Find the work done when a force $\vec{F} = (x^2 - y^2 + 2x)\hat{i}- (2xy + y)\hat{j}$ moves a particle in the xy plane from $(0,0)$ to $(1,1)$ along the parabola $y^2 = x$ . Is the work done different when the path is the straight line $y = x$ ?

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Q. Find the work done when a force

\vec{F} = (x^2 - y^2 + 2x)\hat{i}- (2xy + y)\hat{j}

moves a particle in the xy plane from

(0,0)

(1,1)

along the parabola

y^2 = x

. Is the work done different when the path is the straight line

y = x

Calculate Line Integral: To find the work done, we need to calculate the line integral of the force along the path.
Parameterize Parabola Path: For the parabola path $y^2 = x$ , parameterize the path using $t$ , where $x = t$ and $y = t^2$ . The limits of integration are from $t = 0$ to $t = 1$ .
Substitute Force Components: The force components are $\mathbf{F}_x = x^2 - y^2 + 2x$ and $\mathbf{F}_y = -(2xy + y)$ . Substitute $x$ and $y$ with $t$ and $t^2$ respectively.
Calculate Differential Elements: After substitution, $F_x = t^2 - (t^2)^2 + 2t$ and $F_y = -(2t(t^2) + t^2)$ .
Calculate Work Done for Parabola Path: The differential elements along the path are $dx = dt$ and $dy = 2tdt$ .
Evaluate Integral for Parabola Path: The work done is the integral of $F_xdx + F_ydy$ from $t = 0$ to $t = 1$ .
Simplify Expression for Parabola Path: Calculate the integral $\int_{0}^{1}(t^2 - t^4 + 2t)dt + \int_{0}^{1}-(2t^3 + t^2)(2tdt)$ .
Parameterize Straight Line Path: The first integral becomes $\int(t^2 - t^4 + 2t)dt = [\frac{t^3}{3} - \frac{t^5}{5} + t^2]$ from $0$ to $1$ .
Substitute Force Components for Straight Line Path: The second integral becomes $\int -(2t^3 + t^2)(2tdt) = \int -(4t^4 + 2t^3)dt = [-\frac{4t^5}{5} - t^4]$ from $0$ to $1$ .
Calculate Differential Elements for Straight Line Path: Evaluate both integrals from $0$ to $1$ : $\left[\frac{1}{3} − \frac{1}{5} + 1\right] − \left[−\frac{4}{5} − 1\right]$ .
Calculate Work Done for Straight Line Path: Simplify the expression: $(\frac{1}{3} − \frac{1}{5} + 1) + (\frac{4}{5} + 1) = (\frac{5}{15} + \frac{3}{15} + \frac{15}{15}) + (\frac{12}{15} + \frac{15}{15})$ .
Evaluate Integral for Straight Line Path: Add the fractions: $(\frac{23}{15}) + (\frac{27}{15}) = \frac{50}{15}$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .Calculate the integral $y = x$ $6$ from $y = x$ $7$ to $y = x$ $8$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .Calculate the integral $y = x$ $6$ from $0$ to $1$ .The first integral becomes $y = x$ $7$ from $0$ to $1$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .Calculate the integral $y = x$ $6$ from $y = x$ $7$ to $y = x$ $8$ .The first integral becomes $y = x$ $9$ from $y = x$ $7$ to $y = x$ $8$ .The second integral becomes $t$ $2$ from $y = x$ $7$ to $y = x$ $8$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .Calculate the integral $y = x$ $6$ from $y = x$ $7$ to $y = x$ $8$ .The first integral becomes $y = x$ $9$ from $y = x$ $7$ to $y = x$ $8$ .The second integral becomes $t$ $2$ from $y = x$ $7$ to $y = x$ $8$ .Evaluate both integrals from $y = x$ $7$ to $y = x$ $8$ : $t$ $7$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .Calculate the integral $y = x$ $6$ from $y = x$ $7$ to $y = x$ $8$ .The first integral becomes $y = x$ $9$ from $y = x$ $7$ to $y = x$ $8$ .The second integral becomes $t$ $2$ from $y = x$ $7$ to $y = x$ $8$ .Evaluate both integrals from $y = x$ $7$ to $y = x$ $8$ : $t$ $7$ .Simplify the expression: $t$ $8$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .Calculate the integral $y = x$ $6$ from $y = x$ $7$ to $y = x$ $8$ .The first integral becomes $y = x$ $9$ from $y = x$ $7$ to $y = x$ $8$ .The second integral becomes $t$ $2$ from $y = x$ $7$ to $y = x$ $8$ .Evaluate both integrals from $y = x$ $7$ to $y = x$ $8$ : $t$ $7$ .Simplify the expression: $t$ $8$ .Add the fractions: $t$ $9$ .
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .Calculate the integral $y = x$ $6$ from $y = x$ $7$ to $y = x$ $8$ .The first integral becomes $y = x$ $9$ from $y = x$ $7$ to $y = x$ $8$ .The second integral becomes $t$ $2$ from $y = x$ $7$ to $y = x$ $8$ .Evaluate both integrals from $y = x$ $7$ to $y = x$ $8$ : $t$ $7$ .Simplify the expression: $t$ $8$ .Add the fractions: $t$ $9$ .Compare the work done along the parabola and the straight line: $x = t$ $0$ for the parabola and $x = t$ $1$ for the straight line.
Compare Work Done: Simplify the fraction: $\frac{50}{15} = \frac{10}{3}$ .For the straight line path $y = x$ , parameterize the path using $t$ , where $x = t$ and $y = t$ . The limits of integration are from $t = 0$ to $t = 1$ .The force components are $F_x = t^2 - t^2 + 2t$ and $F_y = -(2t^2 + t)$ . After simplification, $F_x = 2t$ and $F_y = -(2t^2 + t)$ .The differential elements along the path are $y = x$ $1$ and $y = x$ $2$ .The work done is the integral of $y = x$ $3$ from $t = 0$ to $t = 1$ .Calculate the integral $y = x$ $6$ from $y = x$ $7$ to $y = x$ $8$ .The first integral becomes $y = x$ $9$ from $y = x$ $7$ to $y = x$ $8$ .The second integral becomes $t$ $2$ from $y = x$ $7$ to $y = x$ $8$ .Evaluate both integrals from $y = x$ $7$ to $y = x$ $8$ : $t$ $7$ .Simplify the expression: $t$ $8$ .Add the fractions: $t$ $9$ .Compare the work done along the parabola and the straight line: $x = t$ $0$ for the parabola and $x = t$ $1$ for the straight line.Since $x = t$ $0$ is not equal to $x = t$ $1$ , the work done is different for the two paths.