Question

PLEASE HELP ME. I NEED YOU TO EXPLAIN
Find the curvature of r(t)= at the point (1,0,0). How do I know which of that points I need to use? The correct answer must be t=1 but why???

Answer 1

The curvature \(\kappa\) of a vector function \(\vec{r}(t)\) is given by
\[\kappa=\frac{\left\|\vec{r}'(t)\times\vec{r}''(t)\right\|}{\left\|\vec{r}'(t)\right\|}\]
You have
\[\begin{cases}
\vec{r}(t)=\left\langle t^2,\ln t,t\ln t\right\rangle\\\\
\vec{r}'(t)=\left\langle 2t,\dfrac{1}{t},\ln t+1\right\rangle\\\\
\vec{r}(t)=\left\langle 2,-\dfrac{1}{t^2},\dfrac{1}{t}\right\rangle
\end{cases}\]
with
\[\begin{align*}\vec{r}'(t)\times\vec{r}''(t)&=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\
2t&\frac{1}{t}&\ln t+1\\
2&-\frac{1}{t^2}&\frac{1}{t}\end{vmatrix}\\\\
&=\begin{vmatrix}\frac{1}{t}&\ln t+1\\-\frac{1}{t^2}&\frac{1}{t}\end{vmatrix}\vec{i}-\begin{vmatrix}2t&\ln t+1\\
2&\frac{1}{t}\end{vmatrix}\vec{j}+\begin{vmatrix}2t&\frac{1}{t}\\
2&-\frac{1}{t^2}\end{vmatrix}\vec{k}\\\\

&=\left\langle\frac{2+\ln t}{t^2},2\ln t,-\frac{4}{t}\right\rangle
\end{align*}\]
and so on.

Answer 2

As for which value of \(t\) to use: you're asked to find the curvature at the point \((1,0,0)\), which you can use to find \(t\). What value of \(t\) gives
\[\begin{cases}
t^2=1\\
\ln t=0\\
t\ln t=0
\end{cases}~~?\]