Finding curvature for $y = \sin ( -2x )$ at $x =\pi/4$ The answer is 4, but I got 1. I said $r(t)=&lt;t,\sin(-2t)&gt;$ 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.
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.
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
$$\left|\frac{y''}{(1+(y')^2)^{3/2}}\right|=\left|\frac{4}{(1+0^2)^{3/2}}\right|=\color{blue}{4}$$