Motion and Rate of Change Applications

One of the most practical applications of derivatives is analyzing how things change over time. In calculus, we use derivatives to study the motion of objects along a straight line (kinematics) and to solve problems where multiple quantities change in relation to one another (related rates).

Motion Along a Line

When a particle moves along a straight line, its position at any given time $t$ can be modeled by a position function $s(t)$. We can use derivatives to find the particle's velocity and acceleration.

• Position: $s(t)$ represents the location of the object at time $t$.
• Velocity: $v(t) = s'(t)$. The first derivative of position tells us how fast the object is moving and in which direction.
• Acceleration: $a(t) = s''(t) = v'(t)$. The second derivative of position (or the first derivative of velocity) tells us how the velocity is changing.
• Speed: Speed is the absolute value of velocity, $|v(t)|$. It tells us how fast the object is moving regardless of direction.
• At Rest: A particle is momentarily "at rest" or stopped when its velocity is zero, meaning $v(t) = 0$.

Example: Analyzing Motion

Problem: Given the position function $s(t) = t^3 - 6t^2 + 9t$, find when the particle is at rest and determine its acceleration at those times.

Solution:

1. Find the velocity function: Take the first derivative of the position function. $v(t) = s'(t) = 3t^2 - 12t + 9$

2. Find when the particle is at rest: Set the velocity equal to zero and solve for $t$. $3t^2 - 12t + 9 = 0$ $3(t^2 - 4t + 3) = 0$ $3(t - 1)(t - 3) = 0$ The particle is at rest at $t = 1$ and $t = 3$.

3. Find the acceleration function: Take the derivative of the velocity function. $a(t) = v'(t) = 6t - 12$

4. Evaluate acceleration at $t = 1$ and $t = 3$: For $t = 1$: $a(1) = 6(1) - 12 = -6$ For $t = 3$: $a(3) = 6(3) - 12 = 6$

Related Rates of Change

Derivatives represent the instantaneous rate of change of one variable with respect to another. In "related rates" problems, we compute the rate of change of one quantity in terms of the rate of change of another known quantity, usually with respect to time $t$. We achieve this by differentiating an equation relating the quantities using the chain rule.

Example: Related Rates

Problem: The radius of a spherical balloon increases at a rate of $2\text{ cm/s}$. Find the rate of change of the volume when the radius is $r = 5\text{ cm}$.

Solution:

1. Identify the given information: The rate of change of the radius is $\frac{dr}{dt} = 2$. We want to find $\frac{dV}{dt}$ when $r = 5$.

2. Write the relating equation: The volume of a sphere is given by: $V = \frac{4}{3}\pi r^3$

3. Differentiate with respect to time $t$: Using the chain rule, differentiate both sides with respect to $t$. $\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt}$ $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

4. Substitute the known values: Plug in $r = 5$ and $\frac{dr}{dt} = 2$. $\frac{dV}{dt} = 4\pi (5)^2 (2)$ $\frac{dV}{dt} = 4\pi (25) (2) = 200\pi$

The volume is increasing at a rate of $200\pi\text{ cm}^3/\text{s}$.