2011 AP Calculus AB Free Response #5

Allen Tsao The STEM Coach
25 Oct 201807:13
EducationalLearning
32 Likes 10 Comments

TLDRIn this educational video, Alan from Bottle Stem Coach dives into the 2011 AP Calculus AB exam, focusing on questions 5 and 6. The video begins with a problem involving a landfill's solid waste accumulation, modeled by a differential equation. Alan uses the initial condition of 1,400 tons of waste to find the derivative at the start and approximates the waste amount after three months using a tangent line, revealing it as an underestimate due to the concave-up nature of the function. He then proceeds to solve the differential equation using separation of variables, finding an exact solution involving exponential functions. The video concludes with a correction to the initial approximation and an invitation for viewers to engage with the content and seek further assistance through offered platforms like Twitch and Discord.

Takeaways
  • ๐Ÿ“š The video discusses the 2011 AP Calculus AB exam, specifically questions 5 and 6.
  • ๐Ÿ“ˆ The landfill's solid waste amount is modeled by an increasing function W, which satisfies a differential equation over the next 20 years.
  • โž— At T=0, the landfill contains 1,400 tons of solid waste, and the derivative of W at this point is calculated to be 444 tons/year.
  • ๐Ÿ“‰ The tangent line to the graph of W at T=0 is used to approximate the amount of solid waste after three months, resulting in an underestimate due to the concavity of the function.
  • ๐Ÿงฎ The second derivative of W is found to be positive, indicating that the function is concave up, which is why the tangent line approximation is an underestimate.
  • ๐Ÿ” The differential equation DW/DT = 1/25 * (W - 300) is solved using separation of variables, a common method in Calculus AB.
  • ๐ŸŒ The general solution to the differential equation is found to be W = C * e^(1/25 * T) + 300, where C is a constant determined by the initial condition.
  • โœ… By substituting the initial condition W(0) = 1,400, the constant C is determined to be 1,100, leading to the exact solution W = 1,100 * e^(1/25 * T).
  • ๐Ÿ”ข The video contains a minor error regarding the units and the plus 300 part in the solution, which is later corrected.
  • ๐Ÿ“ The presenter uses a step-by-step approach to guide viewers through solving the differential equation and understanding the implications of the concavity on the estimate.
  • ๐ŸŽ“ The video concludes with a reminder that the approach to solving differential equations in AP Calculus AB often involves separation of variables and integration.
  • ๐Ÿ“ข The presenter encourages viewers to engage with the content by leaving comments, liking, or subscribing for more educational content.
Q & A
  • What is the topic of the video?

    -The video is about solving problems from the 2011 AP Calculus AB exam, specifically focusing on questions 5 and 6.

  • What was the initial amount of solid waste in the landfill at the start of 2010?

    -The landfill initially contained 1,400 tons of solid waste at the start of 2010.

  • What is the differential equation that models the total amount of solid waste stored at the landfill?

    -The differential equation is given by dW/dt = 1/(25W - 300), where W is the total amount of solid waste in tons and T is measured in years from the start.

  • How is the slope of the tangent line to the graph of W at T=0 calculated?

    -The slope of the tangent line at T=0 is calculated by taking the derivative of W with respect to T, which results in 1/25 * (1400 - 300) = 44.

  • What is the equation of the tangent line at the point (0, 1400)?

    -The equation of the tangent line is W - 1400 = 44T, which simplifies to W = 44T + 1400.

  • How is the amount of solid waste at the end of the first three months approximated?

    -The amount is approximated by plugging T = 1/4 (which is three months in years) into the tangent line equation, resulting in W = 44 * (1/4) + 1400 = 1411 tons.

  • What does the second derivative of W in terms of W indicate?

    -The second derivative, dยฒW/dTยฒ = 1/(625W - 300)ยฒ, indicates the concavity of the function W. If it's positive, the function is concave up, and if it's negative, it's concave down.

  • Is the approximation of the amount of solid waste at the end of the first three months an overestimate or an underestimate?

    -The approximation is an underestimate because the second derivative is positive, indicating that the function is concave up.

  • How is the particular solution to the differential equation found?

    -The particular solution is found by separating variables and integrating both sides of the differential equation, which leads to W = C * e^(1/25T) + 300, where C is a constant determined by the initial condition.

  • What is the exact solution for W with the given initial condition?

    -The exact solution for W, given the initial condition W(0) = 1400, is W = 1100 * e^(1/25T) + 300.

  • What was the mistake made in the video regarding the final answer for the amount of solid waste?

    -The mistake was forgetting to include the 'plus 300' part in the final equation, which should have been W = 1100 * e^(1/25T) + 300 instead of just 1100 * e^(1/25T).

  • How can viewers get more help with their calculus problems?

    -Viewers can get more help by leaving a comment, liking the video, or subscribing to the channel. Additionally, the presenter offers free homework help on Twitch and Discord.

Outlines
00:00
๐Ÿ˜€ Analyzing the Tangent Line and Second Derivative

In this segment, Alan from Bottle Stem Coach continues dissecting the 2011 AP calculus exam. He focuses on question 5, which deals with a landfill's solid waste accumulation over time. Alan starts by determining the tangent line equation at the beginning point, using the given initial condition. He then delves into the second derivative to determine whether the approximation is an underestimate or overestimate. By analyzing the concavity of the curve, Alan concludes that the approximation is an underestimate.

05:04
๐Ÿ˜€ Solving the Differential Equation

This section involves solving the differential equation provided in the calculus exam. Alan employs the separation of variables technique, isolating variables on each side of the equation. By integrating both sides, he derives the general solution, which includes an arbitrary constant 'C'. Alan then applies the initial condition to determine the exact solution, eventually obtaining the specific solution for the landfill's solid waste accumulation over time.

Mindmap
Keywords
๐Ÿ’กAP Calculus Exam
The AP Calculus Exam is a standardized test administered by the College Board for high school students. It assesses their understanding of calculus concepts. In the video, the host is discussing problems from the 2011 AP Calculus AB exam, which is central to the video's educational theme.
๐Ÿ’กDifferential Equation
A differential equation is a mathematical equation that involves a function and its derivatives. It is used to model various phenomena in physics, engineering, and economics. In the video, the host uses a differential equation to model the amount of solid waste in a landfill over time.
๐Ÿ’กIncreasing Function
An increasing function is a type of function that rises over its domain. In the context of the video, the term is used to describe the function modeling the total amount of solid waste, which is expected to increase over time.
๐Ÿ’กTangent Line
A tangent line is a straight line that touches a curve at a single point without crossing it. It represents the slope of the curve at that point. The host uses the concept of a tangent line to approximate the amount of solid waste at the end of the first three months.
๐Ÿ’กDerivative
In calculus, the derivative of a function measures the rate at which the function changes at a particular point. The host calculates the derivative of the waste function to find the slope of the tangent line at a specific time, which is crucial for the approximation.
๐Ÿ’กSecond Derivative
The second derivative of a function is the derivative of the first derivative. It provides information about the concavity of the function and can indicate whether a function is increasing or decreasing at a faster rate. The host uses the second derivative to determine if the initial estimate is an overestimate or an underestimate.
๐Ÿ’กConcave Up/Down
A graph is said to be concave up if the curve bends in a shape similar to an upside-down U, and concave down if it bends in a shape like a U. The host discusses this concept to interpret the sign of the second derivative and its implication on the accuracy of the estimate.
๐Ÿ’กSeparation of Variables
Separation of variables is a method used to solve differential equations by rearranging terms so that all instances of one variable are on one side of the equation and all instances of the other variable are on the other side. The host uses this technique to solve the given differential equation.
๐Ÿ’กIntegration
Integration is a fundamental operation in calculus, the reverse process of finding a derivative. It is used to find the area under a curve or to solve differential equations. The host applies integration to both sides of the differential equation to find the general solution.
๐Ÿ’กNatural Logarithm
The natural logarithm, often denoted as ln, is the logarithm to the base e. It is used in various mathematical and scientific contexts, including solving differential equations. The host uses the natural logarithm in the process of integrating the differential equation.
๐Ÿ’กExponential Function
An exponential function is a mathematical function of the form f(x) = a^x, where a is a positive real number not equal to 1. In the video, the host uses the exponential function to express the solution to the differential equation, modeling the growth of solid waste over time.
Highlights

Alan is discussing the 2011 AP Calculus AB exam, focusing on questions 5 and 6.

The initial landfill contains 1,400 tons of solid waste, with an increasing function W modeling the total amount of waste.

The differential equation for W over the next 20 years is given as W'(t) = 1/25 * (W - 300), where W is in tons and t is in years.

The derivative at t=0 is calculated to be 444, representing the slope of the tangent line at the start.

The tangent line equation is derived as W - 1400 = 44t, using the slope found.

The amount of solid waste at the end of the first three months is approximated using the tangent line equation.

The second derivative of W is found to be W''(t) = 1/625 * (W - 300), which helps determine if the initial estimate is an over or underestimate.

The second derivative is positive at the initial condition, indicating the function is concave up and the initial estimate is an underestimate.

The differential equation is solved using separation of variables, leading to the general solution W(t) = C * e^(1/25 * t) + 300.

The particular solution to the differential equation with the initial condition W(0) = 1,400 is found to be W(t) = 1100 * e^(1/25 * t).

Alan makes a correction to the previous calculation, emphasizing the importance of including the 'plus 300' part in the equation.

The video concludes with a recap of the solutions and a reminder to check the next video for the final question.

Alan encourages viewers to leave comments, likes, or subscribe for more content and offers free homework help on Twitch and Discord.

The video is part of a series on AP Calculus AB exam preparation, providing detailed walkthroughs of exam questions.

The use of differential equations to model real-world scenarios, such as waste accumulation in a landfill, is demonstrated.

The importance of understanding the concavity of a function to determine the accuracy of linear approximations is emphasized.

Alan's teaching style is interactive and aims to make complex calculus concepts more accessible to students.

The video serves as a resource for students studying for the AP Calculus AB exam, providing insights into solving differential equations.

Transcripts
Rate This

5.0 / 5 (0 votes)

Thanks for rating: