Worked examples · Sheet 2

Work of conservative fields computed with the potential

Theory used: Work of a conservative field Potential by successive integration Characterization of conservative fields

Exercise 1

Compute the work of the field

F (x, y) = (2 x + y, x + 2 y)

along the curve $r (t) = (t, 2 t^{3})$ , $t \in [0, 1]$ .

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

U (x, y) = \int (2 x + y) d x + c (y) = x^{2} + x y + c (y)

U_{y} = x + c^{'} (y) = x + 2 y ⟹ c (y) = y^{2} + k

r (0) = (0, 0), r (1) = (1, 2)

\int_{γ} F \cdot d r = U (1, 2) - U (0, 0) = 1 + 2 + 4 = 7

The direct computation of the line integral along the curve $r (t) = (t, 2 t^{3})$ is possible but laborious, hence it is convenient to check first whether the field is conservative. The cross derivatives both equal $1$ and the domain $R^{2}$ is simply connected, therefore the field is conservative. A potential, constructed by successive integration, is $U (x, y) = x^{2} + x y + y^{2}$ . By the theorem on the work of a conservative field the work is the difference of the values of the potential at the endpoints of the curve: the final point is $r (1) = (1, 2)$ , the initial point is $r (0) = (0, 0)$ , hence the work equals $U (1, 2) - U (0, 0) = 7$ . The shape of the curve between the two endpoints does not affect the result.

Exercise 2

Compute the work of the field

F (x, y) = (e^{x} sin y, e^{x} cos y)

along any piecewise regular curve from $(0, 0)$ to $(ln 2, \frac{π}{2})$ .

\frac{\partial F _{1}}{\partial y} = e^{x} cos y = \frac{\partial F _{2}}{\partial x}

U (x, y) = \int e^{x} sin y d x + c (y) = e^{x} sin y + c (y)

U_{y} = e^{x} cos y + c^{'} (y) = e^{x} cos y ⟹ c (y) = k

\int_{γ} F \cdot d r = U (ln 2, \frac{π}{2}) - U (0, 0) = 2 \cdot 1 - 1 \cdot 0 = 2

The statement asks for the work along an arbitrary curve with assigned endpoints, hence the request makes sense only if the work does not depend on the path, that is, if the field is conservative. The cross derivatives coincide on $R^{2}$ , which is simply connected, therefore the field is conservative and the work is well defined by the endpoints alone. A potential is $U (x, y) = e^{x} sin y$ , since in integrating $e^{x} sin y$ with respect to $x$ the function $sin y$ is a constant and the condition on $U_{y}$ gives $c^{'} (y) = 0$ . At the final point the potential equals $e^{l n 2} sin \frac{π}{2} = 2$ , at the initial point it equals $e^{0} sin 0 = 0$ , hence the work is $2$ .

Exercise 3

Compute the circulation of the field

F (x, y) = (2 x y, x^{2} + 1)

along the circle $x^{2} + y^{2} = 1$ traversed counterclockwise.

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

U (x, y) = x^{2} y + y + k

\oint_{γ} F \cdot d r = 0

The cross derivatives both equal $2 x$ and the domain $R^{2}$ is simply connected, hence the field is conservative; a potential is $U (x, y) = x^{2} y + y$ . By the characterization theorem for conservative fields the circulation of a conservative field along every closed curve is zero, therefore the integral equals $0$ without any computation with a parametrization. The direct computation with $r (t) = (cos t, sin t)$ would lead to the same result after integrating trigonometric functions over $[0, 2 π]$ .