Question

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`

Answer 1

`\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`