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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.