To determine the arc length $S$ of the curve given by parametric equations $x(t)$ and $y(t)$ over the interval $0 \leq t \leq 3$, we follow these steps:

1. Recall the Arc Length Formula for Parametric Curves

The arc length $S$ of a curve defined parametrically as $x(t)$ and $y(t)$ over $a \leq t \leq b$ is:

$S = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt$

2. Differentiate $x(t) = t^2 \sin(t)$ and $y(t) = t^2 \cos(t)$

Using the product rule:

1. For $x(t) = t^2 \sin(t)$: $\frac{dx}{dt} = 2t \sin(t) + t^2 \cos(t)$

2. For $y(t) = t^2 \cos(t)$: $\frac{dy}{dt} = 2t \cos(t) - t^2 \sin(t)$

3. Square and Sum the Derivatives

Now, calculate $\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2$:

$\left( \frac{dx}{dt} \right)^2 = \left( 2t \sin(t) + t^2 \cos(t) \right)^2$ $\left( \frac{dy}{dt} \right)^2 = \left( 2t \cos(t) - t^2 \sin(t) \right)^2$

Adding these two terms simplifies significantly due to the trigonometric identity $\sin^2(t) + \cos^2(t) = 1$. After simplification:

$\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 = 4t^2 + t^4$

4. Integrate to Find $S$

The arc length becomes:

$S = \int_0^3 \sqrt{4t^2 + t^4} \, dt$

Factor $t^2$ out of the square root:

$S = \int_0^3 \sqrt{t^2 (4 + t^2)} \, dt = \int_0^3 t \sqrt{4 + t^2} \, dt$

5. Solve the Integral

To solve $\int t \sqrt{4 + t^2} \, dt$, use substitution:

Let $u = 4 + t^2$, so $du = 2t \, dt$. Then:

$t \, dt = \frac{1}{2} \, du$

Substitute into the integral:

$\int t \sqrt{4 + t^2} \, dt = \frac{1}{2} \int \sqrt{u} \, du$

The integral of $\sqrt{u}$ is:

$\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$

Thus:

$\int t \sqrt{4 + t^2} \, dt = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2}$

Substitute $u = 4 + t^2$ back:

$\int t \sqrt{4 + t^2} \, dt = \frac{1}{3} (4 + t^2)^{3/2}$

6. Evaluate the Definite Integral

Now evaluate from $t = 0$ to $t = 3$:

$S = \left[ \frac{1}{3} (4 + t^2)^{3/2} \right]_0^3$

At $t = 3$:

$(4 + 3^2) = (4 + 9) = 13, \quad (4 + 9)^{3/2} = 13^{3/2}$

At $t = 0$:

$(4 + 0^2)^{3/2} = 4^{3/2} = 8$

Thus:

$S = \frac{1}{3} \left[ 13^{3/2} - 8 \right]$

Final Answer

The exact arc length is:

$S = \frac{1}{3} \left( 13^{3/2} - 8 \right) \, \text{units.}$

Let me know if you need any further clarification or detailed steps!

Related Questions

1. What is the geometric interpretation of arc length in parametric equations?
2. How does the arc length formula change for polar curves?
3. What substitution techniques help solve integrals involving square roots?
4. How do you derive $\sin^2(t) + \cos^2(t) = 1$ in trigonometry?
5. How is arc length calculated for surfaces of revolution?

Tip: To simplify square root integrals, always look for factoring or substitution methods first!