AP CALCULUS AB 2022 Exam Full Solution FRQ#4d
TLDRThe video script presents a related rates problem involving the volume of a cone. Given that the height of the cone decreases at a rate of 2 cm per day, the problem asks to find the rate of change of the cone's volume with respect to time when the radius is 100 cm and the height is 50 cm at T equals three days. The solution involves differentiating the volume formula ( V = 1/3 pi r^2 h ) with respect to time, applying the product rule and chain rule to account for the changing radius and height. The known decrease in height ( dh/dt ) is utilized, and the unknown rate of change of the radius ( dr/dt ) is determined from the given conditions. After substituting the known values and solving, the rate of change of the volume ( dV/dt ) is found to be ( -70000/3 pi ) cubic centimeters per day, indicating a decrease in volume over time.
Takeaways
- 📚 The problem is a related rates problem involving the volume of a cone, which changes over time as its dimensions change.
- ⏳ The rate of change of the height of the cone is given as 2 cm/day, represented as \( \frac{dH}{dt} = -2 \) cm/day.
- 📏 The initial radius (r) of the cone is 100 cm, and the initial height (h) is 50 cm at time T=3 days.
- 🔢 The formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h \), which will be differentiated with respect to time (t).
- 🧮 Differentiating the volume formula with respect to time involves applying the product rule and chain rule, resulting in \( \frac{dV}{dt} = \frac{\pi}{3} (2rh\frac{dr}{dt} + r^2\frac{dH}{dt}) \).
- 🔍 The known values are \( \frac{dH}{dt} = -2 \) cm/day, r = 100 cm, and h = 50 cm, which will be substituted into the differentiated equation.
- 🔑 The rate of change of the radius with respect to time, \( \frac{dr}{dt} \), is not initially known and must be found to solve for \( \frac{dV}{dt} \).
- 🔍 The value of \( \frac{dr}{dt} \) is given later in the transcript as -5 cm/day.
- 🧐 Substituting the known values into the differentiated equation allows for the calculation of \( \frac{dV}{dt} \) at time T=3 days.
- 📉 The final calculation results in \( \frac{dV}{dt} \) being a negative value, indicating that the volume of the cone is decreasing over time.
- 🎓 The overall process demonstrates the application of related rates in calculus to solve real-world problems involving changing dimensions over time.
Q & A
What is the rate of decrease in the height of the cone per day?
-The height of the cone decreases at a rate of two centimeters per day.
What is the radius of the cone at time T equals three days?
-The radius of the cone at time T equals three days is 100 centimeters.
What is the height of the cone at time T equals three days?
-The height of the cone at time T equals three days is 50 centimeters.
What formula is used to calculate the volume of a cone?
-The formula used to calculate the volume of a cone is V = (1/3)πr²h, where r is the radius and h is the height of the cone.
What is the given rate of change of the height of the cone with respect to time, represented as dh/dt?
-The given rate of change of the height of the cone with respect to time, dh/dt, is -2 centimeters per day.
What is the goal of the related rates problem presented in the script?
-The goal of the related rates problem is to find the rate of change of the volume of the cone with respect to time, represented as dv/dt, at T equals three days.
What differentiation technique is used to find dv/dt in the script?
-The differentiation technique used to find dv/dt is implicit differentiation, applying the product rule and the chain rule.
What is the expression for dv/dt after applying the differentiation and before plugging in the values?
-The expression for dv/dt after differentiation and before plugging in the values is dv/dt = (2πrh * dr/dt) + (πr² * dh/dt).
What is the value of dr/dt needed to calculate dv/dt?
-The value of dr/dt needed to calculate dv/dt is -5 centimeters per day, which is given at time T equals three days.
What is the final calculated rate of change of the volume of the cone with respect to time at T equals three days?
-The final calculated rate of change of the volume of the cone with respect to time at T equals three days is -70,000π cubic centimeters per day.
What does the negative sign in the final result for dv/dt indicate?
-The negative sign in the final result for dv/dt indicates that the volume of the cone is decreasing over time.
Why is it necessary to know the rate of change of the radius with respect to time, dr/dt, to solve the problem?
-It is necessary to know dr/dt because the volume of the cone depends on both the radius and the height, and we need to find the rate of change of the volume with respect to time, which involves differentiating both the radius and the height.
Outlines
📚 Calculating the Rate of Change of a Cone's Volume
This paragraph discusses a related rates problem involving a cone. It begins with the given information that the height of the cone decreases at a rate of 2 cm per day. The problem asks to find the rate of change of the cone's volume with respect to time when T equals three days, given the radius is 100 cm and the height is 50 cm. The solution involves differentiating the volume formula V = 1/3 π r^2 h with respect to time, applying the product rule and chain rule to find dV/dt. The known decrease in height (dH/dt = -2 cm/day) is used, and the unknown rate of change of the radius (dr/dt) is determined from the given data at T equals three days, which is -5 cm/day. The final calculation results in the rate of change of volume being -70000π cubic centimeters per day.
🔢 Final Calculation of the Volume Rate of Change
The second paragraph concludes the calculation started in the first. It provides the numerical result of the rate of change of the cone's volume with respect to time. By substituting the known values and the derived dr/dt into the differentiated volume equation, the calculation yields a final result of -50000 - 20000, which simplifies to -70000π cubic centimeters per day. This signifies that the volume of the cone is decreasing at this rate when T is three days.
Mindmap
Keywords
💡Related Rate Problem
💡Cone
💡Height
💡Radius
💡Volume
💡Differentiation
💡Product Rule
💡Chain Rule
💡Derivative
💡Pi
💡Cubic Centimeters per Day
Highlights
The height of the cone decreases at a rate of two centimeters per day.
At time T equals three days, the radius is 100 centimeters and the height is 50 centimeters.
The problem involves finding the rate of change of the volume of the cone with respect to time at T equals three days.
Differentiation of the cone volume formula V = (1/3)πr²h with respect to time is required.
The product rule and chain rule are applied to differentiate the radius and height with respect to time.
The given decrease in height (DH/DT) is two centimeters per day.
The formula for the derivative of volume with respect to time is derived as DV/DT = (2/3)πr²(DH/DT) + (2/3)πrh(Dr/DT).
Parameters for radius (100 cm), height (50 cm), and DH/DT (-2 cm/day) are known and substituted into the formula.
The rate of change of the radius with respect to time (Dr/DT) is still unknown and needs to be determined.
The original table provides Dr/DT at time equals three days, which is -5 cm/day.
Substituting known values and Dr/DT into the formula yields a calculation of the volume rate of change.
The final calculation results in a volume rate of change of -70,000π/3 cubic centimeters per day.
The problem demonstrates a related rates problem-solving approach in calculus.
The use of the chain rule is crucial for differentiating the height with respect to time.
The product rule is essential for differentiating the product of radius and height.
The volume of a cone is given by the formula V = (1/3)πr²h, which is fundamental to the problem.
The problem requires implicit differentiation and application of calculus rules to solve for the rate of change of volume.
The final answer provides insight into how the volume of a cone changes over time with respect to its dimensions.
Transcripts
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