Find the family of curves r=f(varphi) for which the area of…
Find the family of curves $r=f(\varphi)$ for which the area of the sector bounded by the curve and the radii from a fixed point $(r_0,\varphi_0)$ to a variable point $(r,\varphi)$ is proportional to the product $r\varphi$, with proportionality coefficient $k$.
Explanation of Method
Differentiate the area condition $A=kr\varphi$ using $A=\frac12\int r^2\,d\theta$ to obtain a Bernoulli equation in $r(\varphi)$, then substitute $v=1/r$.
Step 1: Differentiate the area condition
From $\tfrac12\int_{\varphi_0}^\varphi r^2\,d\theta=kr\varphi$, differentiating both sides (Leibniz + product rule):
\[\frac12 r^2=k\varphi r'+kr\quad\Longrightarrow\quad r'+\frac{r}{\varphi}=\frac{r^2}{2k\varphi}.\]
Step 2: Substitute $v=1/r$
Bernoulli with $n=2$: $v'-\dfrac{v}{\varphi}=-\dfrac{1}{2k\varphi}$.
Step 3: Solve the linear equation
With $\mu(\varphi)=1/\varphi$: $\dfrac{d}{d\varphi}\!\left(\dfrac{v}{\varphi}\right)=-\dfrac{1}{2k\varphi^2}$, so $\dfrac{v}{\varphi}=\dfrac{1}{2k\varphi}+C$, i.e. $v=\dfrac{1}{2k}+C\varphi$.
Result
Final answer:
\[\boxed{\,r=\frac{2k}{1+C\varphi},\qquad C\in\mathbb{R}}\]
a family of hyperbolic spirals ($C=0$ gives the circle $r=2k$).