ability to apply calculus to model and solve a dynamic physical situation.
The problem requires both conceptual understanding and computational skills in calculus.
Transcripts
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