Problem 5 - 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. $$ \mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{j}+\left(\sin ^{3} t\right) \mathbf{k}, \quad 0 \leq t \leq \pi / 2 $$

Step-by-Step Solution

Answer

Unit tangent vector: \( \mathbf{T}(t) = -\cos t \mathbf{j} + \sin t \mathbf{k} \). Length: \( \frac{3}{2} \).

Step 1: Find the Derivative of the Vector Function

First, find the derivative of the given vector function \( \mathbf{r}(t) = \cos^3(t) \mathbf{j} + \sin^3(t) \mathbf{k} \). The derivative \( \mathbf{r}'(t) \) is given by: \[ \mathbf{r}'(t) = \frac{d}{dt} (\cos^3(t))\mathbf{j} + \frac{d}{dt} ( \sin^3(t)) \mathbf{k}. \]

Step 2: Apply the Chain Rule

Apply the chain rule to differentiate each component:\[ \frac{d}{dt}(\cos^3 t) = 3\cos^2 t (-\sin t) = -3 \cos^2 t \sin t \] and \[ \frac{d}{dt}(\sin^3 t) = 3\sin^2 t \cos t. \] Thus, the derivative is: \[ \mathbf{r}'(t) = -3\cos^2 t \sin t \mathbf{j} + 3\sin^2 t \cos t \mathbf{k}. \]

Step 3: Compute the Magnitude of the Derivative

To find the unit tangent vector, compute the magnitude of \( \mathbf{r}'(t) \):\[ \| \mathbf{r}'(t) \| = \sqrt{(-3\cos^2 t \sin t)^2 + (3\sin^2 t \cos t)^2}. \]

Step 4: Simplify the Magnitude Expression

Simplify \( \| \mathbf{r}'(t) \| \): \[ \| \mathbf{r}'(t) \| = \sqrt{9\cos^4 t \sin^2 t + 9\sin^4 t \cos^2 t}. \]Factor out a common term:\[ = \sqrt{9\cos^2 t \sin^2 t(\cos^2 t + \sin^2 t)}. \]Since \( \cos^2 t + \sin^2 t = 1 \), we have:\[ = \sqrt{9\cos^2 t \sin^2 t} = 3\cos t \sin t. \]

Step 5: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) isthe derivative \( \mathbf{r}'(t) \) divided by its magnitude:\[ \mathbf{T}(t) = \frac{-3\cos^2 t \sin t \mathbf{j} + 3\sin^2 t \cos t \mathbf{k}}{3\cos t \sin t}. \]Simplifying this gives:\[ \mathbf{T}(t) = -\cos t \mathbf{j} + \sin t \mathbf{k}. \]

Step 6: Find the Length of the Curve Segment

The length of the curve from \( t = 0 \) to \( t = \frac{\pi}{2} \) is given by\[ \int_{0}^{\frac{\pi}{2}} \| \mathbf{r}'(t) \| \, dt = \int_{0}^{\frac{\pi}{2}} 3\cos t \sin t \, dt. \]This simplifies to\[ \int_{0}^{\frac{\pi}{2}} \frac{3}{2}\sin 2t \, dt. \]Evaluating this integral, we find\[ \left[-\frac{3}{4}\cos 2t \right]_{0}^{\frac{\pi}{2}} = -\frac{3}{4}(\cos \pi - \cos 0) = \frac{3}{2}. \]

Step 7: Conclusion

Therefore, the unit tangent vector is \( \mathbf{T}(t) = -\cos t \mathbf{j} + \sin t \mathbf{k} \).The length of the curve from \( t = 0 \) to \( t = \frac{\pi}{2} \) is \( \frac{3}{2} \).

Key Concepts

Curve Analysis

Curve analysis involves the study of various properties of curves, usually in a two- or three-dimensional space. A curve is represented as a continuous function and in this context, is typically defined by a vector-valued function. The given vector function is \( \mathbf{r}(t) = (\cos^3 t) \mathbf{j} + (\sin^3 t) \mathbf{k} \) and describes a path traced out in space as \( t \) varies. This curve is parametrized by the variable \( t \), which acts like the navigator moving along the curve from one point to another. Analyzing curves helps us understand how the curve behaves, including changes in direction and speed, and involves several mathematical tools like derivatives, integrals, and arc length computations. These pieces of information are crucial in fields such as physics and engineering where understanding motion and paths are essential.

Vector Calculus

Vector calculus is an extension of calculus that deals with vector-valued functions, which assign vectors to different points in space. It incorporates differentiation and integration over vector fields.In the given problem, the vector \( \mathbf{r}(t) \) represents a curve in 3D space as a function of \( t \). The components of this vector function, \( \cos^3(t) \mathbf{j} \) and \( \sin^3(t) \mathbf{k} \), determine its trajectory. Using vector calculus, you can differentiate \( \mathbf{r}(t) \) to find its derivative \( \mathbf{r}'(t) \), which provides the direction and speed of the curve at any point \( t \). Integrating components of vectors or the magnitude of derivatives helps in calculating parameters like curve length, which illustrates how far a point has traveled along the curve.

Derivative of Vector Functions

Differentiating vector functions involves taking the derivative of each of their component functions separately. For the vector function \( \mathbf{r}(t) = (\cos^3 t) \mathbf{j} + (\sin^3 t) \mathbf{k} \), finding the derivative involved applying the chain rule to each component.

• For \( \cos^3 t \), the derivative is \( -3\cos^2 t \sin t \).
• For \( \sin^3 t \), the derivative is \( 3\sin^2 t \cos t \).

These derivatives lead to \( \mathbf{r}'(t) = -3\cos^2 t \sin t \mathbf{j} + 3\sin^2 t \cos t \mathbf{k} \). This derivative is crucial because it gives the tangent vector to the curve, indicating direction. To find the unit tangent vector \( \mathbf{T}(t) \), divide \( \mathbf{r}'(t) \) by its magnitude \( \|\mathbf{r}'(t)\|\). This normalization process ensures that the vector's magnitude is 1, providing a standard measure of direction without the influence of speed.

Arc Length of Curves

The arc length of a curve gives us the distance a point travels along the curve over a specific interval of the parameter \( t \). Calculating this length involves integrating the magnitude of the derivative of the vector function, \( \|\mathbf{r}'(t)\| \).In this problem, the challenge is to determine the arc length from \( t = 0 \) to \( t = \frac{\pi}{2} \). After finding \( \|\mathbf{r}'(t)\| = 3 \cos t \sin t \), the integral \[ \int_{0}^{\frac{\pi}{2}} 3 \cos t \sin t \ dt \] is computed.Simplifying using trigonometric identities, such as \( \sin(2t) = 2 \sin t \cos t \), helps evaluate the integral effectively. Completing the integral results in the arc length being \( \frac{3}{2} \). This calculation is valuable in understanding how far along the curve a point travels, aiding in various applications like calculating travel time and energy in physics and engineering scenarios.