Problem 1 - Advanced Engineering Mathematics

Question

For the given position function, find the unit tangent. $$ \mathbf{r}(t)=(t \cos t-\sin t) \mathbf{i}+(t \sin t+\cos t) \mathbf{j}+t^{2} \mathbf{k}, t>0 $$

Step-by-Step Solution

Answer

The unit tangent vector \( \mathbf{T}(t) = \left( -\frac{\sin t}{\sqrt{5}} \right) \mathbf{i} + \left( \frac{\cos t}{\sqrt{5}} \right) \mathbf{j} + \frac{2}{\sqrt{5}} \mathbf{k}. \)

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is found by taking the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). Use the derivative \( \frac{d}{dt} \) operator to differentiate each component of \( \mathbf{r}(t) \):\[ \mathbf{v}(t) = \frac{d}{dt}(t \cos t - \sin t) \mathbf{i} + \frac{d}{dt}(t \sin t + \cos t) \mathbf{j} + \frac{d}{dt}(t^2) \mathbf{k}. \]

Step 2: Differentiate Each Component

Compute the derivatives of each component:- For the \( \mathbf{i} \) component: \( \frac{d}{dt}(t \cos t - \sin t) = \cos t - t \sin t - \cos t = -t \sin t. \)- For the \( \mathbf{j} \) component: \( \frac{d}{dt}(t \sin t + \cos t) = \sin t + t \cos t - \sin t = t \cos t. \)- For the \( \mathbf{k} \) component: \( \frac{d}{dt}(t^2) = 2t. \)

Step 3: Combine to Get Velocity Vector

Combine the derivatives to find the velocity vector:\[ \mathbf{v}(t) = (-t \sin t) \mathbf{i} + (t \cos t) \mathbf{j} + (2t) \mathbf{k}. \]

Step 4: Find the Magnitude of Velocity

The magnitude of the velocity vector \( \|\mathbf{v}(t)\| \) is given by:\[ \|\mathbf{v}(t)\| = \sqrt{(-t \sin t)^2 + (t \cos t)^2 + (2t)^2}. \]Simplify the expression by calculating each term:- \((-t \sin t)^2 = t^2 \sin^2 t\)- \((t \cos t)^2 = t^2 \cos^2 t\)- \((2t)^2 = 4t^2\)Combine:\[ \|\mathbf{v}(t)\| = \sqrt{t^2 \sin^2 t + t^2 \cos^2 t + 4t^2}. \]

Step 5: Simplify the Magnitude Expression

Use the identity \( \sin^2 t + \cos^2 t = 1 \) to simplify:\[ \|\mathbf{v}(t)\| = \sqrt{t^2(\sin^2 t + \cos^2 t) + 4t^2} = \sqrt{t^2 + 4t^2} = \sqrt{5t^2}. \]This further simplifies to:\[ \|\mathbf{v}(t)\| = t \sqrt{5}. \]

Step 6: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is defined as \( \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \).Thus,\[ \mathbf{T}(t) = \frac{(-t \sin t) \mathbf{i} + (t \cos t) \mathbf{j} + 2t \mathbf{k}}{t \sqrt{5}} = \left(-\frac{\sin t}{\sqrt{5}}\right)\mathbf{i} + \left(\frac{\cos t}{\sqrt{5}}\right)\mathbf{j} + \left(\frac{2}{\sqrt{5}}\right)\mathbf{k}. \]

Key Concepts

Position Function

In calculus, a position function describes the position of a point in a space determined by time (or another changing variable). For this exercise, the position function \( \mathbf{r}(t) \) is given as:
\[\mathbf{r}(t) = (t \cos t - \sin t) \mathbf{i} + (t \sin t + \cos t) \mathbf{j} + t^2 \mathbf{k}, t > 0\]This vector function shows the position in 3D space — in terms of the standard unit vectors (\( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \)) — as a function of the variable \( t \). Each component correlates to the position along the \( x \), \( y \), and \( z \)-axis respectively. Typically, as \( t \) increases, the point moves through space according to this function.

Velocity Vector

The velocity vector gives us insight into how the position of a point changes over time. To find it, we need to differentiate the position function \( \mathbf{r}(t) \) with respect to \( t \). This differentiation will provide the velocity vector \( \mathbf{v}(t) \):
\[\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t)\]By differentiating each component of the position function, we get the velocity components along the unit vectors:

• For \( \mathbf{i} \): Differentiating \( t \cos t - \sin t \) gives \(-t \sin t\).
• For \( \mathbf{j} \): Differentiating \( t \sin t + \cos t \) leads to \( t \cos t\).
• For \( \mathbf{k} \): Differentiating \( t^2 \) results in \( 2t \).

Putting these together forms the full velocity vector: \[\mathbf{v}(t) = (-t \sin t) \mathbf{i} + (t \cos t) \mathbf{j} + (2t) \mathbf{k}\]This vector represents the rate of change of position over time.

Derivative

The derivative is a fundamental concept in calculus, representing the rate at which one quantity changes with respect to another. In the context of this problem, the derivative helps us move from the position function \( \mathbf{r}(t) \) to the velocity vector \( \mathbf{v}(t) \). Calculating the derivative involves applying the differentiation rules to each component of \( \mathbf{r}(t) \):

• Product rule: Used for terms like \( t \cos t \) and \( t \sin t \).
• Basic derivatives: For \( \sin t \), \( \cos t \), and \( t^2 \).

By precisely following these rules, we gain insight into how quickly the position changes as time progresses, forming the core part of determining the velocity.

Magnitude of Vector

To understand the speed or the "size" of the velocity vector, we calculate its magnitude \( \|\mathbf{v}(t)\| \). The formula to find this is:
\[\|\mathbf{v}(t)\| = \sqrt{(-t \sin t)^2 + (t \cos t)^2 + (2t)^2}\]Breaking it down further:

• \((-t \sin t)^2 = t^2 \sin^2 t\)
• \((t \cos t)^2 = t^2 \cos^2 t\)
• \((2t)^2 = 4t^2\)

After combining all these, we use the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \) to simplify:
\[\|\mathbf{v}(t)\| = \sqrt{5t^2} = t \sqrt{5}\]The magnitude informs us of the speed of motion, which is integral in computing the unit tangent vector.