QuestionWork done by a force: Calculate the work done by a force field $F(x, y)=\left(x^{\wedge} 2\right.$ $\left.+y^{\wedge} 2, x-y\right)$ on an object moving along the curve C : $\mathrm{x}=\mathrm{t}^{\wedge} 2, \mathrm{y}=\mathrm{t}$ from $\mathrm{t}=0$ to t $=1$.

Studdy Solution
Integrate the dot product over the given range of $t$.
$\text{Work} = \int_0^1 (2t^5 + 2t^3 + t^2 - t) \, dt$
Calculate each term separately:
$\int_0^1 2t^5 \, dt = 2 \left[ \frac{t^6}{6} \right]_0^1 = \frac{2}{6} = \frac{1}{3}$
$\int_0^1 2t^3 \, dt = 2 \left[ \frac{t^4}{4} \right]_0^1 = \frac{2}{4} = \frac{1}{2}$
$\int_0^1 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3}$
$\int_0^1 -t \, dt = - \left[ \frac{t^2}{2} \right]_0^1 = - \frac{1}{2}$
Sum up all the terms:
$\text{Work} = \frac{1}{3} + \frac{1}{2} + \frac{1}{3} - \frac{1}{2} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
Solution: The work done by the force field $\mathbf{F}(x, y) = (x^2 + y^2, x - y)$ on an object moving along the curve $x = t^2, y = t$ from $t = 0$ to $t = 1$ is $\frac{2}{3}$.