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