Find the slope of the tangent line to the given polar curve #r=7sintheta, theta=pi/6#?

Find the slope of the tangent line to the given polar curve #r=7sintheta, theta=pi/6#

1 Answer

#\sqrt3#

Explanation:

The given curve: #r=7\sin\theta# represents a circle with center at #(0, 7/2)# & passing through the origin #(0, 0)# whose equation in cartesian coordinates is also given as

#(x-0)^2+(y-7/2)^2=(7/2)^2#

#x^2+y^2-7y=0#

The point of intersection of circle & the line #y=x\tan(\pi/6)=x/\sqrt3#

#x^2+(x/\sqrt3)^2-7(x/\sqrt3)=0#

#4x^2-7\sqrt3x=0#

#x=0, {7\sqrt3}/4#

#y=0, 7/4#

Thus, the points of intersection are #(0, 0)# & #({7\sqrt3}/4, 7/4)#

Slope of tangent to the circle: #x^2+y^2-7y=0#

#\frac{d}{dx}(x^2+y^2-7y)=0#

#2x+2yy'-7y'=0#

#y'=\frac{2x}{7-2y}#

hence, the slope of tangent at the point #({7\sqrt3}/4, 7/4)# to the circle

#y'=(\frac{2\cdot {7\sqrt3}/4}{7-2\cdot 7/4})#

#=\sqrt3#