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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.

James Snyder
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  • Your answer is correct. – NasuSama Jan 13 '14 at 02:29
  • If that is the case, I was also wondering how I might condense it to one equation since other questions in my book ask for "equations" and this one asks for singular. – James Snyder Jan 13 '14 at 03:07
  • @James Snyder: what is the definition of "tangent line" your are using? Because I can't see for the life of me how $x = \cosh t + t; ; y = \sinh t + t; ; z = 2ct \tag{1}$ describes what I am used to calling a line. Am I missing something here? – Robert Lewis Jan 13 '14 at 06:01
  • @RobertLewis: You mean that the parametric equation $x=\cosh(t_0)+t \sinh(t), y= \sinh(t_0)+t \cosh(t), z=ct_0+ct$ is wrong here? – Mikasa Jan 13 '14 at 06:27
  • @ B.S.: to tell the truth, I'm a little confused; if you wrote $x=\cosh(t_0)+t \sinh(t_0), y= \sinh(t_0)+t \cosh(t_0), z=ct_0+ct$ then it would make more sense to me, though I personally would rather use a parameter name other than $t$ for the running parameter along the line. – Robert Lewis Jan 13 '14 at 06:37
  • *ALSO, PLEASE NOTE:* My equation (1) in my previous comment is clearly an error, but too late to edit it now! Should read more like B.S.'s equation! Or the equation in my last comment! – Robert Lewis Jan 13 '14 at 06:39
  • Bollux! Hopefully my doubts and a possible explanation are given in my answer below! – Robert Lewis Jan 13 '14 at 06:45

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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!!!

Robert Lewis
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