To find the unit tangent vector $\mathbf{T}(t)$ of the given curve $\mathbf{r}(t)$, follow these steps:

Given: $\mathbf{r}(t) = \left( 2 \cos^3(t) \right) \mathbf{j} + \left( 2 \sin^3(t) \right) \mathbf{k}$

Step 1: Find the derivative of $\mathbf{r}(t)$

The unit tangent vector $\mathbf{T}(t)$ is found by first determining the velocity vector $\mathbf{r}'(t)$, which is the derivative of $\mathbf{r}(t)$ with respect to $t$:

$\mathbf{r}'(t) = \frac{d}{dt} \left( 2 \cos^3(t) \right) \mathbf{j} + \frac{d}{dt} \left( 2 \sin^3(t) \right) \mathbf{k}$

Using the chain rule: $\frac{d}{dt} \left( 2 \cos^3(t) \right) = 2 \cdot 3 \cos^2(t) \cdot (-\sin(t)) = -6 \cos^2(t) \sin(t)$ $\frac{d}{dt} \left( 2 \sin^3(t) \right) = 2 \cdot 3 \sin^2(t) \cdot \cos(t) = 6 \sin^2(t) \cos(t)$

Thus, the velocity vector $\mathbf{r}'(t)$ is: $\mathbf{r}'(t) = -6 \cos^2(t) \sin(t) \mathbf{j} + 6 \sin^2(t) \cos(t) \mathbf{k}$

Step 2: Find the magnitude of $\mathbf{r}'(t)$

The magnitude $|\mathbf{r}'(t)|$ is: $|\mathbf{r}'(t)| = \sqrt{\left( -6 \cos^2(t) \sin(t) \right)^2 + \left( 6 \sin^2(t) \cos(t) \right)^2}$

Simplifying: $|\mathbf{r}'(t)| = \sqrt{36 \cos^4(t) \sin^2(t) + 36 \sin^4(t) \cos^2(t)}$ $|\mathbf{r}'(t)| = \sqrt{36 \cos^2(t) \sin^2(t) \left( \cos^2(t) + \sin^2(t) \right)}$ Since $\cos^2(t) + \sin^2(t) = 1$, we have: $|\mathbf{r}'(t)| = \sqrt{36 \cos^2(t) \sin^2(t)}$ $|\mathbf{r}'(t)| = 6 |\cos(t) \sin(t)|$

Step 3: Find the unit tangent vector

The unit tangent vector $\mathbf{T}(t)$ is given by: $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$ Thus: $\mathbf{T}(t) = \frac{-6 \cos^2(t) \sin(t)}{6 |\cos(t) \sin(t)|} \mathbf{j} + \frac{6 \sin^2(t) \cos(t)}{6 |\cos(t) \sin(t)|} \mathbf{k}$ $\mathbf{T}(t) = -\frac{\cos(t)}{|\cos(t)|} \mathbf{j} + \frac{\sin(t)}{|\sin(t)|} \mathbf{k}$

Hence, the unit tangent vector is: $\mathbf{T}(t) = -\text{sgn}(\cos(t)) \mathbf{j} + \text{sgn}(\sin(t)) \mathbf{k}$

Would you like any clarifications or further details on any part of this solution?

Here are 5 related questions to explore:

1. How would you find the curvature of a given curve?
2. Can the unit tangent vector ever be zero?
3. What is the significance of the unit tangent vector in physics?
4. How is the principal normal vector related to the tangent vector?
5. What is the geometric interpretation of the velocity vector?

Tip: Always remember that the unit tangent vector represents the direction of motion at a given point on the curve.