Problem 20 - Calculus Early Transcendentals

Question

Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter \(t .\) $$ \mathbf{r}(t)=2 \sin t \mathbf{i}+2 \cos t \mathbf{j}+\tan t \mathbf{k}, \quad t=\pi / 4 $$

Step-by-Step Solution

Answer

The unit tangent vector \(\mathbf{T}(\pi/4) = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} + \frac{\sqrt{2}}{2} \mathbf{k}\).

Step 1: Find the Derivative of the Vector Function

To find the unit tangent vector \(\mathbf{T}(t)\), we first need the derivative of the vector function \(\mathbf{r}(t)\). The derivative is given by \(\mathbf{r}'(t) = \frac{d}{dt}[2 \sin t] \mathbf{i} + \frac{d}{dt}[2 \cos t] \mathbf{j} + \frac{d}{dt}[\tan t] \mathbf{k}\). So, we compute:- \(\frac{d}{dt}[2 \sin t] = 2 \cos t\)- \(\frac{d}{dt}[2 \cos t] = -2 \sin t\)- \(\frac{d}{dt}[\tan t] = \sec^2 t\)Therefore, \(\mathbf{r}'(t) = 2 \cos t \mathbf{i} - 2 \sin t \mathbf{j} + \sec^2 t \mathbf{k}\).

Step 2: Evaluate the Derivative at t = \(\pi/4\)

Next, substitute \(t = \pi/4\) into \(\mathbf{r}'(t)\) to evaluate at that point:- \(2 \cos(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}\)- \(-2 \sin(\pi/4) = -2 \cdot \frac{\sqrt{2}}{2} = -\sqrt{2}\)- \(\sec^2(\pi/4) = (\frac{1}{\cos(\pi/4)})^2 = 2\)Thus, \(\mathbf{r}'(\pi/4) = \sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j} + 2 \mathbf{k}\).

Step 3: Calculate the Magnitude of the Derivative

Find the magnitude of \(\mathbf{r}'(\pi/4)\) using the formula:\[ |\mathbf{r}'(\pi/4)| = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2 + (2)^2} = \sqrt{2 + 2 + 4} = \sqrt{8} = 2\sqrt{2} \]

Step 4: Determine the Unit Tangent Vector \(\mathbf{T}(t)\)

The unit tangent vector \(\mathbf{T}(t)\) is given by the formula:\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \]Therefore, at \(t = \pi/4\):\[ \mathbf{T}(\pi/4) = \frac{1}{2\sqrt{2}} \left( \sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j} + 2 \mathbf{k} \right)\]Simplifying, we get:\[ \mathbf{T}(\pi/4) = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} + \frac{\sqrt{2}}{2} \mathbf{k} \]

Key Concepts

Derivatives

Derivatives are a fundamental concept in vector calculus, representing the rate at which a function changes. In the context of a vector function, such as \(\mathbf{r}(t)\), the derivative \(\mathbf{r}'(t)\) is determined by differentiating each component of the vector separately. This involves calculating how each part of the vector function changes as \(t\) changes.
For instance, if you have \(\mathbf{r}(t) = 2 \sin t \mathbf{i} + 2 \cos t \mathbf{j} + \tan t \mathbf{k}\), you compute the derivatives of each:

• \(\frac{d}{dt}[2 \sin t] = 2 \cos t\)
• \(\frac{d}{dt}[2 \cos t] = -2 \sin t\)
• \(\frac{d}{dt}[\tan t] = \sec^2 t\)

These derivatives give us the new vector \(\mathbf{r}'(t) = 2 \cos t \mathbf{i} - 2 \sin t \mathbf{j} + \sec^2 t \mathbf{k}\), which describes the rate of change of the original vector function at any point \(t\).
By evaluating this derivative at specific points, like \(t = \pi/4\), we can obtain information about the vector's direction and behavior at that moment.

Vector Calculus

Vector calculus is an extension of calculus to vector functions. It brings together the ideas of derivatives and vectors. When dealing with vector functions, such as \(\mathbf{r}(t)\), the key task is to find derivatives that reveal how the function behaves over time or space.
A significant part of vector calculus is finding the unit tangent vector, \(\mathbf{T}(t)\). This vector shows the direction of a vector function at any given point along its path. To find this, we take the derivative of the vector function \(\mathbf{r}'(t)\) and then normalize it:1. Evaluate \(\mathbf{r}'(t)\) at the desired parameter \(t\). 2. Calculate the magnitude of \(\mathbf{r}'(t)\), using the formula: \[ |\mathbf{r}'(t)| = \sqrt{ (2 \cos t)^2 + (-2 \sin t)^2 + (\sec^2 t)^2} \]3. Divide each component of \(\mathbf{r}'(t)\) by this magnitude to get \(\mathbf{T}(t)\).This process highlights the importance of both directional derivatives and normalization in vector calculus.

Trigonometric Functions

Trigonometric functions are often involved in vector calculus, particularly when dealing with curves in 3D space. Functions like \(\sin t\), \(\cos t\), and \(\tan t\) define the components of vectors, making them essential in calculating derivatives and understanding the vector's behavior.
The use of trigonometric functions allows for describing periodic phenomena or rotational movements. For example, with the vector \(\mathbf{r}(t) = 2 \sin t \mathbf{i} + 2 \cos t \mathbf{j} + \tan t \mathbf{k}\), each component corresponds to a familiar trigonometric function:

• \(2 \sin t\) defines oscillations in the \(\mathbf{i}\) direction.
• \(2 \cos t\) describes movements in the \(\mathbf{j}\) direction.
• The \(\tan t\) component in the \(\mathbf{k}\) direction indicates changes related to the angle's tangent, often increasing rapidly.

These functions influence how we find derivatives, as each has specific rules: \(\frac{d}{dt}[\sin t] = \cos t\), \(\frac{d}{dt}[\cos t] = -\sin t\), and \(\frac{d}{dt}[\tan t] = \sec^2 t\). Recognizing and applying these rules is essential for handling vector functions in calculus.