The arc length $S$ of a vector function $\mathbf{r}(t)$ over an interval $a \leq t \leq b$ is given by the formula:

$S = \int_{a}^{b} \|\mathbf{r}'(t)\| \, dt$

Step 1: Compute $\mathbf{r}'(t)$

Given the vector function:

$\mathbf{r}(t) = \langle -4t+3, -3t+5, -2t+5 \rangle$

Differentiate each component with respect to $t$:

$\mathbf{r}'(t) = \langle -4, -3, -2 \rangle$

Step 2: Compute $\|\mathbf{r}'(t)\|$

The magnitude of $\mathbf{r}'(t)$ is:

$\|\mathbf{r}'(t)\| = \sqrt{(-4)^2 + (-3)^2 + (-2)^2}$

$= \sqrt{16 + 9 + 4} = \sqrt{29}$

Since this is constant, the arc length integral simplifies to:

$S = \int_{2}^{5} \sqrt{29} \, dt$

Step 3: Evaluate the Integral

$S = \sqrt{29} \int_{2}^{5} dt$

$S = \sqrt{29} \times (5 - 2)$

$S = 3\sqrt{29}$

Final Answer:

$S = 3\sqrt{29}$

Would you like a step-by-step explanation of any part?

Here are some related questions:

1. How do you find the arc length of a parametric curve?
2. What happens if $\mathbf{r}(t)$ includes trigonometric functions?
3. How does arc length change if velocity is not constant?
4. Can arc length be found in higher dimensions?
5. How does arc length relate to curvature?

Tip: When the magnitude of $\mathbf{r}'(t)$ is constant, arc length computation simplifies to multiplying by the interval length.