AP CALCULUS AB 2022 Exam Full Solution FRQ#2d
TLDRThe video script explains a related rates problem involving a vertical line moving at a constant rate along the base of a solid, changing the area of a cross-sectional square. The goal is to find the rate of change of the area when the line is at x = -0.5. The presenter establishes an expression for the area of the cross section (A(x) = (f(x) - g(x))^2), then differentiates this to find dA/dx. However, since the rate of change is needed with respect to time, the problem is approached as a related rates problem. The derivative at x = 0.5 is found using a graphing calculator, and the known dx/dt (7 units per second) is used to calculate dA/dt. The final answer is a negative value, indicating the area is decreasing at that rate.
Takeaways
- ๐ The problem involves a vertical line in the XY plane that moves horizontally to change the position of a square cross-section, which in turn changes its area.
- ๐ต The vertical line (purple) moves at a constant rate of 7 units per second, which is given as dx/dt.
- ๐ข The cross-section is a square, and its area (A) is represented by the square of the difference between two functions, f(x) - g(x), squared.
- ๐งฎ To find the rate of change of the area (da/dt), we first find da/dx by differentiating the expression for A with respect to x.
- ๐ The differentiation of the area function is done using a graphing calculator to find the derivative at a specific point, x = -0.5.
- ๐ The derivative at x = -0.5 (a') is found to be -1.32455 using the graphing calculator's dy/dx function.
- ๐ด The rate of change of the area with respect to time (da/dt) is then calculated by multiplying the derivative (da/dx) by the given rate of horizontal movement (dx/dt).
- โฑ๏ธ The final calculation for da/dt at x = -0.5, using the values found, results in a value of -9.27 square units per second.
- ๐ The process is framed as a related rates problem, where the rate of change with respect to time is found by relating it to the known rate of change with respect to x.
- ๐ The expression for the area of the cross-section is A(x) = [f(x) - g(x)]^2, which is derived from the side length of the square cross-section.
- ๐ฌ The script demonstrates a step-by-step approach to solving calculus problems involving rates of change and related rates in a geometric context.
Q & A
What is the shape of the cross-sectional area described in the script?
-The cross-sectional area described in the script is a square.
How is the vertical line in the X Y plane moving?
-The vertical line is moving from left to right at a constant rate of 7 units per second.
What is the purpose of the vertical line's movement?
-The movement of the vertical line changes the position of the square, which in turn changes the area of the cross-section.
What is the expression for the area of the cross-section (A) in terms of x?
-The area of the cross-section A is given by the square of the difference between two functions, f(x) and g(x), so A = [f(x) - g(x)]^2.
What is the rate of change of the area with respect to time (dA/dt) that we are trying to find?
-We are trying to find the rate of change of the area when x is at -0.5 seconds.
How is the rate of change of the area with respect to x (dA/dx) related to the rate of change with respect to time (dA/dt)?
-The rate of change of the area with respect to time (dA/dt) is found by multiplying the rate of change with respect to x (dA/dx) by the rate at which x is changing with respect to time (dx/dt).
What is the given rate at which x is changing with respect to time (dx/dt)?
-The given rate at which x is changing with respect to time (dx/dt) is 7 units per second.
How does one find the derivative of the area function at a specific point (dA/dx)?
-The derivative of the area function at a specific point can be found by differentiating the expression for the area and then evaluating it at the given point, or by using a graphing calculator's derivative function.
What is the value of the derivative of the area function at x = -0.5?
-The value of the derivative of the area function at x = -0.5 is -1.324551.
How is the final rate of change of the area with respect to time (dA/dt) calculated?
-The final rate of change of the area with respect to time (dA/dt) is calculated by multiplying the derivative value at x = -0.5 by the rate of change of x with respect to time (dx/dt), which is -1.324551 * 7 units per second.
What is the final calculated rate of change of the area with respect to time (dA/dt) in square units per second?
-The final calculated rate of change of the area with respect to time (dA/dt) is -9.27 square units per second.
What does the negative sign in the rate of change of the area indicate?
-The negative sign in the rate of change of the area indicates that the area is decreasing over time at the point where x is -0.5.
Outlines
๐ Calculating the Rate of Change of a Cross-Sectional Area
This paragraph introduces a geometry problem involving a vertical line moving along the base of a solid in the XY plane, which changes the position of a square cross-section, thereby altering its area. The vertical line moves at a constant rate of 7 units per second. The goal is to find the rate of change of the area of the cross-section when the line is at x = -0.5. To solve this, an expression for the area (A) is established as the square of the side length (s), which is the difference between an upper function (f(x)) and a lower function (g(x)). The task is to find dA/dt, which requires differentiating A with respect to x (dA/dx) and then multiplying by the known dx/dt. The differentiation is done at x = 0.5, and the derivative at this point is found using a graphing calculator, resulting in a slope value that is then multiplied by the rate of movement (dx/dt) to find dA/dt.
๐ Deriving the Final Rate of Change for the Cross-Sectional Area
The second paragraph continues the problem-solving process by finding the derivative (a') at x = -0.5 using the previously established expression from the graphing calculator. With the given dx/dt of seven units per second, the rate of change of the area (dA/dt) is calculated by multiplying the derivative value (a') by dx/dt. The result is a negative value, indicating a decrease in the area over time. The final calculated dA/dt is -9.27 square units per second, which represents the rate at which the area of the cross-section is changing when the vertical line is at x = -0.5.
Mindmap
Keywords
๐กVertical Line
๐กCross-Sectional Square
๐กRate of Change
๐กDifferentiation
๐กRelated Rate Problem
๐กGraphing Calculator
๐กDerivative
๐กConstant Rate
๐กArea Function
๐กExpression
๐กCalculus
Highlights
The vertical line in the XY plane travels from left to right along the base of a solid, changing the position of a cross-sectional square and thus its area
The rate of change of the area of the cross-section above the vertical line is to be found when x is at -0.5
An expression for the area of the cross-section (s^2) is established as the difference between the upper and lower functions squared
Differentiating the area expression with respect to x (da/dx) is required to find the rate of change of the area
The perspective is changed to a related rate problem to find da/dt by first finding da/dx and then using the known dx/dt
The rate of the vertical line moving to the right (dx/dt) is given as 7 units per second
Differentiating the area function at x = 0.5 using a graphing calculator yields a value of -1.32455
The derivative at x = 0.5 (a') represents the slope of the area function at that point
The change in area over time (da/dt) is found by multiplying the a' value at x = -0.5 by the dx/dt rate
The calculated da/dt value is -9.27151 square units per second
The vertical line's movement at a constant rate impacts the area of the cross-section
Finding the rate of change of the area requires differentiating the area expression with respect to x
The area function is graphed and its derivative at a specific point is calculated using a graphing calculator
The known dx/dt rate is used in conjunction with the area function's derivative to find the rate of change of the area
The final calculated da/dt value indicates the rate at which the area of the cross-section changes per second
The problem involves a solid with a cross-sectional square that changes position and area as a vertical line moves along its base
The area of the cross-section is expressed as the square of the difference between the upper and lower functions
Differentiating the area expression with respect to x and using the known dx/dt rate allows finding the rate of change of the area
A graphing calculator is used to graph the area function and calculate its derivative at a specific point
Transcripts
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