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I am a TA in an introductory course to multivariable calculus. As defined in class two curves are said to be equal if their images are equal. Now a problem in their problemset was to prove that two curves in fact were equal. Most tested that the start and endpoints were equal, and their slope were equal. However most if not all forgot to mention that the parametrizations had to be continuous and monotonously increasing.

I came up with the following "counter-example" why it is not enough to check the slope and endpoints.

$\phantom{.}\qquad\qquad\qquad\qquad\quad\quad$ enter image description here

$$x_1(t) = \begin{cases} t & \text{for} & - 1 \leq t \leq 0 \\ 2t& \text{for} & \phantom{-} 0 \leq t \leq 1/2 \end{cases} \quad \text{and} \quad y_1(t) = \begin{cases} t & \text{for} & - 1 \leq t \leq 0 \\ 2t& \text{for} & \phantom{-} 0 \leq t \leq 1/2 \end{cases} \\ $$ Now the parametrization $r_2(t) = \bigl(x(t),y(t)\bigr)$ would pass through the red points in the figure $(-1,1)$ at $t=-1$, $(0,0)$ at $t=0$ and finaly $(1,1)$ at $t=1/2$. Now I want to figure out the parametrization to the green line in the figure, eg finding curve $r_2$ that lie on $y=x$, and pass through the red points at the same time as $r_1$, but it should not equal $r_1$. I was playing around with a variation of sines and cosines, but it was all for nought.

Question: How would I go about defining the green function $r_2(t)$ as a parametrization? Lies on $y=x$, but is unequal to $r_1(t)$.

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1 Answer 1

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Using $x_1,y_1$ as you've defined them, take $x_1(2\sin(\frac{5 \pi}{6}t ))$ and $x_2(2\sin(\frac{5 \pi}{6}t ))$.

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