Finding curvature for $y = \sin ( -2x )$ at $x =\pi/4$

Question

The answer is 4, but I got 1. I said $r(t)=<t,\sin(-2t)>$ and the $|r^\prime(\frac{\pi}{4})|$ is equal to 1. I also got 1 for $T^\prime(\frac{\pi}{4})$ but none of this information matters because I got the question wrong, can someone help me out please? thanks. Also, ideally, use the $\dfrac{|T^\prime|}{|r^\prime|}$ formula.

Perhaps, I shouldn't have translated it into a rectangular form? But I have only been taught how to compute curvature using a rectangular form.

Answer 1

Your function can be parameterized as:

$${\displaystyle \mathbf{r}(t)\,=\,{\begin{pmatrix}t\\\sin(-2t)\end{pmatrix}}}$$

The curvature is given by :

$$\kappa = \frac{\|\mathbf{r}'(t)\times\mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}$$

$${\displaystyle \mathbf{r}'(t)\,=\,{\begin{pmatrix}1\\-2\cos(-2t)\end{pmatrix}}}$$ $${\displaystyle \mathbf{r}''(t)\,=\,{\begin{pmatrix}0\\-4\sin(-2t)\end{pmatrix}}}$$

Setting $x \mapsto \frac{\pi}{4}$ follows:

$${\displaystyle \mathbf{r}'(\frac{\pi}{4})\,=\,{\begin{pmatrix}1\\0\end{pmatrix}}}$$ $${\displaystyle \mathbf{r}''(\frac{\pi}{4})\,=\,{\begin{pmatrix}0\\4\end{pmatrix}}}$$ Now use the formula:

$$\kappa = \frac{\|\mathbf{r}'(\frac{\pi}{4})\times\mathbf{r}''(\frac{\pi}{4})\|}{\|\mathbf{r}'(\frac{\pi}{4})\|^3}=\frac{\left\|\,{\begin{pmatrix}0\\0\\4\end{pmatrix}} \right\|}{\left(0^{2}+1^{2}\right)^{\frac{3}{2}}}=4$$

Answer 2

Alternatively, you can easily compute the curvature of $y=\sin(-2x)$ at $x=\pi/4$ as follows

$y=\sin (-2x)=-\sin(2x)$, $\ y'=-2\cos(2x)$, $\ y''=4\sin(2x)$

Substituting $x=\pi/4$, we get $y'(\pi/4)=0, \ y''(\pi/4)=4$

The curvature of $y=f(x)$ is given by $$\left|\frac{y''}{(1+(y')^2)^{3/2}}\right|=\left|\frac{4}{(1+0^2)^{3/2}}\right|=\color{blue}{4}$$

Answer 3

Straightforward. Successive derivatives are $$y'= - 2 \cos 2x,\; y''= 4 \sin 2x = 4 \quad @ \;x=\frac{\pi}{4}$$

because the denominator in the curvature formula

$$ \dfrac{y''}{(1+y^{'2})^{1.5}}$$

is unity.