Problem 7 - 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. $$ \mathbf{r}(t)=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j}+(2 \sqrt{2} / 3) t^{3 / 2} \mathbf{k}, \quad 0 \leq t \leq \pi $$

Step-by-Step Solution

Answer

Unit tangent vector: \( \mathbf{T}(t) = \frac{(\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j} + \sqrt{2}t^{1/2} \mathbf{k}}{t + 1} \). Length: \( \frac{\pi^2}{2} + \pi \).

Step 1: Find the Derivative of the Curve

To find the unit tangent vector, we first need to find the derivative of the vector function \( \mathbf{r}(t) \). The given vector function is \( \mathbf{r}(t) = (t \cos t) \mathbf{i} + (t \sin t) \mathbf{j} + \left(\frac{2 \sqrt{2}}{3}\right)t^{3/2} \mathbf{k} \). We differentiate each component with respect to \( t \):- The derivative of \( t \cos t \) with respect to \( t \) is \( \cos t - t \sin t \).- The derivative of \( t \sin t \) with respect to \( t \) is \( \sin t + t \cos t \).- The derivative of \( \left(\frac{2 \sqrt{2}}{3}\right)t^{3/2} \) with respect to \( t \) is \( \left(\frac{2 \sqrt{2}}{3}\right)\cdot\frac{3}{2}t^{1/2} = \sqrt{2}t^{1/2} \).Thus, \( \mathbf{r}'(t) = (\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j} + \sqrt{2}t^{1/2} \mathbf{k} \).

Step 2: Find the Magnitude of the Derivative

The next step is to find the magnitude of the derivative \( \mathbf{r}'(t) \) to normalize it. The magnitude \| \mathbf{r}'(t) \| is given by:\[\| \mathbf{r}'(t) \| = \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2 + (\sqrt{2}t^{1/2})^2}\]By expanding and simplifying:- The squared terms become \( \cos^2 t - 2t \cos t \sin t + t^2 \sin^2 t \) and \( \sin^2 t + 2t \sin t \cos t + t^2 \cos^2 t \).- When added, these expand to \( 1 + t^2 \) (after using the identity \( \sin^2 t + \cos^2 t = 1 \) and canceling terms).The third term is \( 2t \).Thus, \( \| \mathbf{r}'(t) \| = \sqrt{t^2 + 2t + 1} = t + 1 \).

Step 3: Compute the Unit Tangent Vector

Now we find the unit tangent vector \( \mathbf{T}(t) \) by normalizing \( \mathbf{r}'(t) \):\[\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{(\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j} + \sqrt{2}t^{1/2} \mathbf{k}}{t + 1}\]This gives us the expression for the unit tangent vector.

Step 4: Calculate the Length of the Curve

To find the length of the curve from \( t = 0 \) to \( t = \pi \), we integrate the magnitude of the derivative:\[L = \int_{0}^{\pi} \| \mathbf{r}'(t) \| \, dt = \int_{0}^{\pi} (t + 1) \, dt\]Evaluating this integral:\[L = \left[\frac{t^2}{2} + t\right]_{0}^{\pi} = \left( \frac{\pi^2}{2} + \pi \right) - (0 + 0) = \frac{\pi^2}{2} + \pi\]This is the length of the indicated portion of the curve.

Key Concepts

Derivative of Vector Function

The derivative of a vector function reveals how the function changes at each point, giving us the direction of the curve. To differentiate a vector function, we find the derivative of each coordinate function separately. In this exercise, the vector function is given as \( \mathbf{r}(t) = (t \cos t) \mathbf{i} + (t \sin t) \mathbf{j} + \left(\frac{2 \sqrt{2}}{3}\right)t^{3/2} \mathbf{k} \). Each component needs to be differentiated with respect to \( t \):

• The \( i \)-component, \( t \cos t \), has a derivative of \( \cos t - t \sin t \).
• The \( j \)-component, \( t \sin t \), becomes \( \sin t + t \cos t \).
• For the \( k \)-component, \( \left(\frac{2 \sqrt{2}}{3}\right)t^{3/2} \), the derivative is \( \sqrt{2}t^{1/2} \).

By putting these derivatives together, we have \( \mathbf{r}'(t) = (\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j} + \sqrt{2}t^{1/2} \mathbf{k} \). This derivative vector describes the direction in which the curve is moving at every point \( t \). In essence, these components help us understand the rate of change and direction of the curve at any given moment.

Magnitude of Derivative

Finding the magnitude of the derivative of a vector function is an important step for normalizing the vector, which is necessary for determining the unit tangent vector. The magnitude is like the length of the derivative vector. It shows how quickly the curve is changing at any point along its path. To find the magnitude, denoted as \( \| \mathbf{r}'(t) \| \), we use the formula: \[\| \mathbf{r}'(t) \| = \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2 + (\sqrt{2}t^{1/2})^2}\] We simplify this by using some trigonometric identities. Adding the squares of the first two terms and simplifying using \( \sin^2 t + \cos^2 t = 1 \), we find it simplifies to \( 1 + t^2 \). The third term becomes \( 2t \). Therefore, the overall magnitude simplifies to: \[\| \mathbf{r}'(t) \| = \sqrt{t^2 + 2t + 1} = t + 1\]Having the magnitude makes it possible to compute the unit tangent vector, which requires normalizing the derivative to make it a vector of length 1.

Length of Curve

The length of a curve is the total distance traveled along the curve between two points. For vector functions such as these, the length can be calculated by integrating the magnitude of the derivative over the given interval. This curvature computation tells us how extensive the path of the curve is without actually plotting it. In the exercise, the length \( L \) is calculated from \( t = 0 \) to \( t = \pi \) as follows: \[L = \int_{0}^{\pi} \| \mathbf{r}'(t) \| \, dt = \int_{0}^{\pi} (t + 1) \, dt\] This integral is straightforward and results in: \[L = \left[\frac{t^2}{2} + t\right]_{0}^{\pi} = \left( \frac{\pi^2}{2} + \pi \right) - (0 + 0) = \frac{\pi^2}{2} + \pi\]Thus, the length of the curve from \( t = 0 \) to \( t = \pi \) is \( \frac{\pi^2}{2} + \pi \). This length gives us the actual distance along the curve's path, which is a vital characteristic of the curve and important in various applications involving curve analysis.