I have a task which says the following:

A planar curve is given by $$x=\cos(t)+t,\\y=t^2+2t+1.$$ I had to calculate for which value of the parameter $t$ does the curve pass through the point $P=(1,1)$, which i calculated to being $0$.

The next question is: What is the curvature of the curve at the point P?

With the different possible answers: a)$\frac{1}{\sqrt{5}}$, b)$\frac{4}{5\sqrt{5}}$, c)$\frac{4}{{5}}$, d)$\frac{3}{5\sqrt{5}}$, e)$\frac{5}{3\sqrt{5}}$.

Bonus info: Results list says the correct answer is b.

To solve this I have tried using the theory connected to determining said curvature, which says if we have a curve: $r(t)=\langle x,y\rangle$, the curvature is found using kappa:

$$\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|} \tag{1}\label{kappa}$$, where $\vec T(t)=\frac{\vec r'(t)}{\|\vec r'(t)\|} \tag{2}\label{Toft}.$

So I tried differenciating $r(t)$ and let $t=0$ so I could insert into equation (\ref{Toft}):

$r(t)= \langle \cos(t)+t,t^2+2t+1 \rangle$

$r'(t)= \langle -\sin(t)+1,2t+2 \rangle$

$r'(0)= \langle -\sin(0)+1 , 2\cdot0+2 \rangle \rightarrow \langle 1,2 \rangle$

$\vec T(t)=\frac{\langle 1,2 \rangle}{\sqrt{1^2 + 2^2}} \rightarrow \frac{\langle 1,2 \rangle}{\sqrt{5}} \rightarrow \Bigl\langle \frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}} \Bigr\rangle$

I then calculated $\|\vec T'(0)\|$ so it could be inserted into equation (\ref{kappa})

$\|\vec T'(0)\| = \sqrt{\Bigl(\frac{1}{\sqrt{5}}\Bigr)^2 + \Bigl(\frac{2}{\sqrt{5}}\Bigr)^2} \rightarrow \sqrt{\frac{1^2}{\sqrt{5}^2} +\frac{2^2}{\sqrt{5}^2} } \rightarrow \sqrt{ \frac{1}{5} + \frac{4}{5}} \rightarrow \sqrt{\frac{5}{5}} \rightarrow \sqrt{1} = 1 $

Finally I insert into (\ref{kappa}): $$\kappa = \frac{1}{\sqrt{5}}$$ see that I get one of the possible answers, then check the results list and see that I am getting the wrong result.. Anyone who knows what I am doing wrong and possibly can help me out? Thanks in advance =)


The correct answer is $\frac1{4\sqrt5}$, since\begin{align}\kappa(0)&=\frac{|x'(0)y''(0)-x''(0)y'(0)|}{\sqrt{x'(0)^2+y'(0)^2}^3}\\&=\left.\frac{-2 (1 + t) \cos(t) - 2 (1 - \sin(t))}{\bigl(4(1 + t)^2 +(1 -\sin(t))^2\bigr)^{\frac32}}\right|_{t=0}\\&=\frac1{4\sqrt5}.\end{align}

What you are doing wrong is that you computed $\vec T$ only at $0$ and, with that information alone, you cannot possibly compute $\vec T'(0)$.

  • $\begingroup$ Arh I see, thank you +1 $\endgroup$
    – JayFreddy
    Jul 23 '18 at 10:14
  • $\begingroup$ Around here, when we write “+1” in a comment, that means that we are upvoting the answer, but you didn't do that. $\endgroup$ Jul 23 '18 at 10:16
  • $\begingroup$ Yeah, I would, but I can't because I have less than 15 reputation.. sry 'bout that :/ $\endgroup$
    – JayFreddy
    Jul 23 '18 at 10:52
  • $\begingroup$ No problem, but then you shouldn't have written “+1” in your comment. $\endgroup$ Jul 23 '18 at 11:02

$1 = t^2 + 2t + 1 \iff t = 0$ or $t = -2$


$x'(t) = -\sin t + 1$, $x''(t) = -\cos t$, $y'(t) = 2t + 2$, $y''(t) = 2$,

and one value is

$ k(0) = \left| \dfrac{ x'(0)y''(0) - y'(0)x''(0)}{( x'^2(0) + y'^2(0) )^\frac{3}{2}} \right| = \left| \dfrac{ 1\cdot 2 - 2 \cdot (-1)}{( 1 + 4 )^\frac{3}{2}} \right| = \frac{4}{5\sqrt{5}} $

and another value

$ k(-2) \overset{y'(-2) = 0}{=}\left| \dfrac{ y''(-2)}{x'^2(-2)} \right|= \dfrac{2}{(-\sin (-2) + 1)^2} $

is invalid because $\cos(-2) - 2 \neq 1$



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