Problem 3 - Calculus III

Question

Given the following position functions, find the velocity, acceleration, and speed in terms of the parameter t. $$ \mathbf{r}(t)=\left\langle 3 \cos t, 3 \sin t, t^{2}\right\rangle $$

Step-by-Step Solution

Answer

The velocity is \( \langle -3 \sin t, 3 \cos t, 2t \rangle \), acceleration is \( \langle -3 \cos t, -3 \sin t, 2 \rangle \), and speed is \( \sqrt{9 + 4t^2} \).

Step 1: Identify the position function

We are given the position function \( \mathbf{r}(t) = \langle 3 \cos t, 3 \sin t, t^{2} \rangle \). This function describes the position in space as a vector, depending on the parameter \( t \).

Step 2: Find the velocity function

The velocity function \( \mathbf{v}(t) \) is found by taking the derivative of the position function \( \mathbf{r}(t) \) with respect to \( t \). Using the derivative rules, we have:\[ \mathbf{v}(t) = \frac{d}{dt}\langle 3 \cos t, 3 \sin t, t^{2} \rangle = \langle -3 \sin t, 3 \cos t, 2t \rangle \]

Step 3: Find the acceleration function

The acceleration function \( \mathbf{a}(t) \) is found by taking the derivative of the velocity function \( \mathbf{v}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt}\langle -3 \sin t, 3 \cos t, 2t \rangle = \langle -3 \cos t, -3 \sin t, 2 \rangle \]

Step 4: Find the speed

Speed is the magnitude of the velocity vector. The magnitude of \( \mathbf{v}(t) = \langle -3 \sin t, 3 \cos t, 2t \rangle \) is calculated as follows:\[ ||\mathbf{v}(t)|| = \sqrt{(-3 \sin t)^2 + (3 \cos t)^2 + (2t)^2} = \sqrt{9(\sin^2 t + \cos^2 t) + 4t^2} = \sqrt{9 + 4t^2} \]This uses the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \).

Key Concepts

Velocity Function

In vector calculus, understanding the velocity function is crucial for analyzing how an object's position changes over time. The velocity function, denoted as \( \mathbf{v}(t) \), is derived from the position function \( \mathbf{r}(t) \) by taking its derivative with respect to the parameter \( t \). This function provides a vector that indicates the object's direction and speed at any point in time.

• The direction of the velocity vector is tangent to the path of the object.
• The magnitude of this vector tells us how fast the object is moving at any given instant.

In the original exercise, we started with the position function \( \mathbf{r}(t) = \langle 3 \cos t, 3 \sin t, t^2 \rangle \). By deriving each component with respect to \( t \), the velocity vector became \( \mathbf{v}(t) = \langle -3 \sin t, 3 \cos t, 2t \rangle \). This derivative reveals both the changes in horizontal and vertical motion, as well as changes along the third component.

Acceleration Function

The acceleration function measures how the velocity of an object changes over time. It's calculated as the derivative of the velocity function. By understanding acceleration, we can predict the future motion path and analyze forces acting on an object. The acceleration vector gives us valuable insights in different dimensions:

• The direction of the acceleration vector indicates where the object is speeding up or slowing down.
• The magnitude of this vector shows how quickly the velocity is changing.

In the given problem, the velocity was expressed as \( \mathbf{v}(t) = \langle -3 \sin t, 3 \cos t, 2t \rangle \). Differentiating each component of this velocity function, we get the acceleration function \( \mathbf{a}(t) = \langle -3 \cos t, -3 \sin t, 2 \rangle \). This suggests that the acceleration in the third dimension is constant, while the components involving sine and cosine indicate periodic accelerations.

Parametric Equations

Parametric equations are powerful tools in calculus and physics and are used to express motion paths and trajectories. They separate the coordinates of a vector function into distinct equations, each driven by a common parameter, typically \( t \). This provides a flexible way to represent curves in different dimensions:

• They define a set of equations for each coordinate, allowing complex paths in space to be outlined more simply.
• These types of equations make analyzing movement direction and speed straightforward, as they can be easily differentiated to find velocity and acceleration.

In our exercise, the position vector function \( \mathbf{r}(t) = \langle 3 \cos t, 3 \sin t, t^2 \rangle \) is defined parametric, with \( t \) as the parameter. Each function component—associated with x, y, and z coordinates—enables a clear understanding of how an object moves through space over time.

Speed in Vector Functions

Speed is a scalar quantity that provides the magnitude of a vector function's velocity, indicating how fast an object is moving regardless of its direction. The speed is found by calculating the magnitude of the velocity vector. Unlike velocity, speed doesn’t tell us the direction of the motion:

• It’s obtained through the Euclidean norm, essentially the length of the velocity vector.
• Knowing the speed, we can compare motions even if the directions are different.

To find the speed in our exercise, we looked at the velocity function \( \mathbf{v}(t) = \langle -3 \sin t, 3 \cos t, 2t \rangle \). The speed was then calculated as \( ||\mathbf{v}(t)|| = \sqrt{9 + 4t^2} \), which stems from summing the squares of the velocity components. This expression provides a simple metric for assessing how quickly the object travels along its path.