https://file.aiquickdraw.com/mathbot/file/1726770374042167hsd6m.jpg

The question from the image is asking to find the length of a curve, given its parametric equations:

\[
x = t^3 - 3t, \quad y = 3t^2
\]

The parameter \( t \) ranges from 0 to 1. Let's calculate the length of the curve using the formula for the length of a curve in parametric form:

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

### Step 1: Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

\[
\frac{dx}{dt} = \frac{d}{dt}(t^3 - 3t) = 3t^2 - 3
\]
\[
\frac{dy}{dt} = \frac{d}{dt}(3t^2) = 6t
\]

### Step 2: Set up the integral

Now substitute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) into the formula for the length of the curve:

\[
L = \int_{0}^{1} \sqrt{(3t^2 - 3)^2 + (6t)^2} \, dt
\]
\[
L = \int_{0}^{1} \sqrt{(9t^4 - 18t^2 + 9) + (36t^2)} \, dt
\]
\[
L = \int_{0}^{1} \sqrt{9t^4 + 18t^2 + 9} \, dt
\]
\[
L = \int_{0}^{1} \sqrt{9(t^4 + 2t^2 + 1)} \, dt
\]
\[
L = \int_{0}^{1} 3\sqrt{(t^2 + 1)^2} \, dt
\]
\[
L = \int_{0}^{1} 3(t^2 + 1) \, dt
\]

### Step 3: Compute the integral

\[
L = 3 \int_{0}^{1} (t^2 + 1) \, dt
\]
\[
L = 3 \left[\frac{t^3}{3} + t\right]_{0}^{1}
\]
\[
L = 3 \left[\frac{1^3}{3} + 1 - 0\right] = 3 \left(\frac{1}{3} + 1\right)
\]
\[
L = 3 \times \frac{4}{3} = 4
\]

Thus, the length of the curve is \(L = 4\).

Would you like more details or have any questions?

Here are five related questions:
1. What is the formula for arc length in polar coordinates?
2. How can you compute the length of a curve if the function is given in Cartesian form?
3. Can you find the arc length of a circle in parametric form?
4. What is the physical interpretation of the length of a curve?
5. How does changing the parameterization of a curve affect its arc length?

**Tip:** When dealing with parametric equations, always differentiate both \(x(t)\) and \(y(t)\) to find the components for the arc length formula.

Find the length of the curve whose equation is given in parametric form: x = t^3 - 3t, y = 3t^2; t from 0 to 1.

Find the length of a curve with parametric equations x = t^3 - 3t and y = 3t^2 from t = 0 to t = 1. This problem explores the concept of arc length for parametric curves, using calculus to derive the solution.

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