1
$\begingroup$

Let $c : [a,b] \longrightarrow \Bbb R^3$ be a differentiable function given by $c(t) = \left (x(t),y(t),z(t) \right ),$ $a \leq t \leq b$ which traces out a curve in the three dimensional space as $t$ varies from $a$ to $b.$ Let $\overrightarrow {r}(t)$ be the position vector of $c(t)$ with respect to the origin for $a \leq t \leq b.$ Let $t_0 \in [a,b].$ Let $l(t)$ be a parametrization of the tangent line to the curve $c$ at $c(t_0)$ for $c \leq t \leq d.$ Let $\overrightarrow {s} (t)$ be the position vector of the $l(t)$ with respect to the origin. Then the equation of the tangent line to the curve $c$ at $c(t_0)$ is given by $$\overrightarrow {s} (t) = \overrightarrow {r} (t_0) + (t-t_0)\ \overrightarrow {r'} (t_0)$$ where $\overrightarrow {r'} (t_0) = \left (x'(t_0),y'(t_0),z'(t_0) \right ).$

I know that the vector $\overrightarrow {r'} (t_0)$ determines the direction of the tangent line to the curve $c$ at $c(t_0)$ and owing to that reason the vector $\overrightarrow {r'} (t_0)$ is called the tangent vector to the curve $c$ at $c(t_0).$

Now lets move back where we studied straight lines in the three dimensional space in the context of solid geometry. We know that any straight line can be specified entirely by a point and a vector such that the line is passing through the given point and parallel to the given vector which is called the support of the line. The way we formulated the equation of a line through vector algebra in the following way.

Let $L$ be a line in the three dimensional space which passes through the point $M_0(x_0,y_0,z_0)$ and parallel to the vector $\overrightarrow a = l \hat i + m \hat j + n \hat k.$ Let $M(x,y,z)$ be any point on the line line except $M_0.$ Then we can easily see that the vector $\overrightarrow {MM_0} = (x-x_0) \hat i + (y-y_0) \hat j + (z-z_0) \hat k$ is collinear to the vector $\overrightarrow a.$ Therefore there exists non-zero scalar $s$ such that $\overrightarrow {MM_0} = s \overrightarrow a$ which simplfies three equations $$\begin{align*} x & = x_0 + s l\\ y & = y_0 + sm\\ z & = z_0 + sn \end{align*}$$ These three equations together represent the equation of the line $L$ in parametric form where $s$ is a parameter. As $s$ varies from $-\infty$ to $\infty$ the points $(x,y,z)$ trace out line $L.$

Now coming back to the case of tangent line we now have a point $c(t_0)$ through which the line tangent to the curve $c$ passes and we also have a direction vector namely $\overrightarrow {r'} (t_0).$ I know then by my concept of three dimensional solid geometry what I have just discussed above that the tangent line to the curve is completely specified by the point $c(t_0)$ and the direction vector $\overrightarrow {r'} (t_0).$ So the equation of the tangent line is given by $$\overrightarrow {s} (t) = \overrightarrow {r} (t_0) + \alpha \ \overrightarrow {r'} (t_0)$$ where $\alpha$ is a parameter just as before. How do I get back from here the equation of the tangent line in the prescribed format?

Any help or valuable suggestion regarding this will be highly appreciated.

Thank you very much for your valuable time for reading.

$\endgroup$
  • 2
    $\begingroup$ Couldn't $\alpha$ as a parametric variable just be the $t - t_0$ used in the tangent line prescribed format, or am I missing something about what you're asking about? $\endgroup$ – John Omielan Dec 28 '19 at 8:10
  • $\begingroup$ This came up in another question not too long ago. I still see no compelling reason to offset the parameter by $t_0$. $\endgroup$ – amd Dec 28 '19 at 8:14
  • $\begingroup$ @amd Wouldn't offsetting the parameter by $t_0$ be useful so $\overrightarrow {s} (t_0) = \overrightarrow {r} (t_0)$? I assume this is the most likely reason why it's done. $\endgroup$ – John Omielan Dec 28 '19 at 8:20
  • $\begingroup$ @mathmaniac In your equation $\overrightarrow {s} (t) = \overrightarrow {r} (t_0) + \alpha \ \overrightarrow {r'} (t_0)$, the LHS has a parameter of $t$ and, thus, is assumed to vary based on $t$. However, there's no explicit $t$ used on the RHS. As such, I assume the $\alpha$ is meant to be a function of $t$. If so, what function do you propose for it, if not the original $t - t_0$ which is used? $\endgroup$ – John Omielan Dec 28 '19 at 8:29
  • $\begingroup$ @JohnOmielan There’s a bit more to it than just that: why use $\vec{r'}(t_0)$ specifically instead of, say, a unit vector pointing in the same direction? This particular parameterization also serves as the best linear approximation to the curve at $\vec r(t_0)$. It would be nice if textbooks would make that clear instead of simply declaring that this is “the” parameterization of the tangent line. $\endgroup$ – amd Dec 28 '19 at 8:37
2
$\begingroup$

This came up in another question not too long ago. The use of the word “the” is a poor choice here (as it is in the all too common usage “the eigenvector.”) There is obviously an infinite number of parameterizations of the tangent line. If all you’re interested in is the tangent line per se, there’s no particular reason to choose the parameterization in the quote block over another. You can certainly offset the parameter by some constant without changing the line that’s being described: replace $\alpha$ by $t-t_0$ to go from your parameterization to the other one. Why stop there, though? One might just as well normalize the tangent vector $\vec {r'}(t_0)$ for an arclength parameterization of the line—the parameter represents distance along it. If there’s any parameterization that deserves to be called “the” parameterization of a curve, one could argue that it’s that one.

That said, the parameterization $\vec r(t_0)+(t-t_0)\vec {r'}(t_0)$ used in the quoted text has the virtue of being a linear approximation (in fact, the best one in a specific technical sense) to the curve at $r(t_0)$. It’s the equivalent in $\mathbb R^3$ of approximating a differentiable function $f:\mathbb R\to\mathbb R$ at $x_0$ as $f(x)\approx f(x_0)+(x-x_0)f'(x_0)$. This also explains why the same parameter is used for both the tangent line and the original curve, which otherwise would be a dubious practice. It would’ve been nice had the author explained this particular choice as “the” parameterization of the tangent line. Perhaps that’s done elsewhere in the text.

$\endgroup$
  • $\begingroup$ Do you want to mean "... best linear approximation in small enough neighbourhood of $t_0$"? Isn't it so? Also I think $\alpha(t)$ should be taken in such a way that $\alpha$ is a bijection which satisfies the initial condition if there is any. Just in the case solid geometry we take $\alpha (s) = s$ which is indeed a bijective function of one real variable $s$ without having any initial condition. But in this case the initial condition is that $l(t_0) = c(t_0).$ So the parameter of the real variable should be such which satisfies the initial condition which we are implicitly assuming. Right? $\endgroup$ – math maniac. Dec 28 '19 at 8:52
  • $\begingroup$ No, I don’t mean that. To be sure, the precise definition of “best linear approximation” does involve being able to make the error arbitrarily small by choosing a small enough neighborhood of $t_0$, but the qualification you propose is unnecessary. There’s also no need per se to require that $\alpha$ be a bijection. Just as the parameterization $(\cos t,\sin t)$, $t\in\mathbb R$ of the unit circle traces the circle out an infinite number of times, $r(t_0)+tan(s)\vec{r'}(t_))$, $s\in\mathbb R\setminus\{\pi/2+n\pi\}$ is a perfectly valid, if not particularly useful, parameterization of the line. $\endgroup$ – amd Dec 28 '19 at 18:21
  • $\begingroup$ But @amd then at least you require surjectivity of the parameteric varables. Right? Even in case of your example the parametric variable $s \mapsto \tan s,$ $s \in \Bbb R \setminus \left \{\frac {\pi} {2} + n \pi\ :\ n \in \Bbb Z \right \}$ has the property that it is surjective inspite of not being injective which is somewhat unnecessary here as you have rightly pointed out. $\endgroup$ – math maniac. Dec 29 '19 at 6:02
  • $\begingroup$ I think the parametric variable you have introduced above is not perfectly valid in this case as it is clear from the context that $l(t_0) = c(t_0).$ Do yo agree that your parametrization doesn't reflect this matter? I think in this case you have to be bit careful. Because the prescribed equation of a plane while writing without using vectors just becomes $$l(t) = c(t_0) + \alpha (t) c'(t_0).$$ In this case if I take $\alpha (t) = t - t_0,$ $t \in \Bbb R$ then we indeed get $l(t_0) = c(t_0)$ as per our requirement. But your equation doesn't follow the initial condition. $\endgroup$ – math maniac. Dec 29 '19 at 6:13
  • $\begingroup$ The initial condition is kind of hidden here or may be you can say that it is implicitly assumed. $\endgroup$ – math maniac. Dec 29 '19 at 6:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.