Find the equation of the normal to y = 1/sin(2x) at the point where x = pi/4

Guest Aug 11, 2017

#1**0 **

\(x=\dfrac{\pi}{4}\\ \therefore y = \dfrac{1}{\sin\left(2\cdot\dfrac{\pi}{4}\right)}=1\\ \dfrac{dy}{dx}=(\csc(2x))'=-2\csc(2x)\cot(2x)\\ \dfrac{dy}{dx}|_{x=\frac{\pi}{4}}=-2(1)(0)=0\\ y-1=0(x-\dfrac{\pi}{4})\\ y=1\\ \therefore\text{The equation of the normal to }y=\dfrac{1}{\sin(2x)}\text{ when }x=\pi/4\text{ is }\boxed{y=1}\)

I think this is the first time I do an application question :P

MaxWong
Aug 11, 2017

#5**+2 **

Find the equation of the normal to y = 1/sin(2x) at the point where x = pi/4

We can write this as

y = [sin(2x) ]^{-1}

When x = pi/4, y = 1 / sin (2 (pi/4)) = 1 / sin (pi/2) = 1

y ' = -2 [ sin (2x)]-2 * [ cos(2x)] = [-2 cos (2x)] / [ sin (2x)] ^{2} = -2cot (2x)csc(2x)

At pi/4....the slope of the tangent line = -2 [ cot (2(pi/4) ) csc (2(pi/4) )=

-2 [ cot (pi/2)] [csc(pi/2)] = 0

So the equation of the tangent line is ... y - 1 = 0 ( x - pi/4) → y = 1

So...the slope of a line normal to this will be undefined.....and the equation of the line will be x = pi/4

Here's a graph to show this : https://www.desmos.com/calculator/qyxqzuupbs

CPhill
Aug 11, 2017