Problem 8 - Thomas Calculus

Question

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(t \sin t+\cos t) \mathbf{i}+(t \cos t-\sin t) \mathbf{j}, \quad \sqrt{2} \leq t \leq 2 $$

Step-by-Step Solution

Answer

The unit tangent vector is \( \mathbf{T}(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} \). The length of the curve is 1.

Step 1: Find the derivative of the curve

The first step is to find the derivative of the vector function \( \mathbf{r}(t)=(t \sin t+\cos t) \mathbf{i}+(t \cos t-\sin t) \mathbf{j} \). The derivative, \( \mathbf{r}'(t) \), gives the velocity vector. Use the product rule and chain rule where necessary for each component.- The derivative of the \( x \)-component, \( t \sin t + \cos t \), is \( \sin t + t \cos t - \sin t = t \cos t \).- The derivative of the \( y \)-component, \( t \cos t - \sin t \), is \( \cos t - t \sin t - \cos t = -t \sin t \).So, \( \mathbf{r}'(t) = (t \cos t) \mathbf{i} + (-t \sin t) \mathbf{j} \).

Step 2: Compute the magnitude of the velocity vector

To find the unit tangent vector, we need the magnitude of \( \mathbf{r}'(t) \), which is given by \( \| \mathbf{r}'(t) \| = \sqrt{(t \cos t)^2 + (-t \sin t)^2} \). Simplify to get the magnitude:- \( (t \cos t)^2 = t^2 \cos^2 t \)- \( (-t \sin t)^2 = t^2 \sin^2 t \)- Thus, \( \| \mathbf{r}'(t) \| = \sqrt{t^2 (\cos^2 t + \sin^2 t)} = \sqrt{t^2} = |t| \).Since \( t \) is within \( \sqrt{2} \leq t \leq 2 \), \( |t| = t \).

Step 3: Find the unit tangent vector

The unit tangent vector, \( \mathbf{T}(t) \), is \( \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{(t \cos t) \mathbf{i} + (-t \sin t) \mathbf{j}}{t} \).Simplify to get: \( \mathbf{T}(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} \).This is the unit tangent vector at any point \( t \).

Step 4: Set up the integral for the length of the curve portion

We find the length of the curve from \( t = \sqrt{2} \) to \( t = 2 \) using the arc length formula: \[ L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \].Here, \( a = \sqrt{2} \) and \( b = 2 \). We computed \( \| \mathbf{r}'(t) \| = t \) in Step 2, so the integral becomes: \[ L = \int_{\sqrt{2}}^{2} t \, dt \].

Step 5: Evaluate the integral

Compute the integral \( \int_{\sqrt{2}}^{2} t \, dt \) as follows:- The antiderivative of \( t \) is \( \frac{t^2}{2} \).- Evaluate from \( t = \sqrt{2} \) to \( t = 2 \):\[ \left. \frac{t^2}{2} \right|_{\sqrt{2}}^{2} = \frac{2^2}{2} - \frac{(\sqrt{2})^2}{2} = \frac{4}{2} - \frac{2}{2} = 2 - 1 = 1 \].Thus, the length of the curve is \( 1 \).

Key Concepts

Unit Tangent Vector

When studying curves through vector calculus, a vital concept is the unit tangent vector. This vector provides insight into the direction in which a curve is heading at any given point. To find the unit tangent vector at a point on a curve, we begin by determining the derivative of the vector function, which gives us the velocity vector. The velocity vector indicates the direction and rate of change of the position vector over time.Once you have the velocity vector, calculate its magnitude. The magnitude is essentially the "length" or "size" of the vector and is found using the formula: \[ \| \mathbf{r}'(t) \| = \sqrt{(x'(t))^2 + (y'(t))^2} \] where \( x'(t) \) and \( y'(t) \) are the derivatives of the respective components of the position vector.Finally, the unit tangent vector \( \mathbf{T}(t) \) is obtained by dividing the velocity vector by its magnitude:\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \]This yields a vector of length 1 that points in the direction the curve is moving, providing a standardized way of expressing direction.

Arc Length

Arc length is a measure of the distance along a curve, and it's an important concept in vector calculus. To compute the arc length of a curve represented by a vector function \( \mathbf{r}(t) \), first identify the portion of the curve you want to measure. In our example, this was between \( t = \sqrt{2} \) and \( t = 2 \).The formula for arc length is:\[ L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \]Where:- \( L \) is the total length of the curve.- \( a \) and \( b \) represent the bounds of the interval for the parameter \( t \).- \( \| \mathbf{r}'(t) \| \) is the magnitude of the derivative of the vector function, calculated as described in the unit tangent vector section.To find the arc length, integrate the magnitude of the derivative over the interval \([a, b]\). This process sums up infinitesimal distances along the curve to provide the total length. For the example provided, the result was an arc length of 1, showing how integration is used to piece together these small lengths. Understanding this concept helps with analyzing real-world problems involving any path or trajectory.

Derivative of Vector Functions

In vector calculus, the derivative of vector functions is a core concept that informs much of our analysis. The derivative describes how the vector function \( \mathbf{r}(t) \) changes as the parameter \( t \) changes.To compute the derivative of a vector function:

• Take the derivative of each component of the vector separately. In our example, this involves using rules of differentiation like the product rule and chain rule.
• Apply these rules to find \( x'(t) \) and \( y'(t) \), which are the derivatives of the \( i \)- and \( j \)-components, respectively.

The result is a new vector \( \mathbf{r}'(t) \) that represents the direction and speed at which the position on the curve is changing. This derivative vector, also known as the velocity vector, is vital for understanding motion patterns and behavior of objects in both physics and engineering contexts.By learning how to apply differentiation techniques to vector functions, you gain powerful tools to tackle complex problems involving dynamic systems and advanced motion analysis.