Question

How do you find the points of horizontal tangency of `r=asinthetacos^2theta`?

Answer 1

Horizontal tangents occur when `dy/dx=0`.

For polar equations, `dy/dx=(dy//d theta)/(dx//d theta)` where `x=rcostheta` and `r=sintheta`. Then, `dy/dx=(d/(d theta)rsintheta)/(d/(d theta)rcostheta)`.

So, horizontal tangents occur when `dy/dx=0`, which is the same as when `dy/(d theta)=0`, or when `d/(d theta)rsintheta=0`.

Here `r=asinthetacostheta`, so `y=rsintheta=asin^2thetacos^2theta`.

We can simplify this by noting that `y=1/4a(4sin^2thetacos^2theta)=1/4a(2sinthetacostheta)^2=1/4asin^2 2theta`.

Differentiating this with the chain rule, we see that `dy/(d theta)=1/4a(2sin^1 2theta)(cos2theta)(2)=asin2thetacos2theta`.

`dy/(d theta)=0` then when either `sin2theta=0` or `cos2theta=0`, which occur at `theta=(kpi)/4,kinZZ`.