Derivatives and Related Rates

Study Notes

Derivatives, a fundamental concept in calculus, enable us to analyze changing quantities and relationships. They help us understand how a function's output changes in response to variations in its inputs. In this article, we'll focus on derivatives in the context of related rates, which is a specific application of these concepts.

Derivatives

A derivative, denoted by (f'(x)) or (dy/dx), is the instantaneous rate of change of a function with respect to its arguments. For instance, (f'(x)) represents the slope of the tangent line to the graph of (f(x)) at any given point (x).

To calculate derivatives, we apply the following rules:

Power Rule: If (f(x) = x^n), then (f'(x) = nx^{n-1}).
Sum Rule: If (f(x) = g(x) + h(x)), then (f'(x) = g'(x) + h'(x)).
Product Rule: If (f(x) = g(x)h(x)), then (f'(x) = g'(x)h(x) + g(x)h'(x)).
Chain Rule: If (f(x) = g(h(x))), then (f'(x) = g'(h(x))h'(x)).

Related rates is a method that uses derivatives to analyze the relationship between changing quantities. When two or more quantities are related, the related rates method helps us find the instantaneous rate of change of one quantity in terms of the instantaneous rates of change of the other quantities.

For example, consider a cylinder with a radius of (r) and a height of (h). If the cylinder is rolling and its circumference increases at a rate of (dr/dt = 0.2) units per second, and its height increases at a rate of (dh/dt = 0.5) units per second, we want to find (dh/dr).

Using the related rates method, we note that the circumference and radius are related by (2\pi r). Since the circumference is increasing at a rate of 0.2 units per second, we have (2\pi \frac{dr}{dt} = 0.2). Dividing both sides by (2\pi), we get (\frac{dr}{dt} = \frac{0.2}{2\pi}).

The volume of the cylinder is given by (V = \pi r^2h). We want to find the rate of change of (h) with respect to (r), which is (dh/dr). To do this, we'll differentiate both sides of the volume equation with respect to time (t):

[ \frac{dV}{dt} = \pi (2rh\frac{dr}{dt} + r^2\frac{dh}{dt}) ]

Since the volume increases at a rate of (\frac{dV}{dt} = 0.5\pi) units per second, we can substitute this and the expressions for (\frac{dr}{dt}) and (dh/dt) we found earlier:

[ 0.5\pi = \pi (0.2r\frac{dr}{dt} + r^2\frac{dh}{dt}) ]

Solving for (\frac{dh}{dr}), we get

[ \frac{dh}{dr} = \frac{0.5 - 0.2r\frac{dr}{dt}}{r^2} = \frac{0.5 - 0.2(0.2)\frac{0.2}{2\pi}}{r^2} = \frac{0.5}{r^2} ]

This is the instantaneous rate at which the height of the cylinder changes with respect to its radius.

Applications and Uses

Related rates problems are useful in many practical situations, including:

Motion and kinematics: Analyzing the relationships between the position, velocity, and acceleration of objects in motion.
Geometry: Finding quantities such as the rate of change of a cone's volume as its base radius changes.
Engineering and physics: Calculating the rate of change of forces, pressures, and other variables in various systems.

Related rates problems provide us with a powerful tool to analyze the relationships between changing quantities in a wide variety of contexts. By understanding derivatives and how to apply them to related rates problems, we can better understand and model the world around us.

Description

Explore the fundamental concepts of derivatives and related rates in calculus. Learn how to calculate derivatives using rules such as the Power Rule, Sum Rule, Product Rule, and Chain Rule. Delve into related rates problems to analyze changing quantities and relationships between variables in various real-world scenarios.