MHB Bebe's question at Yahoo Answers (Curv. and torsion)

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The discussion centers on computing the curvature and torsion of the curve r(t) = <sinh t, cosh t, t> at the point (0,1,0). The first derivatives at t=0 yield values necessary for these calculations, specifically, the first derivative is (1, 0, 1) and the second derivative is (0, 1, 0). The curvature at this point is calculated to be 1/2 using the cross product of the first two derivatives. The torsion can be computed using the third derivative and the previously calculated derivatives, though the exact value is not provided in the discussion. The computations for both curvature and torsion are straightforward and follow established formulas.
Fernando Revilla
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Here is the question:

Consider the curve r(t)=<sinh t, cosh t, t>, where sinh t= (e^t- e^-t)/2 and cosh t= (e^t+ e^-t)/2. Compute the curvature and torsion of r(t) at the point (0,1,0).
[Hint: it may be helpful to know that sinh^2 (t) +1= cosh^2 (t) for all t]

Here is a link to the question:

Consider the curve r(t)=<sinh t, cosh t, t>? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Bebe,

We have:

$$\begin{aligned}&\vec{r}(t)=(\sinh t,\cosh t,t) \Rightarrow\vec{r}(0)=(0,1,0)\\&\frac{d\vec{r}}{dt}=\left (\cosh t,\sinh t,1\right)\Rightarrow\frac{d\vec{r}}{ dt }(0)=\left(1, 0,1\right)\\&\frac{d^2\vec{r}}{dt^2}=\left(\sinh t,\cosh t,0\right)\Rightarrow \frac{d^2\vec{r}}{dt^2}(0)=(0,1,0)\\&\frac{d^3\vec{r}}{dt^3}=\left(\cosh t,\sinh t,0\right)\Rightarrow \frac{d^3\vec{r}}{dt^3}(0)=(1,0,0)\end{aligned}$$ Using a well-known formula, the curvature at $(0,1,0)$ is: $$\kappa (0)=\dfrac{\left |\dfrac{d\vec{r}}{dt}(0)\times \dfrac{d^2\vec{r}}{dt^2}(0)\right |}{\left |\dfrac{d\vec{r}}{dt}(0)\right |^3}=\dfrac{\left |(1,0,1)\times (0,1,0)\right |}{\left |(1,0,1)\right |^3}=\ldots=\dfrac{1}{2}$$ The torsion is: $$\tau (0)=\dfrac{\left[\dfrac{d\vec{r}}{dt}(0),\dfrac{d^2\vec{r}}{dt^2}(0),\dfrac{d^3\vec{r}}{dt^3}(0)\right]}{\left(\dfrac{d\vec{r}}{dt}(0)\times \dfrac{d^2\vec{r}}{dt^2}(0)\right)^2}=\ldots$$ Easily you can complete the computations.
 
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