The work done on a particle enacted to move from (0,0)(0,0) to (2,4)(2,4) along the parabola y=x2𝑦=𝑥2 by the force F––(t)=x2i–+y2j–𝐹_(𝑡)=𝑥2𝑖_+𝑦2𝑗_ is given by

The work done by a force on a particle moving along a path is given by the line integral of the force along the path. In this case, the force is F(t) = x^2 i + y^2 j and the path is along the parabola y = x^2 from (0,0) to (2,4).

Step 1: Parameterize the path We can parameterize the path along the parabola y = x^2 from (0,0) to (2,4) by setting x = t and y = t^2. This gives us the parameterization r(t) = t i + t^2 j for 0 <= t <= 2.

Step 2: Compute the derivative of the path The derivative of r(t) with respect to t is r'(t) = i + 2t j.

Step 3: Substitute the parameterization into the force Substituting the parameterization into the force gives F(t) = t^2 i + t^4 j.

Step 4: Compute the dot product F(t) . r'(t) The dot product of F(t) and r'(t) is F(t) . r'(t) = t^2 * 1 + t^4 * 2t = t^2 + 2t^5.

Step 5: Integrate the dot product from 0 to 2 The work done by the force is the integral of the dot product from 0 to 2. This gives us the integral from 0 to 2 of (t^2 + 2t^5) dt = [1/3 t^3 + 1/3 t^6] from 0 to 2 = 8/3 + 64/3 = 72/3 = 24 units of work.