Usage of implicit differentiation in this case

Question

I have been doing one exercise problem, and there is something unclear about it. Namely, the 1st approach and its solution are clear, but the implicit way is unclear. I am uncertain how did they get the equations marked by the red rectangle, how did they get it by using implicit differentiation?

Answer 1

First, compute the derivative of $x\left(r, \theta\right)=r\cos\theta$ with respect to $\theta$ (treating left-hand side as an independent variable, not as a function depending on $\theta$):

• First, let us differentiate polar form expression for $x$ with respect to $y$: $$ \begin{aligned} x&=r\cos\theta & \implies && \dfrac{\partial}{\partial y}\left[x\right] &= \dfrac{\partial}{\partial y} \left[r\cos\theta\right] = \dfrac{\partial r}{\partial y}\cos\theta -r\sin\theta \cdot \dfrac{\partial \theta}{\partial y} \\ && && \dfrac{\partial x}{\partial y} &= r_y\cos\theta - r\sin\theta\cdot \theta_y \\ && && 0 &= r_y\cos\theta - r\sin\theta\cdot\theta_y \\ && \implies && &\boxed{\theta_y = \dfrac{r_y\cos\theta}{r\sin\theta}=\dfrac{\frac{r_y}{r}x}{y} = \dfrac{r_y}{r}\cdot\dfrac{x}{y}} \end{aligned} $$
• Second, differentiate $r^2 = x^2+y^2$ with respect to $y$: $$ \begin{aligned} r^2 &= x^2+y^2 &\implies &&\dfrac{\partial}{\partial y} \left[r^2 \right]&=\dfrac{\partial}{\partial y} \left[x^2+y^2\right] = 2\cdot 0 + 2y \\ && && 2rr_y &= 0+2y \\ &&\implies && &\boxed{r_y = \dfrac{y}{r}} \end{aligned} $$
• Third, combine two above results with the formula $x^2+y^2=r^2$ to get the final answer:

$$ \theta_y = \dfrac{r_y}{r}\cdot\dfrac{x}{y} = \dfrac{\frac{y}{r}}{r}\cdot\dfrac{x}{y} = \dfrac{x}{r^2} $$

Thus we finally get

$$ \boxed{\theta_y = \dfrac{x}{x^2+y^2}} $$ $\hspace{64ex}$Q.E.D.