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

Derivatives Applied to Motion

Position, Velocity, and Acceleration

In calculus, the motion of an object is often described using its position function s(t) and its velocity function v(t). The velocity is the rate of change of position with respect to time, and acceleration is the rate of change of velocity with respect to time.

• Position Function: $s(t)$ gives the location of an object at time $t$.

• Velocity Function: $v(t) = \frac{ds}{dt}$ is the derivative of the position function.

• Acceleration Function: $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ is the derivative of the velocity function.

Key Formulas:

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

• Average Velocity: $v_{\text{avg}} = \frac{\Delta s}{\Delta t}$

• Instantaneous Velocity: $v(t) = \frac{ds}{dt}$

• Speed: $|v(t)|$

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

Projectile Motion

Projectile motion problems use position functions that model vertical motion under gravity. The general form is:

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

• Where $s_0$ is the initial height, $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity (usually $9.8\,\text{m/s}^2$).

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

Derivatives Applied to Acceleration

Understanding Acceleration

Acceleration is the rate of change of velocity over time. It is the derivative of the velocity function and the second derivative of the position function.

• $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

Example: Given $v(t) = t^2 - 3t^2 + 2t$, find the change in velocity, average acceleration, and instantaneous acceleration over a given interval.

Practice with Acceleration

• Given $s(t) = 12t - t^2$, find $a(5)$.

• Given a velocity graph, acceleration is zero where the slope of the velocity graph is zero (i.e., at local maxima or minima).

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 $\frac{dy}{dx}$.

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

• Apply the chain rule when differentiating terms involving $y$ (treat $y$ as a function of $x$).

• Solve for $\frac{dy}{dx}$.

Example: For $x^2 + y^2 = 49$, differentiate both sides to get $2x + 2y\frac{dy}{dx} = 0$, then solve for $\frac{dy}{dx}$.

Implicit Differentiation Practice

• 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.

• Use implicit differentiation with respect to time ($t$).

• Identify all variables and their rates of change.

• Plug in known values and solve for the desired rate.

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

Changing Geometries

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

• Draw a diagram and label variables.

• Write an equation relating the variables.

• Differentiation both sides with respect to time.

• Solve for the desired rate.

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 $0.8\,\text{cm}$?

Real World Applications

Related rates are used to solve real-world problems involving moving objects, changing volumes, and other dynamic systems.

• Example: As an ice cube melts, each side changes at $-3\,\text{mm/min}$. Find the rate of change of the volume when the side is $0.9\,\text{cm}$.

• Example: A 15-foot plank slides down a pole; find the rate at which the bottom moves away from the pole.

• Example: Two cars leave an intersection at different times and speeds; find the rate at which the distance between them changes.

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 useful for estimating function values close to $a$.

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

Linear Approximation Practice

• Approximate $f(5.1)$ for $f(x) = x^2 + 1$ using linearization at $a = 5$.

• Approximate $f(16.01)$ for $f(x) = \sqrt{x + 12}$ using linearization at $a = 16$.

Summary Table: Derivatives in Motion

Quantity	Function	Derivative	Physical Meaning
Position	$s(t)$	$\frac{ds}{dt}$	Velocity
Velocity	$v(t)$	$\frac{dv}{dt}$	Acceleration
Acceleration	$a(t)$	--	Rate of change of velocity

How To: Solve Related Rates Problems

1. Draw and label a picture of the scenario.

2. Identify all variables and their rates of change.

3. Write an equation relating the variables.

4. Differentiation both sides with respect to time.

5. Plug in known values and solve for the desired rate.

Additional info: Some context and explanations have been expanded for clarity and completeness, as the original slides were concise and sometimes fragmentary.