AP Calculus BC 4.4 Introduction to Related Rates Study Notes - New Syllabus
AP Calculus BC 4.4 Introduction to Related Rates Study Notes- New syllabus
AP Calculus BC 4.4 Introduction to Related Rates Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Calculate related rates in applied contexts.
Key Concepts:
- Introduction to Related Rates
Introduction to Related Rates
Introduction to Related Rates
Related rates problems involve finding the rate at which one quantity changes with respect to time, based on its relationship with other quantities that also change over time. These problems typically arise in geometry, physics, engineering, and real-world applications where multiple variables are changing simultaneously.
If two or more variables are related through an equation, and each variable is changing with time, then we can differentiate both sides of the equation with respect to time \( t \), using the Chain Rule.
General Strategy for Solving Related Rates Problems:
- Identify all given quantities and rates (known values and derivatives).
- Identify the rate you need to find.
- Write an equation relating the variables involved.
- Differentiating both sides of the equation with respect to time \( t \) (use implicit differentiation).
- Substitute the known values and solve for the required rate.
Notation:
- \( \dfrac{dx}{dt} \) is the rate of change of \( x \) with respect to time.
- \( \dfrac{dy}{dt} \) is the rate of change of \( y \) with respect to time.
Example:
The radius of a circle is increasing at a rate of 3 cm/s.
Find the rate at which the area of the circle is increasing when the radius is 5 cm.
▶️Answer/Explanation
Let \( A \) be the area and \( r \) be the radius. The area of a circle is given by:
\( A = \pi r^2 \)
Differentiating both sides with respect to time \( t \):
\( \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt} \)
Given: \( \dfrac{dr}{dt} = 3 \) cm/s, and \( r = 5 \) cm
Substitute:
\( \dfrac{dA}{dt} = 2\pi(5)(3) = 30\pi \)
So, the area is increasing at \( 30\pi \approx 94.25 \) cm²/s when the radius is 5 cm.
Example:
A 10-foot ladder is leaning against a vertical wall. The bottom of the ladder slides away from the wall at a rate of 2 ft/s.
How fast is the top of the ladder sliding down the wall when the bottom is 6 ft from the wall?
▶️Answer/Explanation
Let \( x \) be the distance from the wall to the bottom of the ladder, and \( y \) be the height of the top of the ladder. By the Pythagorean Theorem:
\( x^2 + y^2 = 10^2 = 100 \)
Differentiating both sides with respect to time \( t \):
\( 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0 \)
Simplify:
\( x\dfrac{dx}{dt} + y\dfrac{dy}{dt} = 0 \)
Given: \( x = 6 \), \( \dfrac{dx}{dt} = 2 \), and we need \( \dfrac{dy}{dt} \)
First find \( y \):
\( y = \sqrt{100 – x^2} = \sqrt{100 – 36} = \sqrt{64} = 8 \)
Substitute into the equation:
\( 6(2) + 8\dfrac{dy}{dt} = 0 \Rightarrow 12 + 8\dfrac{dy}{dt} = 0 \)
\( \dfrac{dy}{dt} = -\dfrac{12}{8} = -1.5 \)
The top of the ladder is sliding down at 1.5 ft/s when the bottom is 6 ft from the wall.