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Work of a force.

Definition 8.11   Suppose that the vector field $\overrightarrow{F} $ represents a force throughout a region $\mathcal{D}$ in the 3-dimensional space. Let $\mathcal{C}$ be a smooth curve in $\mathcal{D}$.

The work done by this force over the curve $\mathcal{C}$ is equal to the integral

\begin{displaymath}W = \int_{\mathcal{C}} \overrightarrow{F}\cdot d \overrightarrow{r}\end{displaymath}

where $\overrightarrow{r} $ represents the (variable) radius vector from the origin to a point on $\mathcal{C}$.

Assume that the curve $\mathcal{C}$ is given by the parametrization $\overrightarrow{r} (t)= g_1(t) \overrightarrow{i} + g_2(t) \overrightarrow{j} + g_3(t) \overrightarrow{k} $, where $a \leq t \leq b$. Then $\overrightarrow{T} (t)= g'_1(t)\overrightarrow{i} +g'_2(t)\overrightarrow{j} +g'_3(t)\overrightarrow{k} $ is a direction vector of the tangent to $\mathcal{C}$ at the coresponding point.

\fbox{
\begin{minipage}{10cm}
\centerline{\underline{Various ways for computing...
... &= \int _{t=a}^{t=b} ( M dx + n dy + P dz ) \; dt
\end{align*} \end{minipage} }

Example 8.12   Let $\mathcal{C}$ be the curve with parametrization $\overrightarrow{r} (t)= \cos t \overrightarrow{i} + \cos 2t \overrightarrow{j} , \; 0 \leq t \leq \pi$. Take a force (= a vector field) $\overrightarrow{F} (x,y) = (x-y) \overrightarrow{i} + (x+y) \overrightarrow{j} $.

Example 8.13  



Noah Dana-Picard
2001-05-30