To find the length along the curve $\mathbf{r}(t) = \langle 1 + t^2, 2 - t^2, 3 + t^3 \rangle$ from the point $(1,2,3)$ to $(2,1,4)$, we need to follow these steps:

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Step 1: Find the parameter values for the points

1. At $(1,2,3)$: $x = 1 + t^2, \quad y = 2 - t^2, \quad z = 3 + t^3$ Set each component to the corresponding coordinates: $1 = 1 + t^2 \implies t^2 = 0 \implies t = 0$ Therefore, $t = 0$ for the point $(1,2,3)$.

2. At $(2,1,4)$: $x = 1 + t^2 \implies 2 = 1 + t^2 \implies t^2 = 1 \implies t = 1 \quad (\text{we ignore negative roots})$ Therefore, $t = 1$ for the point $(2,1,4)$.

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Step 2: Compute the length formula

The formula for the arc length of a curve is: $L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt$ where $a = 0$, $b = 1$, and $\mathbf{r}'(t)$ is the derivative of $\mathbf{r}(t)$.

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Step 3: Compute $\mathbf{r}'(t)$

$\mathbf{r}'(t) = \frac{d}{dt} \langle 1 + t^2, 2 - t^2, 3 + t^3 \rangle = \langle 2t, -2t, 3t^2 \rangle$ The magnitude of $\mathbf{r}'(t)$ is: $\| \mathbf{r}'(t) \| = \sqrt{(2t)^2 + (-2t)^2 + (3t^2)^2} = \sqrt{4t^2 + 4t^2 + 9t^4}$ $= \sqrt{8t^2 + 9t^4} = \sqrt{t^2(8 + 9t^2)} = t\sqrt{8 + 9t^2}$

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Step 4: Compute the definite integral

$L = \int_{0}^{1} t\sqrt{8 + 9t^2} \, dt$ Let $u = 8 + 9t^2$, so $du = 18t \, dt$.

When $t = 0$, $u = 8$.
When $t = 1$, $u = 17$.

Thus, $L = \frac{1}{18} \int_{8}^{17} \sqrt{u} \, du$ The integral of $\sqrt{u}$ is: $\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$ Applying the bounds: $L = \frac{1}{18} \times \frac{2}{3} \left[ u^{3/2} \right]_{8}^{17}$ $L = \frac{1}{27} \left[ 17^{3/2} - 8^{3/2} \right]$ Numerically: $17^{3/2} \approx 70.248, \quad 8^{3/2} = 22.627$ $L \approx \frac{1}{27} (70.248 - 22.627) \approx \frac{47.621}{27} \approx 1.764$

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Final Answer:

$L \approx 1.764 \, \text{units}$

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Would you like me to go deeper into the substitution process or provide additional examples?

Related Questions

1. How do you find the arc length if the parameter range isn't explicitly given?
2. Can we derive arc length for curves in polar coordinates?
3. What happens when velocity becomes zero in the arc-length formula?
4. How does parameter substitution simplify arc-length integrals?
5. How is arc length calculated numerically in more complex cases?

Tip: Always check boundary conditions carefully to ensure substitution aligns with the problem's limits.