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