Deduce the equation of the tangent line to the curve defined by
the equations x=cosh(t), y=sinh(t), and z=ct
I have somewhat of a good grip on the definition of a tangent line, but the lack of a given point is throwing me off compared to other examples I have looked at. If
r(t)=(cosh(t), sinh(t), ct)
r'(t)=(sinh(t), cosh(t), c)
Then would the line be given by
x=cosh(t)+t sinh(t), y= sinh(t)+t cosh(t), z=2ct ?
If so how do I reduce it to a single equation as is implied by the question? There is always solving for the t (outside of the hyperbolic function part) and setting all three equal to each other (in the canonical form) but there is no guarantee that the denominator in that form would not be 0 without a given point.
Thanks for the help.
Look, gang, I must be missing something serious, or have a critical screw loose, because I am having an inordinately hard time getting the gist as stated of the seemingly straightforward question. Seriously, I kid you not. So if what I say below is off the mark, please set me straight!
Not too put too fine a point on it, but I believe that, despite NasuSama's comment, the given formulas
$x = \cosh t + t \sinh t; \; y = \sinh t + t \cosh t; \; z = 2ct \tag{1}$
do not describe any tangent line to the curve
$\mathbf r(t) = (\cosh t, \sinh t, ct), \tag{2}$
simply because (1) is not the equation of any line. But before going any further, I would like to say that we can obtain a solution with one equation if we allow it to be a vector equation. I'll take this as the intent of the question, and show how to derive a vector equation for the tangent line to the given curve through any point this curve.
For any $t_0 \in \Bbb R$, the point $\mathbf r(t_0)$ given by taking $t = t_0$ in (2) is a point on the curve; the tangent vector to the curve at this point is clearly
$\mathbf r'(t_0) = (\sinh t_0, \cosh t_0, c). \tag{3}$
The tangent line to this curve at the point given by $t = t_0$ is, as I am given to understand it, the line through the point $\mathbf r(t) = (\cosh t_0, \sinh t_0, ct_0)$ whose tangent vector is given by (3). To write an equation for the points on such a line, we need to introduce a second variable $r \in \Bbb R$ which parametrizes that line along its extent, just a $t$ parametrizes a given point on such a line via (2). If $\mathbf l$ is a point on the line, then the vector $\mathbf l - \mathbf r(t_0)$ must be collinear with $\mathbf r'(t_0)$, whence
$\mathbf l - \mathbf r(t_0) = r \mathbf r'(t_0) \tag{4}$
for some $r \in \Bbb R$, whence we can write the vector equation of the line as
$\mathbf l(r) = \mathbf r(t_0) + r \mathbf r'(t_0) = (\cosh t_0, \sinh t_0, ct_0) + r (\sinh t_0, \cosh t_0, c), \tag{5}$
which depends on two parameters, $t_0$ for the point on the curve, and $r$ for the point on the resulting line. As $t_0, r$ vary over $\Bbb R$, (5) describes all points on all lines tangent to the curve $\mathbf r(t)$; holding $t_0$ fixed, we obtain all points on a given tangent line.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!