# Thread: Consider the the cardioid

1. ## Consider the the cardioid

Consider the the cardioid r=1+sin(θ)

find dy/dx. I'm a little unsure where to start here, I'm guessing convert it which, I got x^2 + y^2= (1+sin(θ))^2, am I going in the right direction? Any guidance would be appreciated. Thanks

2. OK so I did a little more work, and figure I should times both sides by r instead of squaring it.
which gave me:

r^2= r + rsin(t)

x^2+y^2=r+y

r= (x^2+y^2-y)

r^2= (x^2+y^2-y)^2

x^2+y^2= (x^2+y^2-y)^2

But I'm still stuck on how to find the derivative of this? [dy/dx] = [dy/dt / dx/dt ]

3. You are going in the right direction!...

... because is $\displaystyle \theta = \tan^{-1} \frac{y}{x}$, y can be expressed as function of x in implicit form as...

$\displaystyle \displaystyle f(x,y)= x^{2} + y^{2} - \{ 1 + \sin (\tan^{-1} \frac{y}{x})\} ^{2} = 0$ (1)

... so that is...

$\displaystyle \displaystyle \frac{dy}{dx} = - \frac{f_{x} (x,y)}{f_{y} (x,y)}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. in which one was I going in the right direction...lol, my first or second post?

5. still stuck...any guidance would be appreciated. thanks