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In the following let $\gamma : \mathbb R \to \mathbb R^2$ denote a smooth curve.

While trying to derive the equation for the tangent line at the point $\gamma (t)$ I got confused:

Observation 1 I made was that the tangent vector at $\gamma (t)$ is given by $(\gamma_1'(t), \gamma_2'(t))$.

Observation 2 that I made was that for the tangent line at $\gamma(t)$, the $x$-intercept is equal to $\gamma_1(t)$ and the $y$-intercept is equal to $\gamma_2(t)$.

Then from observation 1 I derived that, since the slope of the tangent line is equal to the slope of the tangent vector, the slope of the tangent line is $m = {\gamma_2'(t) \over \gamma_1'(t)}$.

From observation 2 I derived that the slope of the tangent line is equal to $m = {\gamma_2(t) \over \gamma_1(t)}$.

The problem is that if my reasoning was correct these two would have to be equal. But a simple example like e.g. $\gamma(t) = (t^2, t^3)$ shows that this is not true.

What am I doing wrong?

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    $\begingroup$ It is not clear what your observation 2 mean. $\endgroup$ – user99914 May 2 '15 at 6:46
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    $\begingroup$ Observation 2 is wrong. $(\gamma_1(t),\gamma_2(t))$ are the coordinates of the point on the curve, not the intercepts. $\endgroup$ – zed111 May 2 '15 at 6:51
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Observation 2 is not correct. The tangent vector is $(\gamma'_1(t), \gamma'_2(t))$, but this is relative to the point $(\gamma_1(t), \gamma_2(t))$, so the tangent line has equation $$\gamma'_1(t)(y - \gamma_2(t)) = \gamma'_2(t)(x - \gamma_1(t)).$$ This line has $x$-intercept when $y = 0$, i.e., $$x = -\frac{\gamma'_1(t)}{\gamma'_2(t)}\gamma_2(t) + \gamma_1(t),$$ which is not what you claimed.

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