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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: $$\kappa (1)=\dfrac{\left |\dfrac{d\vec{r}}{dt}(1)\times \dfrac{d^2\vec{r}}{dt^2}(1)\right |}{\left |\dfrac{d\vec{r}}{dt}(1)\right |^3}=\dfrac{\left |(2,1,1)\times (2,-1,1)\right |}{\left |(2,-1,1)\right |^3}=\ldots $$
 
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