AP Calculus BC 4.5 Solving Related Rates Problems Study Notes - New Syllabus
AP Calculus BC 4.5 Solving Related Rates Problems Study Notes- New syllabus
AP Calculus BC 4.5 Solving Related Rates Problems Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- The derivative can be used to solve related rates problems; that is, finding a rate at which one quantity is changing by relating it to other quantities whose rates of change are known.
Key Concepts:
- Solving Related Rates Problems
Solving Related Rates Problems
Solving Related Rates Problems
The derivative can be used to solve related rates problems situations in which the rate of change of one quantity is related to the rate of change of another. These types of problems often appear in geometry, physics, and real-life applications.
In a related rates problem, you are typically given a rate of change (like \( \frac{dx}{dt} \)) and asked to find another rate of change (like \( \frac{dy}{dt} \)), using a relationship between the variables (often a formula or equation).
General Steps for Solving Related Rates Problems
- Draw a diagram if applicable.
- Assign variables to quantities that change over time.
- Write an equation relating those variables.
- Differentiate both sides with respect to time \( t \).
- Substitute all known values (including rates).
- Solve for the unknown rate.
Example:
A 10-foot ladder is leaning against a vertical wall. The bottom of the ladder is sliding away from the wall at a rate of 3 ft/sec.
How fast is the top of the ladder sliding down the wall when the bottom is 6 ft from the wall?
▶️Answer/Explanation
- \( x \): distance from the wall to the bottom of the ladder (ft)
- \( y \): height of the top of the ladder above the ground (ft)
- \( \frac{dx}{dt} = 3 \) ft/sec (given)
- \( \frac{dy}{dt} = ? \) (to find)
Use the Pythagorean Theorem
Because the ladder length is constant: \( x^2 + y^2 = 10^2 = 100 \)
Differentiate with respect to \( t \)
\( 2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0 \)
Substitute known values
- \( x = 6 \), so use Pythagorean Theorem to find \( y \): \( y = \sqrt{100 – 36} = \sqrt{64} = 8 \)
- Plug into the differentiated equation: \( 2(6)(3) + 2(8)\frac{dy}{dt} = 0 \)
- \( 36 + 16 \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = \frac{-36}{16} = -2.25 \)
Final Answer: The top of the ladder is sliding down the wall at a rate of 2.25 ft/sec.
Example:
Water is draining from a conical tank at a rate of 2 cubic meters per minute. The tank is 4 meters high and has a radius of 2 meters at the top.
How fast is the water level dropping when the water is 3 meters deep?
▶️Answer/Explanation
- \( V \) = volume of water in the tank (m³)
- \( h \) = height of the water at time \( t \) (m)
- \( r \) = radius of the water surface at time \( t \) (m)
The volume of a cone is given by:
\( V = \frac{1}{3} \pi r^2 h \)
Relate radius and height using similar triangles
The tank has radius 2 m and height 4 m, so: \( \frac{r}{h} = \frac{2}{4} = \frac{1}{2} \Rightarrow r = \frac{h}{2} \)
Substitute into volume formula
\( V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi}{12} h^3 \)
Differentiate with respect to time
\( \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt} \)
Given \( \frac{dV}{dt} = -2 \) and \( h = 3 \):
\( -2 = \frac{\pi}{4} (3)^2 \cdot \frac{dh}{dt} = \frac{9\pi}{4} \cdot \frac{dh}{dt} \)
\( \frac{dh}{dt} = \frac{-8}{9\pi} \approx -0.283 \) m/min
Final Answer: The water level is dropping at approximately 0.283 m/min when the water is 3 meters deep.
Example:
Sand is falling from a conveyor belt to form a conical pile.
The radius of the base of the cone is always half the height. If the height of the cone is increasing at a rate of 2 cm/min when the height is 12 cm, at what rate is the volume of the cone increasing?
- 48π cm³/min
- 96π cm³/min
- 144π cm³/min
- 192π cm³/min
▶️Answer/Explanation
Volume of a cone is: \( V = \frac{1}{3} \pi r^2 h \)
Since \( r = \frac{1}{2}h \), substitute: \( V = \frac{1}{3} \pi \left( \frac{1}{2}h \right)^2 h = \frac{1}{3} \pi \cdot \frac{1}{4}h^2 \cdot h = \frac{\pi}{12} h^3 \)
Differentiate: \( \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt} \)
Plug in \( h = 12 \), \( \frac{dh}{dt} = 2 \): \( \frac{dV}{dt} = \frac{\pi}{4} \cdot 144 \cdot 2 = 72\pi \cdot 2 = 144\pi \) cm³/min
Correct Answer: C) 144π cm³/min