Problem 67 - Calculus III

Question

Reparameterize the following functions with respect to their arc length measured from t=0 in direction of increasing t. $$ \mathbf{r}(t)=2 t \mathbf{i}+(4 t-5) \mathbf{j}+(1-3 t) \mathbf{k} $$

Step-by-Step Solution

Answer

Reparameterized function: \( \mathbf{r}(s) = \frac{2s}{\sqrt{29}} \mathbf{i} + \left(\frac{4s}{\sqrt{29}} - 5\right) \mathbf{j} + \left(1 - \frac{3s}{\sqrt{29}}\right) \mathbf{k} \).

Step 1: Compute the Derivative

To begin, compute the derivative of \( \mathbf{r}(t) \) with respect to \( t \). This gives us the velocity vector: \( \mathbf{r}'(t) = \frac{d}{dt}(2t) \mathbf{i} + \frac{d}{dt}(4t - 5) \mathbf{j} + \frac{d}{dt}(1 - 3t) \mathbf{k} = 2 \mathbf{i} + 4 \mathbf{j} - 3 \mathbf{k} \).

Step 2: Find the Speed

The speed \( v(t) \) is the magnitude of the velocity vector: \( v(t) = \| \mathbf{r}'(t) \| = \sqrt{(2)^2 + (4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \).

Step 3: Express Arc Length Function

Since arc length \( s \) is given by \( s = \int_0^t \| \mathbf{r}'(u) \| \, du \), substitute and calculate: \( s = \int_0^t \sqrt{29} \, du = \sqrt{29} t \).

Step 4: Solve for Reparameterization Function

We set \( s = \sqrt{29} t \), and solve for \( t \) in terms of \( s \): \( t = \frac{s}{\sqrt{29}} \).

Step 5: Reparameterize the Function

Substitute \( t = \frac{s}{\sqrt{29}} \) back into the original equation: \[ \mathbf{r}(s) = 2 \left( \frac{s}{\sqrt{29}} \right) \mathbf{i} + (4 \left( \frac{s}{\sqrt{29}} \right) - 5) \mathbf{j} + (1 - 3 \left( \frac{s}{\sqrt{29}} \right)) \mathbf{k} \].

Key Concepts

Velocity Vector

When analyzing the motion of an object described by a vector function, it's crucial to understand the concept of a velocity vector. This is the derivative of the function with respect to time and indicates how the position is changing at any given instant. For a vector function \( \mathbf{r}(t) = 2t \mathbf{i} + (4t - 5) \mathbf{j} + (1 - 3t) \mathbf{k} \), taking the derivative with respect to \( t \) will yield the velocity vector.
The process involves differentiating each component of the vector function separately:

• For \( 2t \mathbf{i} \), the derivative is \( 2 \mathbf{i} \).
• For \( (4t - 5) \mathbf{j} \), the derivative is \( 4 \mathbf{j} \).
• For \( (1 - 3t) \mathbf{k} \), the derivative is \(-3 \mathbf{k} \).

Thus, the complete velocity vector is \( \mathbf{r}'(t) = 2 \mathbf{i} + 4 \mathbf{j} - 3 \mathbf{k} \).
This vector gives both the direction and the rate of change of the object's position.

Speed Calculation

The speed of an object at any point in time can be determined by finding the magnitude of the velocity vector. This provides a scalar value that represents how fast the object is moving, regardless of direction.
To calculate the speed, apply the formula for the magnitude of a vector:
\[ v(t) = \| \mathbf{r}'(t) \| = \sqrt{(2)^2 + (4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \]
This value, \( \sqrt{29} \), is a constant speed. This indicates the object moves with a uniform speed, given that the magnitude of the velocity vector does not change over time.

Derivative of Vector Functions

The derivative of a vector function measures how the vector itself changes as the input variable changes. In our exercise, we differentiate \( \mathbf{r}(t) \) with respect to \( t \), resulting in the velocity vector.
Differentiation involves the following steps:

• Taking the derivative of each component of the vector individually.
• Applying basic derivative rules, like the power rule, to each term.

The result of this differentiation is a new vector, which tells us the rate at which each component of the original vector function is changing.
This derivative is foundational because it opens the door to understanding further physical concepts such as acceleration. Acceleration is simply the derivative of the velocity vector over time, showing another layer of change.

Vector Magnitude

Understanding the magnitude of a vector is essential in physics and engineering. It provides the length, or size, regardless of the vector's direction.
For the velocity vector \( \mathbf{r}'(t) = 2 \mathbf{i} + 4 \mathbf{j} - 3 \mathbf{k} \), its magnitude is calculated using:\[ \| \mathbf{v} \| = \sqrt{(2)^2 + (4)^2 + (-3)^2} = \sqrt{29} \]
Magnitude is an application of the Pythagorean theorem in three dimensions, summing the squares of each component before taking the square root. It is crucial for understanding the true pace of movement and forms a stepping stone for more complex vector operations, such as normalization or dot products.
Remember, magnitude is always a non-negative value and gives insight into the vector's overall effect or reach.