Problem 75 - Calculus 3

Question

Given \(\quad \mathbf{r}(t)=\langle t+\cos t, t-\sin t\rangle\) find the velocity and the speed at any time.

Step-by-Step Solution

Answer

Velocity: \(\langle 1 - \sin t, 1 - \cos t \rangle\), Speed: \(\sqrt{(1 - \sin t)^2 + (1 - \cos t)^2}\)."

Step 1: Understand the problem

You need to find the velocity and the speed of a given vector function \(\mathbf{r}(t)=\langle t+\cos t, t-\sin t\rangle\). The velocity is the derivative of \(\mathbf{r}(t)\) with respect to \(t\), and the speed is the magnitude of the velocity vector.

Step 2: Calculate the Velocity

To find the velocity, differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\). The derivative of \(t+\cos t\) is \(1 - \sin t\), and the derivative of \(t-\sin t\) is \(1 - \cos t\). Therefore, the velocity vector \(\mathbf{v}(t)\) is given by \(\mathbf{v}(t) = \langle 1 - \sin t, 1 - \cos t \rangle\).

Step 3: Calculate the Speed

The speed is the magnitude of the velocity vector. Calculate the magnitude using the formula \(\sqrt{(v_x)^2 + (v_y)^2}\), where \(v_x\) and \(v_y\) are the components of the velocity vector \(\mathbf{v}(t)\). Therefore, the speed is \(\sqrt{(1 - \sin t)^2 + (1 - \cos t)^2}\).

Key Concepts

Velocity Vector

In vector calculus, the concept of a velocity vector is crucial for understanding motion along a path defined by a vector function. A velocity vector represents the rate of change of position with respect to time, essentially capturing both the direction and the speed of a moving object in a vector field.
For a given vector function \( \mathbf{r}(t) = \langle t + \cos t, t - \sin t \rangle \), determining the velocity vector involves differentiating each of its components with respect to time \( t \).
This provides a new vector function that reveals how the position changes instantaneously at any point in time.

• The derivative of the first component \( t + \cos t \) is \( 1 - \sin t \).
• The derivative of the second component \( t - \sin t \) is \( 1 - \cos t \).

The resulting velocity vector \( \mathbf{v}(t) = \langle 1 - \sin t, 1 - \cos t \rangle \) reflects both the direction and rate of motion. Remember that the velocity vector can point in different directions at different times, indicating changes in the trajectory of the motion.

Speed Calculation

Calculating the speed of an object described by a vector function involves finding the magnitude of its velocity vector. Speed is a scalar quantity, meaning it only has magnitude without direction. It tells us how fast an object is moving along its path, independent of the direction.
To compute the speed from the velocity vector \( \mathbf{v}(t) = \langle 1 - \sin t, 1 - \cos t \rangle \), use the formula for magnitude:
\[ \text{Speed} = \sqrt{(v_x)^2 + (v_y)^2} \]

• Plug in the components \( v_x = 1 - \sin t \) and \( v_y = 1 - \cos t \).
• The formula becomes: \( \sqrt{(1 - \sin t)^2 + (1 - \cos t)^2} \).

This expression gives the speed at any time \( t \). It's important to note that while the velocity vector provides both the rate and direction of travel, speed focuses solely on the rate—how fast the object is moving along the path.

Derivative of Vector Functions

The process of taking the derivative of vector functions is similar to differentiating regular, scalar functions, but it is applied to each component of the vector independently. This is a vital technique in vector calculus, particularly when dealing with position, velocity, and acceleration vector functions.
For a vector function like \( \mathbf{r}(t) = \langle f(t), g(t) \rangle \), the derivative \( \mathbf{r}'(t) \) is found by differentiating each individual component function. This yields a new vector function representing the velocity, essentially the instantaneous rate of change of each component.

• In our example, \( f(t) = t + \cos t \) is differentiated to \( f'(t) = 1 - \sin t \).
• Similarly, \( g(t) = t - \sin t \) becomes \( g'(t) = 1 - \cos t \).

Understanding how to derive vector functions is essential as it lays the groundwork for analyzing and solving various physics and engineering problems involving motion. This concept is foundational for interpreting dynamic systems in higher dimensions.