Problem 1 - 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. $$ \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{j}+\sqrt{5} t \mathbf{k}, \quad 0 \leq t \leq \pi $$

Step-by-Step Solution

Answer

The unit tangent vector is \( \left( -\frac{2}{3} \sin t \right) \mathbf{i} + \left( \frac{2}{3} \cos t \right) \mathbf{j} + \left( \frac{\sqrt{5}}{3} \right) \mathbf{k} \). The length of the curve is \( 3\pi \).

Step 1: Find the Velocity Vector

To find the unit tangent vector, first calculate the velocity vector by taking the derivative of the position vector \( \mathbf{r}(t) \). Given \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5} t \mathbf{k} \), compute \( \mathbf{r}'(t) \): \[ \mathbf{r}'(t) = \frac{d}{dt}[(2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5} t \mathbf{k}] = (-2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + \sqrt{5} \mathbf{k} \].

Step 2: Compute the Magnitude of the Velocity Vector

Next, find the magnitude of the velocity vector \( \mathbf{r}'(t) \). The magnitude is given by: \[ \| \mathbf{r}'(t) \| = \sqrt{(-2 \sin t)^2 + (2 \cos t)^2 + (\sqrt{5})^2} \] \[ = \sqrt{4\sin^2 t + 4\cos^2 t + 5} \] \[ = \sqrt{4(\sin^2 t + \cos^2 t) + 5} \] \( \sin^2 t + \cos^2 t = 1 \), so \[ \| \mathbf{r}'(t) \| = \sqrt{9} = 3 \].

Step 3: Find the Unit Tangent Vector

Now, divide the velocity vector \( \mathbf{r}'(t) \) by its magnitude to get the unit tangent vector \( \mathbf{T}(t) \): \[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{(-2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + \sqrt{5} \mathbf{k}}{3} \] \[ = \left( -\frac{2}{3} \sin t \right) \mathbf{i} + \left( \frac{2}{3} \cos t \right) \mathbf{j} + \left( \frac{\sqrt{5}}{3} \right) \mathbf{k} \].

Step 4: Find the Length of the Curve

Finally, compute the length of the curve over the interval \( [0, \pi] \) using the formula for arc length: \[ L = \int_{0}^{\pi} \| \mathbf{r}'(t) \| \, dt \] Since \( \| \mathbf{r}'(t) \| = 3 \), the integral becomes: \[ L = \int_{0}^{\pi} 3 \, dt = 3t \bigg|_{0}^{\pi} \] \[ = 3\pi - 3(0) = 3\pi \].

Key Concepts

Velocity Vector

The velocity vector is a crucial concept when studying curves in space. It represents the rate at which a particle changes its position over time. To find the velocity vector for a curve given by a position vector function, we need to calculate the derivative of the position vector with respect to time.

• In this exercise, the position vector is given as \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5} t \mathbf{k} \).
• The velocity vector, denoted as \( \mathbf{r}'(t) \), is the derivative of this position vector.
• By differentiating each component of \( \mathbf{r}(t) \) with respect to \( t \), we find that \( \mathbf{r}'(t) = (-2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + \sqrt{5} \mathbf{k} \).

The velocity vector thus gives us a new vector showing how fast and in which direction the position changes at any point \(t\) over the specified interval.

Arc Length

Arc length represents the distance traveled by a particle along a curve from one point to another. To compute it, we integrate the magnitude of the velocity vector over the range of interest.

• The magnitude of the velocity vector \( \| \mathbf{r}'(t) \| \) in this exercise was found to be constant at 3.
• This is derived from the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), simplifying to a constant value.
• The arc length \( L \) over the interval \([0, \pi]\) is calculated as:\( L = \int_{0}^{\pi} 3 \, dt = 3t \bigg|_{0}^{\pi} = 3\pi \).

This integral shows us that the length of the curve from \( t = 0 \) to \( t = \pi \) is \(3\pi\), giving direct insight into the total distance covered along the curve.

Position Vector

A position vector provides the location of a point in space at any given time \( t \). It is often expressed in terms of its components along the standard basis vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).

• In this case, the position vector is \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5} t \mathbf{k} \).
• This vector describes a point rotating around a circle with radius 2 in the \(xy\)-plane and simultaneously moving linearly in the \(z\)-direction.
• The position vector traces out a helical path as \( t \) varies from 0 to \( \pi \).

Understanding the position vector helps to visualize how the particle moves in a three-dimensional space over time and provides the starting point for calculating velocity and arc length.

Derivative

The derivative is a fundamental tool in calculus used to determine the rate at which one quantity changes with respect to another. In the context of this exercise, the derivative helps us find the velocity vector.

• Taking the derivative of the position vector \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5} t \mathbf{k} \) gives us \( \mathbf{r}'(t) \).
• The process involves differentiating each component separately:
  • The derivative of \(2 \cos t\) is \(-2 \sin t\).
  • The derivative of \(2 \sin t\) is \(2 \cos t\).
  • The derivative of \(\sqrt{5} t\) is \(\sqrt{5}\).
• Therefore, \( \mathbf{r}'(t) = (-2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + \sqrt{5} \mathbf{k} \).

The derivative defines how quickly the position changes at any given instant and is essential for determining the unit tangent vector and the arc length.