What is the slope of th line tangent to the polar curve r = 2(theta) at the point theta = pi/2

Somebody help me please..

To parametric
x=rcosT
y=rsinT
The position vector P is
P= rcos T i+ r sinT j

P= 2TcosT i+2TsinT j

Tangent , dP/dT = (- 2T sinT +2cosT) i + (2TcosT+2sinT) j

At T=pi/2 , cos pi/2=0 and sinpi/2=1

dP/dT = -pi i +2 j ( It is a director vector of the tangent line )

Slope m= 2/(-pi) = -(2/pi)

dr/d(theta) = 2

So we can make a tiny right triange at that point, (Theta,r) = (Pi/2, Pi):

Base = r(dTheta)
Height = r* 2dr

At that point, dr is equal to dy, so the slope is simply:

dy/dx = -2