2012 AP Calculus AB Free Response #5
TLDRIn this educational video, Alan from Bothell Stem Coach dives into question number five of the 2012 AP Calculus exam. The question explores the weight gain of a baby bird, which is proportional to the difference between its adult weight and its current weight. With an initial weight of 20 grams, Alan uses differential equations to model the bird's growth, ultimately finding that the bird gains weight faster when it weighs 40 grams compared to 70 grams. He also discusses the shape of the bird's growth curve, explaining why it is concave down based on the second derivative. The video concludes with solving the differential equation using separation of variables, resulting in an exponential function that describes the bird's weight over time. Alan's clear and detailed explanation makes complex calculus concepts accessible to viewers, encouraging further engagement with the subject.
Takeaways
- 🐿 The baby bird's weight gain rate is proportional to the difference between its adult weight and current weight.
- 📈 The bird starts at 20 grams, and the question explores its weight gain at different points: 40 grams and 70 grams.
- 🔢 At 40 grams, the bird's weight gain rate (dB/dt) is faster (12 grams/unit time) compared to when it weighs 70 grams (6 grams/unit time).
- 📉 The differential equation governing the weight gain is derived and solved using separation of variables.
- 📖 Integration and a U-substitution are used to solve the differential equation, leading to the natural log function.
- 🔧 The second derivative of the bird's weight is calculated to determine the concavity of the growth curve.
- 👀 The graph of the bird's weight is shown to be concave down throughout its growth, conflicting with an example graph which is concave up.
- 📌 The initial condition B(0) = 20 grams is utilized to solve for the constant in the equation, refining the bird's growth model.
- 📗 The final equation for the bird's weight over time is B(t) = 100 - 80e^(-t/5).
- 📱 The video promotes further engagement through calls to like, comment, and subscribe, and offers additional homework help through Twitch and Discord.
Q & A
What is the relationship between the rate at which a baby bird gains weight and its current weight?
-The rate at which a baby bird gains weight is proportional to the difference between its adult weight and its current weight.
What differential equation describes the weight gain of the baby bird?
-The weight gain of the baby bird is described by the differential equation dy/dt = k(100 - y), where y is the weight of the bird, t is time, and k is the constant of proportionality.
Explain why the bird is gaining weight faster when it weighs 40 grams compared to when it weighs 70 grams.
-The rate of weight gain, represented by dB/dt, is calculated to be 12 at 40 grams and 6 at 70 grams. Since 12 is greater than 6, the bird is gaining weight faster at 40 grams.
What is the second derivative of the weight function B(t) in terms of B?
-The second derivative of the weight function B(t) is -1/25(100 - B), which indicates concavity.
Why does the graph of the bird's weight resemble a concave-up shape despite the negative second derivative?
-Although the second derivative is negative, indicating concavity downwards, a portion of the graph appears concave up. This discrepancy suggests a flaw in the reasoning or calculation.
How is separation of variables used to find the particular solution to the differential equation?
-Separation of variables involves isolating variables on different sides of the equation, integrating, and solving for the constant. In this case, the equation is separated into variables involving weight and time, integrated, and the constant determined using the initial condition.
What is the particular solution to the given differential equation with the initial condition B(0) = 20?
-The particular solution to the differential equation is B(t) = 100 - 80e^(t/5), where t is time.
Why is it unnecessary to solve for the constant C immediately after integrating the equation?
-Solving for the constant immediately after integration is unnecessary because it can be determined later using the initial condition. It's more efficient to solve for C at a later stage.
What value does the constant C take on in the particular solution?
-The constant C takes on the value of 80 in the particular solution.
What conclusions can be drawn from the analysis of the given differential equation and its solutions?
-The analysis reveals insights into the bird's weight gain, including its rate of change and concavity. It also demonstrates the application of mathematical concepts such as differential equations and separation of variables in solving real-world problems.
Outlines
📚 AP Calculus Exam Question Analysis
In this paragraph, Alan from Bothell Stem Coach is discussing AP Calculus 2012 exam, specifically question number five. The question involves a differential equation where the rate at which a baby bird gains weight is proportional to the difference between its adult weight and its current weight. The initial condition is given as 20 grams. Alan solves the differential equation using separation of variables, finding the relationship between the bird's weight (B) and time (T). He then uses the first and second derivatives to explain why the bird gains weight faster at 40 grams than at 70 grams and why the graph of (B) cannot resemble a certain shape, as the second derivative indicates the graph should be concave down. The solution to the differential equation is provided using the initial condition to find the constant (C), resulting in B = 100 - 80e^{(150t)}.
📈 Derivative Insights and Final Solution
The second paragraph continues the discussion on the AP Calculus exam question. Alan explains that the bird gains weight faster when it weighs 40 grams due to a higher first derivative at that weight. He also clarifies that the graph of (B) should be concave down, which contradicts a portion of the graph being concave up. The final solution to the differential equation is reiterated as B = 100 - 80e^{(150t)}. Alan concludes by inviting viewers to the next video where they will continue working through the AP exam. He also encourages viewers to engage with the content by commenting, liking, or subscribing and mentions that he offers free homework help on Twitch and Discord.
Mindmap
Keywords
💡AP Calculus
💡Differential Equation
💡Proportional
💡Initial Condition
💡Derivative
💡Second Derivative
💡Separation of Variables
💡Natural Logarithm
💡Concave Down
💡Constant of Integration
💡Exponential Function
Highlights
The rate at which a baby bird gains weight is proportional to the difference between its adult weight and its current weight.
The bird's initial weight is 20 grams.
A differential equation is used to model the bird's weight gain over time.
The solution to the differential equation is found using the initial condition B(0) = 20.
The bird gains weight faster when it weighs 40 grams compared to when it weighs 70 grams.
The rate of weight gain (derivative) at 40 grams is 12 grams/time, and at 70 grams is 6 grams/time.
The second derivative of B with respect to time is negative, indicating the graph of B(t) is concave down.
The weight gain graph should not resemble a graph that is concave up, as the second derivative is always negative.
The differential equation is solved using separation of variables.
Integration of the differential equation leads to a natural logarithm function.
The constant C in the solution is determined using the initial condition.
The final solution for the bird's weight B(t) is expressed as B = 100 - 80e^(150t).
The bird's weight increases more rapidly at first, then slows down as it approaches its adult weight.
The video provides a step-by-step explanation of solving the differential equation related to the bird's weight gain.
The presenter offers free homework help on twitch and discord for further assistance.
The video is part of a series on AP Calculus exam solutions.
Transcripts
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