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For part a), I know that $\int(F\cdot ds)$ = $\int F(c(t)\cdot c'(t)) dt$.

So for my first path, I use points A and B to get a vector:

$$v = \langle 1-0,0-0,2-0 \rangle = \langle 1,0,2\rangle$$

Then, I get the following parametric equations:

$x = t, y =0, z = 2t$.

I can use this as my $c(t)$, but how do I find the bounds of my integral?

  • $\begingroup$ You have used the wrong tags. This is no a vector analysis problem, and it is a homework exercise. $\endgroup$ – Gabriel Sandoval Oct 17 '17 at 19:27
  • $\begingroup$ It's for my vector analysis class.... $\endgroup$ – KM9 Oct 17 '17 at 19:31


You can write the parametric equation of the segment between $A=(x_A,y_A,z_A)$ and $B=(x_B,y_B,z_B)$ as: $$ (x,y,z)=(x_A,y_A,z_A)+t(x_A-x_B,y_A-y_B,z_A-z_B) $$ with $t \in [0,1]$


$$\int_\mathbf\sigma\mathbf F\cdot d\mathbf r = \int_{\mathbf\sigma_1}\mathbf F\cdot d\mathbf r +\int_{\mathbf\sigma_2}\mathbf F\cdot d\mathbf r$$

Where $\mathbf\sigma_1$ is the segment from $A$ to $B$, and $\mathbf\sigma_2$ is the segment from $B$ to $C$. Using dummy parameterizations, $$\mathbf\sigma_1 (t) = \langle t,0,2t\rangle \\ \mathbf\sigma_2 (t) = \langle 0,2t,-2t\rangle$$ In both cases, the velocity vectors $\mathbf\sigma_1' (t)$ and $\mathbf\sigma_2' (t)$ are constant. And clearly, $0 \leq t\leq 1$

  • $\begingroup$ Why wouldn't $\mathbf\sigma_2 (t) = \langle 0,2t,-2t\rangle$ be $\langle 1,2t,2-2t\rangle$ $\endgroup$ – KM9 Oct 17 '17 at 19:39
  • $\begingroup$ Because the vector $\vec{BC}=\langle 1-1,2-0,0-2 \rangle$ $\endgroup$ – Gabriel Sandoval Oct 17 '17 at 21:08
  • $\begingroup$ Right, but should the parametric equation be $\mathbf\sigma_2 (t)$ = (x0,y0,z0) + tv? $\endgroup$ – KM9 Oct 17 '17 at 21:32
  • $\begingroup$ It shouldn't, but it could if it works. Just, be careful with the limits of $t$. Remember that there are many parameterizations for the same curve. $\endgroup$ – Gabriel Sandoval Oct 17 '17 at 21:42

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