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

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The discussion centers on finding the curvature of the curve defined by r(t) = t^2 i + ln(t) j + t ln(t) k at the point (1,0,0). The first derivative of the curve at t=1 is calculated as (2,1,1), and the second derivative is (2,-1,1). Using the curvature formula, the cross product of the derivatives is computed to find the curvature at the specified point. The calculation involves determining the magnitudes of the vectors involved. The final curvature value is derived from these computations.
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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: $$\kappa (1)=\dfrac{\left |\dfrac{d\vec{r}}{dt}(1)\times \dfrac{d^2\vec{r}}{dt^2}(1)\right |}{\left |\dfrac{d\vec{r}}{dt}(1)\right |^3}=\dfrac{\left |(2,1,1)\times (2,-1,1)\right |}{\left |(2,-1,1)\right |^3}=\ldots $$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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