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The question is to find the curvature of the curve $r(t)=\langle t^2, \ln{t}, t\ln{t}\rangle$ at point $(1, 0, 0).$

I've found $r^{\prime} = \langle 2t, 1/t, \ln{t} + 1 \rangle$ and $r^{\prime\prime} = \langle 2, -t^{-2}, 1/t \rangle$ and got $$\lvert r^{\prime}\rvert =\sqrt{4t^2 +1/t^2 +\ln^{2}{t} + 2\ln{t} + 1}$$

The cross product I got use in for $$\frac{|r^{\prime}\times r^{\prime\prime}|}{|r^{\prime}|}$$ wasn't much less of a complex mess to deal with: $$(1/4 - ( \ln{t} + 1)(1/2\sqrt{t}))\mathbf{i} - (t-2\ln{t}-2)\mathbf{j} + (2t/2\sqrt{t}-1)\mathbf{k}$$

Am I doing these right? I don't see anything you could do with all this. The answer is supposed to be $\frac{1}{7}\sqrt{\frac{19}{14}}.$

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    $\begingroup$ I have edited your typesetting (using \langle and \rangle instead of < and > makes things much more readable). I have also replaced an $x$ with a $\times.$ I was unsure whether or not to simplify the $2t/2\sqrt{t}$ to just $\sqrt{t}$ though; did you mean something like $\frac{2t}{2\sqrt{t}-1}?$ Similarly, it is a little unclear whether $1/2\sqrt{t}$ is meant to be $\frac{1}{2}\sqrt{t}$ or $\frac{1}{2\sqrt{t}}.$ Regardless, you should remember that you aren't interested in the curvature in general; you only care about what's going on at $t=1.$ $\endgroup$ – Will R Sep 23 '16 at 3:41
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    $\begingroup$ Also, I think the formula for curvature that you are thinking of is $$\frac{\lvert r^{\prime}\times r^{\prime\prime}\rvert}{\lvert r^{\prime} \rvert^{3}}.$$ Are you sure the answer is supposed to be $\frac{1}{7}\sqrt{\frac{19}{14}}?$ $\endgroup$ – Will R Sep 23 '16 at 3:56
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    $\begingroup$ Double checked, looked at the wrong problem's answer, you're right, that is incorrect. $\endgroup$ – windy401 Sep 23 '16 at 4:18
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Starting from $r(t) = (t^{2},\ln{t},t\ln{t}),$ we have $$r^{\prime}(t) = (2t,1/t,1+\ln{t})\quad\text{and}\quad r^{\prime\prime}(t) = (2,-1/t^{2},1/t).$$

Hence, $$r^{\prime}(1) = (2,1,1)\quad\text{and}\quad r^{\prime\prime}(1) = (2,-1,1).$$ Now we have $$\lvert r^{\prime}(1)\times r^{\prime\prime}(1) \rvert = \lvert (2,0,-4) \rvert = \sqrt{20}$$ and $$\lvert r^{\prime}(1) \rvert = \lvert (2,1,1) \rvert = \sqrt{6}.$$ Therefore the curvature at $(1,0,0)$ appears to be $$\frac{\lvert r^{\prime}(1)\times r^{\prime\prime}(1) \rvert}{\lvert r^{\prime}(1) \rvert^{3}} = \frac{\sqrt{30}}{18}.$$

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  • $\begingroup$ So when finding curvature given a vector and a point you just plug in the x value if the point given as soon as you get the derivatives? $\endgroup$ – windy401 Sep 23 '16 at 4:20
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    $\begingroup$ @windy401: In this case, there's no reason not to do so. If the question were asking you for a general expression for $\kappa(t)$ then, of course, there would be no specific value of $t$ to plug-in and you might have some tedious computations to do. But yes, if you're given a specific point, like $(1,0,0),$ then plugging in the corresponding $t$ value as soon as possible (so, once you're done differentiating) can only simplify calculations and make life easier. $\endgroup$ – Will R Sep 23 '16 at 4:24
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    $\begingroup$ Oh, I seen now, it's the t that gives the points when put in the original r(t). Ok. Thanks. $\endgroup$ – windy401 Sep 23 '16 at 4:25

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