The arc length of a curve given by parametric equations can be found using the following formula:

$L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt$

Here, the parametric equations are:

• $x = \sqrt{t}$
• $y = 3t - 3$

We need to calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

1. $\frac{dx}{dt} = \frac{d}{dt}(\sqrt{t}) = \frac{1}{2\sqrt{t}}$
2. $\frac{dy}{dt} = \frac{d}{dt}(3t - 3) = 3$

Now, substitute these into the arc length formula:

$L = \int_{0}^{1} \sqrt{\left( \frac{1}{2\sqrt{t}} \right)^2 + (3)^2} \, dt$ $L = \int_{0}^{1} \sqrt{\frac{1}{4t} + 9} \, dt$

Let's compute this integral to find the arc length.The arc length of the curve given by the parametric equations $x = \sqrt{t}$ and $y = 3t - 3$ over the interval $0 \leq t \leq 1$ is approximately 3.250 (rounded to three decimal places).

Would you like more details, or do you have any questions?

Here are five related questions:

1. How would the arc length change if the parametric equation for $y$ were different?
2. What is the geometric significance of this arc length for the curve?
3. How can you calculate the arc length if given polar coordinates instead of parametric ones?
4. Can the arc length formula be applied to higher-dimensional curves?
5. What is the difference between arc length and distance between two points?

Tip: When solving arc length problems, always check the bounds of the parameter $t$ and confirm that the integrand remains continuous and non-negative within those bounds.