T(t)≠0 for all values of t and the tangent line at any given point of the curve always passes through point D. Show that r represents a straight line Let $r: \mathbb{R}->\mathbb{R}^3$ be a curve in arc-length parametrization such that $T(t)$ does not equal to zero for all values of $t$. Assume that the tangent line at any given point $r(t)$ of the curve always passes through a point $D$. Show that $r$ represents a straight line.
Here is my thought process on approaching this question:
For a line to be straight, the curvature $k(t$) has to be $0$.
Since it has been established that the tangent line of the curve at any given point passes through point $D$ and that $T(t)$ does not equal to $0$, the curve has to be linear.
Since the curve is linear, the tangent line, which is the derivative of the curve, has to be constant.
Because I have to show that the tangent line is straight, I have to show that the curvature of this tangent line is $0$. The tangent line $T(t)$ for $r(t)$ is zero and the derivative of $T(t)$ is also zero, therefore the curvature of $r(t)$ is zero.
Can someone please tell me if I am right?
 A: It's hard for me to tell our OP Calvin Chen's arguments are right or not; they seem a little unclear to me.
That being said, here's how I would do it:
Given that $r(s)$ is parametrized by arc-length, the unit tangent vector $T(s)$ is given by
$T(s) = r'(s); \tag 1$
since the tangent line to any point on the curve passes through $D$, and the vector joining $D$ and $r(s)$ is $r(s) - D$, we must have
$T(s) = c(s)(r(s) - D) \tag 2$
for any value of $s$; (2) simply says the tangent vector to the curve at $s$ is collinear with $r(s) - D$, which must be the case if he tangent line contains $D$.  Combining (1) and (2) we obtain
$r'(s) = c(s)(r(s) - D), \tag 3$
which in fact is easy to solve, given an initial value of $r(s)= r(s_0)$ at $s = s_0$. If we define
$r_1(s) = r(s) - D, \tag 4$
then
$r_1'(s) = r'(s), \tag 5$
so the equation (3) becomes
$r_1'(s) = c(s) r_1(s); \tag 6$
the solution to (6) is
$r_1(s) = \exp \left ( \displaystyle \int_{s_0}^s c(u) \; du \right ) r_1(s_0); \tag 7$
now if we back-substitute (4) we have
$r(s) - D = \exp \left ( \displaystyle \int_{s_0}^s c(u) \; du \right ) (r(s_0) - D), \tag 8$
or 
$r(s) = \exp \left ( \displaystyle \int_{s_0}^s c(u) \; du \right ) (r(s_0) - D) + D; \tag 9$
we can determine $c(s)$ explicitly using (1) and (9), which yields
$T(s) = r'(s) = c(s) \exp \left ( \displaystyle \int_{s_0}^s c(u) \; du \right ) (r(s_0) - D); \tag{10}$
since $T(s)$ is of unit length we have
$1 = \Vert T(s) \Vert = \vert c(s) \vert  \exp \left ( \displaystyle \int_{s_0}^s c(u) \; du \right ) \Vert r(s_0) - D \Vert, \tag{11}$ 
which prohibits $c(s) = 0$; thus $c(s)$ is of fixed sign; if $c(s) > 0$ then
(11) yields
$c(s) = \Vert r(s_0) - D \Vert^{-1} \exp \left ( \displaystyle -\int_{s_0}^s c(u) \; du \right ), \tag{12}$
whence
$c'(s) = -c(s) \Vert r(s_0) - D \Vert^{-1} \exp \left ( \displaystyle -\int_{s_0}^s c(u) \; du \right ) = -c^2(s); \tag{13}$
also from (12)
$c(s_0) = \Vert r(s_0) - D \Vert^{-1}, \tag{14}$
the solution to (13) with initial condition (14) is
$c(s) = \dfrac{1}{(s - s_0) + \Vert r(s_0) - D \Vert}; \tag{15}$
we have
$\displaystyle \int_{s_0}^s c(u) \; du = \int_{s_0}^s \dfrac{du}{(u - s_0) + \Vert r(s_0) - D \Vert}$
$= \ln ((s - s_0) + \Vert r(s_0) - D \Vert) - \ln(\Vert r(s_0) - D \Vert) = \ln \dfrac{(s - s_0) + \Vert r(s_0) - D \Vert}{\Vert r(s_0) - D \Vert}, \tag{16}$
and then
$\exp \left ( \displaystyle \int_{s_0}^s c(u) \; du \right ) = \dfrac{(s - s_0) + \Vert r(s_0) - D \Vert}{\Vert r(s_0) - D \Vert}, \tag{17}$
and the equation (9) for $r(s)$ thus takes the form
$r(s) = \dfrac{(s - s_0) + \Vert r(s_0) - D \Vert}{\Vert r(s_0) - D \Vert}(r(s_0) - D) + D, \tag{18}$
which is clearly the equation of a line.  Also, it is easily seen that
$r(s_0) = \dfrac{\Vert r(s_0) - D \Vert}{\Vert r(s_0) - D \Vert}(r(s_0) - D) + D = r(s_0) - D + D = r(s_0), \tag{19}$
that is, (18) is consistent with our initital value of $r(s)$.  We may also find $s_D$ such that $r(s_D) = D$, writing
$D = r(s_D) = \dfrac{(s_D - s_0) + \Vert r(s_0) - D \Vert}{\Vert r(s_0) - D \Vert}(r(s_0) - D) + D, \tag{20}$
we see that
$ \dfrac{(s_D - s_0) + \Vert r(s_0) - D \Vert}{\Vert r(s_0) - D \Vert}(r(s_0) - D) = 0, \tag{21}$
which in turn implies
$(s_D - s_0) + \Vert r(s_0) - D \Vert = 0, \tag{22}$
or
$s_D = s_0 - \Vert r(s_0) - D \Vert; \tag{23}$
this shows the line $r(s)$ passes through the two points $r(s_0)$ and $D$.  Finally, from (18),
$r'(s) = \dfrac{r(s_0) - D}{\Vert r(s_0) - D \Vert}, \tag{24}$
and thus
$\Vert r'(s) \Vert = \dfrac{\Vert r(s_0) - D \Vert}{\Vert r(s_0) - D \Vert} = 1, \tag{25}$
consistent with our assumption that $s$ is the arc-length of $r(s)$ and that $r'(s) = T(s)$ the unit tangent vector to $r(s)$.
I believe the case $c(s) < 0$ is quite similar but don't have time to work out the details at the moment.  Maybe in a little while . . . 
So concludes our derivation and discussion of the curve $r(s)$.
In closing, we note that the a priori unknown factor $c(s)$ must be included in (2) to accomodate the possibility that, in general $\Vert r(s) - D \Vert \ne 1$ and is varying with $s$.  The essence of the solution is then the determination of $c(s)$.
