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Mr.Ask's question at Yahoo! Answers (curvature)

Fernando Revilla

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

Fernando Revilla

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

We have: $$\begin{aligned}&\vec{r}(t)=(t^2,\log t,t\log t)\Rightarrow\vec{r}(1)=(1,0,0)\\&\frac{d\vec{r}}{dt}=\left(2t,\dfrac{1}{t},1+\log t\right)\Rightarrow\frac{d\vec{r}}{dt}(1)=\left(2,1,1\right)\\&\frac{d^2\vec{r}}{dt^2}=\left(2,-\dfrac{1}{t^2},\dfrac{1}{t}\right)\Rightarrow \frac{d^2\vec{r}}{dt^2}(1)=\left(2,-1,1\right)&\end{aligned}$$ Using a well-known formula, the curvature at $(1,0,0)$ is: $$\kappa (1)=\dfrac{\left |\dfrac{d\vec{r}}{dt}(1)\times \dfrac{d^2\vec{r}}{dt^2}(1)\right |}{\left |\dfrac{d\vec{r}}{dt}(1)\right |^3}=\dfrac{\left |(2,1,1)\times (2,-1,1)\right |}{\left |(2,-1,1)\right |^3}=\ldots $$