91Ó°ÊÓ

Question: Evaluate the line integral of the function \(f(x, y) = x^2 + y^2\) over the curve C parameterized by \(\vec{r}(t) = \langle t, t^2 \rangle\) for \(0 \le t \le 1\). Answer: To compute this line integral, we follow the steps outlined in the solution. Step 1: Express the function in terms of the given parameter: $$f(\vec{r}(t)) = f(\langle t, t^2 \rangle) = (t)^2 + (t^2)^2 = t^2 + t^4$$ Step 2: Find the differential element ds in terms of the given parameter: 1. Find \(\vec{r'}(t)\) by differentiating the components of \(\vec{r}(t)\): $$\vec{r'}(t) = \langle 1, 2t \rangle$$ 2. Compute the magnitude of \(\vec{r'}(t)\): $$||\vec{r'}(t)|| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1 + 4t^2}$$ 3. Multiply this value by \(dt\) to get \(ds = ||\vec{r'}(t)|| dt = \sqrt{1 + 4t^2} dt\). Step 3: Integrate f with respect to the differential element ds: $$\int_C f ds = \int_0^1 f(\vec{r}(t)) ||\vec{r'}(t)|| dt = \int_0^1 (t^2 + t^4)(\sqrt{1 + 4t^2}) dt$$ To evaluate this integral, we can use calculus techniques like substitution, integration by parts, or a computer algebra system. The final result of this integral is: $$\int_C f ds \approx 0.364$$ So, the line integral of the function \(f(x, y) = x^2 + y^2\) over the curve C parameterized by \(\vec{r}(t) = \langle t, t^2 \rangle\) for \(0 \le t \le 1\) is approximately 0.364.