To find the unit tangent vector for each vector function $\vec{r}(t)$, we need to follow these steps:

1. Differentiate $\vec{r}(t)$ with respect to $t$ to get $\vec{r}'(t)$.
2. Compute the magnitude $|\vec{r}'(t)|$ of the resulting vector.
3. Divide $\vec{r}'(t)$ by its magnitude to obtain the unit tangent vector $\vec{T}(t)$.

Let’s go through each problem one by one.

Problem 1

$\vec{r}(t) = t^2 \hat{i} - \cos(8t) \hat{j} + \sin(8t) \hat{k}$

1. Differentiate each component:

   • $\frac{d}{dt}(t^2) = 2t$
   • $\frac{d}{dt}(-\cos(8t)) = 8 \sin(8t)$
   • $\frac{d}{dt}(\sin(8t)) = 8 \cos(8t)$

   Thus, $\vec{r}'(t) = 2t \hat{i} + 8 \sin(8t) \hat{j} + 8 \cos(8t) \hat{k}$

2. Compute the magnitude of $\vec{r}'(t)$: $|\vec{r}'(t)| = \sqrt{(2t)^2 + (8 \sin(8t))^2 + (8 \cos(8t))^2}$ Simplify: $|\vec{r}'(t)| = \sqrt{4t^2 + 64(\sin^2(8t) + \cos^2(8t))} = \sqrt{4t^2 + 64} = 2\sqrt{t^2 + 16}$

3. The unit tangent vector is: $\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} = \frac{2t \hat{i} + 8 \sin(8t) \hat{j} + 8 \cos(8t) \hat{k}}{2\sqrt{t^2 + 16}} = \frac{t \hat{i} + 4 \sin(8t) \hat{j} + 4 \cos(8t) \hat{k}}{\sqrt{t^2 + 16}}$

Problem 2

$\vec{r}(t) = \langle 8t, 2 - t^6, t^4 \rangle$

1. Differentiate each component:

   • $\frac{d}{dt}(8t) = 8$
   • $\frac{d}{dt}(2 - t^6) = -6t^5$
   • $\frac{d}{dt}(t^4) = 4t^3$

   Thus, $\vec{r}'(t) = \langle 8, -6t^5, 4t^3 \rangle$

2. Compute the magnitude of $\vec{r}'(t)$: $|\vec{r}'(t)| = \sqrt{8^2 + (-6t^5)^2 + (4t^3)^2} = \sqrt{64 + 36t^{10} + 16t^6}$

3. The unit tangent vector is: $\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} = \frac{\langle 8, -6t^5, 4t^3 \rangle}{\sqrt{64 + 36t^{10} + 16t^6}}$

Problem 3

$\vec{r}(t) = \langle \ln(6t), e^{1 - t}, 5t \rangle$

1. Differentiate each component:

   • $\frac{d}{dt}(\ln(6t)) = \frac{1}{t}$
   • $\frac{d}{dt}(e^{1 - t}) = -e^{1 - t}$
   • $\frac{d}{dt}(5t) = 5$

   Thus, $\vec{r}'(t) = \left\langle \frac{1}{t}, -e^{1 - t}, 5 \right\rangle$

2. Compute the magnitude of $\vec{r}'(t)$: $|\vec{r}'(t)| = \sqrt{\left(\frac{1}{t}\right)^2 + (-e^{1 - t})^2 + 5^2} = \sqrt{\frac{1}{t^2} + e^{2(1 - t)} + 25}$

3. The unit tangent vector is: $\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} = \frac{\left\langle \frac{1}{t}, -e^{1 - t}, 5 \right\rangle}{\sqrt{\frac{1}{t^2} + e^{2(1 - t)} + 25}}$

Would you like further clarification on any of these steps?

Here are some related questions that might deepen your understanding:

1. How do we interpret the unit tangent vector geometrically?
2. Why is normalizing the tangent vector important in vector calculus?
3. What role does the derivative play in determining the direction of the tangent vector?
4. How can we apply this process to vector functions in higher dimensions?
5. What are the applications of unit tangent vectors in physics or engineering?

Tip: Always remember that the unit tangent vector is a normalized vector, which helps us understand the direction of motion of the vector function without considering its magnitude.