To evaluate the integral of $y \, ds$ over the curve $C$ parameterized by $\mathbf{r}(t) = \langle 2t^2, 2t \rangle$ for $-3 \leq t \leq 0$, we proceed as follows:

Step 1: Recall the formula

The line integral of $y \, ds$ is given by: $\int_C y \, ds$ where $ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt$.

Step 2: Compute derivatives of $x(t)$ and $y(t)$

From the parameterization $\mathbf{r}(t) = \langle 2t^2, 2t \rangle$, we have: $x(t) = 2t^2 \quad \text{and} \quad y(t) = 2t.$ The derivatives are: $\frac{dx}{dt} = 4t \quad \text{and} \quad \frac{dy}{dt} = 2.$

Step 3: Compute $ds$

Using the formula for $ds$: $ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt,$ we substitute the derivatives: $ds = \sqrt{(4t)^2 + 2^2} \, dt = \sqrt{16t^2 + 4} \, dt = 2\sqrt{4t^2 + 1} \, dt.$

Step 4: Express the integral

The integral becomes: $\int_C y \, ds = \int_{-3}^{0} y(t) \cdot ds = \int_{-3}^{0} 2t \cdot 2\sqrt{4t^2 + 1} \, dt.$ Simplify: $\int_C y \, ds = 4 \int_{-3}^{0} t \sqrt{4t^2 + 1} \, dt.$

Step 5: Solve the integral

Let $u = 4t^2 + 1$, so that $du = 8t \, dt$. This substitution gives: $t \, dt = \frac{1}{8} \, du.$ When $t = -3$, $u = 4(-3)^2 + 1 = 37$.
When $t = 0$, $u = 4(0)^2 + 1 = 1$.

The integral becomes: $4 \int_{-3}^{0} t \sqrt{4t^2 + 1} \, dt = 4 \int_{37}^{1} \sqrt{u} \cdot \frac{1}{8} \, du.$ Simplify: $= \frac{1}{2} \int_{37}^{1} u^{1/2} \, du.$

Step 6: Compute the antiderivative

The antiderivative of $u^{1/2}$ is: $\int u^{1/2} \, du = \frac{2}{3} u^{3/2}.$ Evaluate: $\frac{1}{2} \int_{37}^{1} u^{1/2} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{37}^{1}.$ Simplify: $= \frac{1}{3} \left[ u^{3/2} \right]_{37}^{1} = \frac{1}{3} \left( 1^{3/2} - 37^{3/2} \right).$

Step 7: Final result

$\int_C y \, ds = \frac{1}{3} \left( 1 - 37^{3/2} \right).$

This is the evaluated integral. Let me know if you have questions or need clarification!

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Expanding Understanding:

1. What is the geometric meaning of the integral $\int_C y \, ds$?
2. How would the calculation differ if $C$ was parameterized in a different direction?
3. How do you compute $\sqrt{u}^{3/2}$ for other complex cases?
4. Could the substitution method change for other parabola-like curves?
5. How is this integral related to physical concepts like arc length or mass distribution?

Tip: Always verify parameter limits when performing a substitution to ensure consistency with the original integral bounds.