Simmons 7.10 & 7.11: Find Curves Intersecting at Angle pi/4

[yucheng]

Homework Statement

N/A

Relevant Equations

N/A

>10. Let a family of curves be integral curves of a differential equation $y^{\prime}=f(x, y) .$ Let a second family have the property that at each point $P=(x, y)$ the angle from the curve of the first family through $P$ to the curve of the second family through $P$ is $\alpha .$ Show that the curves of the second family are solutions of the differential equation
$$
y^{\prime}=\frac{f(x, y)+\tan \alpha}{1-f(x, y) \tan \alpha}
$$
>11. Use the result of the preceding problem to find the curves that form the angle $\pi / 4$ with

>(a) all straight lines through the origin;

>(b) all circles $x^{2}+y^{2}=c^{2}$

>(Simmons, problem 7.10 and 7.11)

I was trying to work out problem 11. The solutions given are
11.(a) $r=c e^{\theta}$
11.(b) $r=c e^{-\theta}$

My solution:
11.
$y=mx$, $f(x,y)=m$, $\tan \frac{\pi}{4}=1$,thus $$\frac{dy}{dx}=\frac{m+1}{1-m}=C_1$$
$y=r \sin\theta$, $x=r \cos\theta$, $dy= \sin\theta dr$, $dx=-r \sin\theta d\theta$, substituting into our differential equation,

\begin{align}
\frac{dy}{dx} &= C_1\\
dy &= C_1 dx\\
\sin\theta dr &= C_1 (-r \sin\theta d\theta)\\
\frac{1}{r} dr &= C_2 d\theta\\
\log r &= C_2 \theta\\
r &= Ce^{\theta} \blacksquare\\
\end{align}

>Edit:
I realized I made an illegal move: it should be $\log r = C_2 \theta +C_3$, thus $r=Ce^{C_2\theta}$, which is umm...

12.
$x^2+y^2=c^2$, $r=c$, $y=r \sin\theta$, $x=r \cos\theta$
\begin{align}
\frac{dy}{dx}&=-\frac{x}{\sqrt{c^2-x^2}}\\
&=-\frac{\cos\theta}{\sin\theta}\\
\end{align}

Substituing into the equation from problem 10,

\begin{align}
\frac{dy}{dx}&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}\\
\frac{dy}{dx}&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}\\
dy&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}dx\\
dr&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}(-r)d\theta\\
\end{align}

How am I supposed to continue? Can I pull off the same trick as before? Thanks in advance.

P.S. I read this post [Express the differential equation that solves the below problem in polar form](https://math.stackexchange.com/questions/909582/express-the-differential-equation-that-solves-the-below-problem-in-polar-form)

 

[pasmith]

Your starting point is $$\frac{dy}{d\theta} = \frac{1 + f(x,y)}{1 - f(x,y)} \frac{dx}{d\theta}$$ where the chain rule gives $$\begin{align*} \frac{dx}{d\theta} &= \frac{dr}{d\theta} \cos \theta - r \sin \theta, \\ \frac{dy}{d\theta} &= \frac{dr}{d\theta}\sin \theta + r\cos \theta. \end{align*}$$

For part (a): The straight line through the origin which passes through $(x,y)$ has $f(x,y) = \frac{y}{x}$.

For part (b): You already found $f(x,y) = -\cot \theta$. Proceed from here with the above expressions.