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

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The curvature of the curve defined by the vector function r(t) = t² i + ln(t) j + t ln(t) k at the point (1,0,0) is calculated using the formula κ(1) = |(dr/dt)(1) × (d²r/dt²)(1)| / |(dr/dt)(1)|³. The first derivative at t=1 is (2,1,1) and the second derivative is (2,-1,1). The cross product of these vectors is computed to find the curvature value, which is a critical aspect of understanding the curve's geometric properties.

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Here is the question:

find the curvature of the curve r(t)= t^2 i + ln(t) j + t ln(t) K at the point (1,0,0)

Here is a link to the question:

Find the curvature of the curve? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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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: $$\kappa (1)=\dfrac{\left |\dfrac{d\vec{r}}{dt}(1)\times \dfrac{d^2\vec{r}}{dt^2}(1)\right |}{\left |\dfrac{d\vec{r}}{dt}(1)\right |^3}=\dfrac{\left |(2,1,1)\times (2,-1,1)\right |}{\left |(2,-1,1)\right |^3}=\ldots $$
 

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