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hivesaeed4
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If $${\tmmathbf{r}}$$ and $${\tmmathbf{s}}$$ are piecewise smooth paths, which have the same graph, then they are said to be equivalent paths.
They either trace out a set of points in the same direction, or in the opposite direction.
If they trace out a curve $${C}$$ in the same direction
$${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$ $${{\int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$
If they trace out a curve $${C}$$ in the opposite direction
$${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$$${{- \int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$
Evaluate the path integral $${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t}$$ where
$${\tmmathbf{F}=\tmmathbf{i}}$$
and
$${C}$$ is the parabola $${y = x^2}$$ traced out by $${\tmmathbf{r}}$$ so that the path goes from (0,0) to (1,1).
What's the significance of mentioning that C is a parabola y=x^2? I mean can't we do without it?
They either trace out a set of points in the same direction, or in the opposite direction.
If they trace out a curve $${C}$$ in the same direction
$${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$ $${{\int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$
If they trace out a curve $${C}$$ in the opposite direction
$${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$$${{- \int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$
Evaluate the path integral $${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t}$$ where
$${\tmmathbf{F}=\tmmathbf{i}}$$
and
$${C}$$ is the parabola $${y = x^2}$$ traced out by $${\tmmathbf{r}}$$ so that the path goes from (0,0) to (1,1).
What's the significance of mentioning that C is a parabola y=x^2? I mean can't we do without it?