Worked examples · Sheet 3

Work of non-conservative fields by direct computation

Theory used: Work and circulation Irrotational field (closed field)Characterization of conservative fields

Exercise 1

Compute the work of the field

F (x, y) = (y, x^{2})

along the segment from $(0, 0)$ to $(1, 1)$ .

\frac{\partial F _{1}}{\partial y} = 1, \frac{\partial F _{2}}{\partial x} = 2 x

r (t) = (t, t), t \in [0, 1], r^{'} (t) = (1, 1)

F (r (t)) \cdot r^{'} (t) = t \cdot 1 + t^{2} \cdot 1 = t + t^{2}

\int_{γ} F \cdot d r = \int_{0}^{1} (t + t^{2}) d t = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}

The cross derivatives equal $1$ and $2 x$ and do not coincide, hence the field is not conservative and the work must be computed with the definition. The segment is parametrized by $r (t) = (t, t)$ for $t \in [0, 1]$ , with derivative $r^{'} (t) = (1, 1)$ . Substituting the parametrization into the components of the field one obtains $F (r (t)) = (t, t^{2})$ , and the scalar product with $r^{'} (t)$ equals $t + t^{2}$ . The integral of this function over $[0, 1]$ gives $\frac{5}{6}$ .

Exercise 2

Compute the work of the field

F (x, y) = (y^{2}, - x)

along the arc of the parabola $y = x^{2}$ from $(0, 0)$ to $(1, 1)$ .

r (t) = (t, t^{2}), t \in [0, 1], r^{'} (t) = (1, 2 t)

F (r (t)) \cdot r^{'} (t) = t^{4} \cdot 1 + (- t) \cdot 2 t = t^{4} - 2 t^{2}

\int_{γ} F \cdot d r = \int_{0}^{1} (t^{4} - 2 t^{2}) d t = \frac{1}{5} - \frac{2}{3} = - \frac{7}{15}

The field is not conservative, since the derivative of $y^{2}$ with respect to $y$ equals $2 y$ and the derivative of $- x$ with respect to $x$ equals $- 1$ . The arc of the parabola is parametrized by $x = t$ and $y = t^{2}$ for $t \in [0, 1]$ , hence $r^{'} (t) = (1, 2 t)$ . The components of the field along the curve are $F_{1} = (t^{2})^{2} = t^{4}$ and $F_{2} = - t$ , therefore the integrand is $t^{4} - 2 t^{2}$ . The work is negative: along this arc the field does resistive work, since on average it forms an obtuse angle with the direction of motion.

Exercise 3

Compute the circulation of the field

F (x, y) = (- y, x)

along the circle $x^{2} + y^{2} = 1$ traversed counterclockwise, and deduce that the field is not conservative.

r (t) = (cos t, sin t), t \in [0, 2 π], r^{'} (t) = (- sin t, cos t)

F (r (t)) \cdot r^{'} (t) = (- sin t) (- sin t) + cos t \cdot cos t = 1

\oint_{γ} F \cdot d r = \int_{0}^{2 π} 1 d t = 2 π

On the unit circle the field equals $F (r (t)) = (- sin t, cos t)$ , which coincides with the tangent vector $r^{'} (t)$ : the field is at every point tangent to the circle and agrees with the direction of travel. The scalar product is therefore $sin^{2} t + cos^{2} t = 1$ and the circulation equals $2 π$ . By the characterization theorem a conservative field has zero circulation along every closed curve; since this circulation is nonzero, the field is not conservative on $R^{2}$ . The same result is reached with the cross derivatives, which equal $- 1$ and $1$ .