What is the equation of the line that is normal to the polar curve $f(theta)=sin(2theta+pi) -theta$ at $theta = pi/4$ ?

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Answer 1 · 2017-05-19T07:56:59

$(y-(-4sqrt(2)-pi sqrt(2))/8)=(pi/(8+pi))(x-(-4sqrt(2)-pi sqrt(2))/8)$

Find $r$ when $theta=pi/4$
$r=sin(2(pi/4)+pi) -theta=(-4-pi)/4$

Find $(dr) /(d theta):$
$r=sin(2 theta + pi) -theta$
$(dr) /(d theta)=cos(2 theta + pi) (2)-1=2cos(2 theta + pi) -1$

Substitute $theta=pi/4$ :
$=2cos(2(pi/4)+pi)-1=-1$

Find the derivative:
$dy/(dx)=((-1)sin (pi/4) +((-4-pi)/4)cos (pi/4))/((-1)cos(pi/4) -((-4-pi)/4) sin (pi/4))=-8/pi-1$

Find the respective $x$ and $y$ .
$x=rcos(theta)=((-4-pi)/4) (sqrt(2)/2)=(-4sqrt(2)-pi sqrt(2))/8$

$y=rsin(theta)=((-4-pi)/4) (sqrt(2)/2)=(-4sqrt(2)-pi sqrt(2))/8$

The slope of a normal line is the negative reciprocal of the derivative.
$-1/(-8/pi-1)=pi/(8+pi)$

Putting it all together:
$(y-y_1)=m(x-x_1)$
$(y-(-4sqrt(2)-pi sqrt(2))/8)=(pi/(8+pi))(x-(-4sqrt(2)-pi sqrt(2))/8)$

What is the equation of the line that is normal to the polar curve f(theta)=sin(2theta+pi) -thetaf(theta)=sin(2theta+pi) -theta at theta = pi/4theta = pi/4?