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I get how to answer the qs below, the problem is actually finding path $2$ $ \left( 2, 0, 0 \right) $ to $ \left( 0, 1, 0 \right) $ I get $(2-t)i+tj$

yet the answer for path 2 is...

$$ (2-t)i+(t/2)j $$

Don't understand why, any help would be appreciated.


Some-more context...

Let $G$ be the vector field given by

$$ G = 2{y i} + x^{2}{j} + z k $$

Evaluate the line integral

$$ I = \oint_{c} G \cdot dr $$

where $C$ is given by the three sides of the triangle with verticies $ \left( 0, 0, 0 \right) $, $ \left( 2, 0, 0 \right) $ and $ \left( 0, 0, 0 \right) $, and the integration is preformed in the following direction: from $ \left( 0, 0, 0 \right) $ to $ \left( 2, 0, 0 \right) $ then to $ \left( 0, 1, 0 \right) $ and finally back to $ \left( 0, 0, 0 \right) $. You may evaluate the integral $I$...

enter image description here

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  • $\begingroup$ I’d suggest double-checking your calculations. You have $2-t$ instead of $2(1-t)$, so it looks to me like you made a simple error along the way. That aside, there are many different parameterizations possible—there’s not one “right” parameterization. $\endgroup$ – amd May 11 at 18:04
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A parametric equation for a line, whose 2 points $\overrightarrow a$ and $\overrightarrow b$ are given, is - $$\overrightarrow r = \overrightarrow a + (\overrightarrow b - \overrightarrow a)k $$

Putting in the 2 points given, we get (writing in coordinate form) - \begin{align} \overrightarrow r &= (2,0,0) + ((0,1,0)-(2,0,0))k \\ &= (2,0,0) + (-2,1,0)k \end{align} Rescaling $ k = \frac{t}{2}$ to match with your values, we get - $$\overrightarrow r = (2-t)\hat i + \frac{t}{2}\hat j$$

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  • $\begingroup$ @amd My bad. 'A' is more appropriate. $\endgroup$ – Ishan Deo May 11 at 18:14
  • $\begingroup$ got it thanks mate $\endgroup$ – Anon Ymous May 11 at 18:32
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The parametrisation of the line between two points

$$ A = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}, B = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$$

is given by

$$\gamma(t) = A + (B - A)t$$

although other parametrisations are possible.

In particular, in your case, this yields

$$\gamma(t) = \begin{pmatrix} 2 \\ 0 \\0 \end{pmatrix} + (\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 0\end{pmatrix})t = \begin{pmatrix} 2 - 2t \\ t \\ 0 \end{pmatrix}$$

which is equivalent with your second parametrisation.

The main problem with your first parametrisation is that your second point does not lie on the path. You likely intended this to happen at $t = 1$; however, try entering that value in your parametrisation.

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  • $\begingroup$ I meant it to be interpreted as 'This is always a parametrisation of said line'. The fact that I referred to equivalence in my answer already implies this. $\endgroup$ – Alexander Geldhof May 11 at 18:50
  • $\begingroup$ No. It is interpretable both ways. If I say '2 is always a natural number' then I am not excluding 2 being a real number. If I say that the given parametrisation is always given by $\gamma(t)$, I am not excluding it being given by another formula. $\endgroup$ – Alexander Geldhof May 11 at 18:59
  • $\begingroup$ I'm going to edit this, since if one person interpretes it differently, OP might too. $\endgroup$ – Alexander Geldhof May 11 at 19:06

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