Calculus Study Guide: Derivatives in Motion, Implicit Differentiation, Related Rates, and Linear Approximation

Derivatives Applied to Velocity

Position, Velocity, and Motion

In calculus, the motion of an object over time is often described using its position function and velocity function. The position function, typically denoted as s(t) or x(t), gives the location of an object at time t. The velocity function, v(t), represents the rate of change of position with respect to time.

• Velocity is the derivative of position: $v(t) = s'(t)$

• Displacement over an interval: $\Delta s = s(t_{\text{final}}) - s(t_{\text{initial}})$

• Average velocity over $[a, b]$: $v_{\text{avg}} = \frac{s(b) - s(a)}{b - a}$

• Instantaneous velocity at $t = a$: $v(a) = s'(a)$

• Speed is the magnitude of velocity: $\text{speed} = |v(t)|$

Example: If $s(t) = t^2$, then $v(t) = 2t$.

Derivatives Applied to Acceleration

Acceleration as a Second Derivative

Acceleration is the rate of change of velocity with respect to time. It is the derivative of the velocity function and the second derivative of the position function.

• $a(t) = v'(t) = s''(t)$

• Change in velocity: $\Delta v = v(t_{\text{final}}) - v(t_{\text{initial}})$

• Average acceleration: $a_{\text{avg}} = \frac{v(t_{\text{final}}) - v(t_{\text{initial}})}{t_{\text{final}} - t_{\text{initial}}}$

• Instantaneous acceleration at $t = a$: $a(a) = v'(a)$

Example: If $v(t) = t^2 - 3t^2 + 2t$, then $a(t) = 2t - 6t + 2$.

Projectile Motion Applications

Projectile motion problems often use the position function for vertical motion under gravity:

• $s(t) = s_0 + v_0 t - \frac{1}{2}gt^2$

• $v(t) = v_0 - gt$

• $a(t) = -g$ (where $g \approx 9.8$ m/s2)

Example: For $s(t) = 5 + 20t - 4.9t^2$, find vertical velocity at $t = 3$ s, time of maximum height, and maximum height.

Implicit Differentiation

Implicit vs. Explicit Differentiation

When a function is not given explicitly as $y = f(x)$, but rather as a relation involving $x$ and $y$, we use implicit differentiation to find $\frac{dy}{dx}$.

• Differentiating both sides with respect to $x$

• Applying the chain rule: $\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}$

• Solving for $\frac{dy}{dx}$

Example: For $x^2 + y^2 = 49$, $\frac{dy}{dx} = -\frac{x}{y}$.

Practice Problems

• Find $\frac{dy}{dx}$ for $5x^2 + y^3 = 12$

• Find $\frac{dy}{dx}$ for $xy = \sin(y)$

Related Rates

Introduction to Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to time, given information about the rate of change of another related quantity. These problems often require implicit differentiation with respect to time.

• Take $\frac{d}{dt}$ of both sides of the equation

• Isolate the target rate of change

• Plug in known values and solve

Example: If $y = x^2$ and $\frac{dx}{dt} = 2$, find $\frac{dy}{dt}$ when $x = 4$.

Changing Geometries

Many related rates problems involve geometric shapes whose dimensions change over time, such as cubes, spheres, and triangles.

• Draw a diagram of the scenario

• Identify equations relating all variables

• Differentiation with respect to time

• Isolate the desired rate

• Substitute known values

Example: A cube grows at a rate of $2 \text{ cm}^3/\text{s}$. How fast is the side length increasing when the side is $0.8$ cm?

Real World Applications

Related rates are used to solve real-world problems involving rates of change, such as melting ice cubes, sliding ladders, and moving vehicles.

• Identify the changing quantities and their relationships

• Apply implicit differentiation with respect to time

• Interpret the result in the context of the problem

Example: An ice cube melts so that each side changes at $-3$ mm/min. Find the rate of change of the volume when the side is $0.9$ cm.

Linear Approximation

Linearization and Tangent Lines

Linear approximation uses the tangent line at a point to approximate the value of a function near that point. The linearization $L(x)$ of $f(x)$ at $x = a$ is:

• $L(x) = f(a) + f'(a)(x - a)$

• This is the equation of the tangent line to $f(x)$ at $x = a$

• Linearization is most accurate near $x = a$

Example: For $f(x) = x^2$ at $x = 1$, $L(x) = 1 + 2(x - 1)$

Practice Problems

• Use $L(x)$ at $a = 5$ for $f(x) = x^2 + 1$ to approximate $f(5.1)$

• Use $L(x)$ at $a = 16$ for $f(x) = \sqrt{x + 12}$ to approximate $f(16.01)$

Summary Table: Derivatives in Motion

Quantity	Function	Derivative	Physical Meaning
Position	$s(t)$	$s'(t)$	Velocity
Velocity	$v(t)$	$v'(t)$	Acceleration
Acceleration	$a(t)$	$a'(t)$	Jerk (rate of change of acceleration)

Additional info: The notes above expand on the provided slides and images, filling in standard calculus context and definitions for clarity and completeness.