Show that two cardioids $r=a(1+\cos\theta)$ and $r=a(1-\cos\theta)$ are at right angles.

Question

Show that two cardioids $r=a(1+\cos\theta)$ and $r=a(1-\cos\theta)$ are at right angles.

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$\frac{dr}{d\theta}=-a\sin\theta$ for the first curve and $\frac{dr}{d\theta}=a\sin\theta$ for the second curve but i dont know how to prove them perpendicular.

Answer 1

Since $(x,y)=(r\cos\theta,r\sin\theta)$, the 2 curves can be given parametrically as

\begin{align} \vec{r}_1 &= \big(a(1+\cos\theta)\cos\theta,a(1+\cos\theta)\sin\theta\big) \\ \vec{r}_2 &= \big(a(1-\cos\theta)\cos\theta,a(1-\cos\theta)\sin\theta\big) \end{align}

Their tangent vectors are

\begin{align} \vec{r}_1' &= \big(a(-\sin\theta - \sin2\theta),a(\cos\theta + \cos2\theta) \big) \\ \vec{r}_2' &= \big(a(-\sin\theta + \sin2\theta),a(\cos\theta - \cos2\theta) \big) \end{align}

Then $$ \vec{r}_1' \cdot \vec{r}_2' = a^2(\sin^2\theta - \sin^2 2\theta) + a^2(\cos^2\theta - \cos^2 2\theta) = 0 $$

i.e. $\vec{r}_1'\perp \vec{r}_2' $

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Alternatively, you can use

$$ \frac{dy}{dx} = \frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}} = \frac{\frac{dr}{d\theta}\cos\theta - r\sin\theta}{\frac{dr}{d\theta}\sin\theta + r\cos\theta} $$

to compute the tangent slopes that way

Answer 2

Solution 1 : take a look at the following figure :

Representing the two cardioids we are working on and 2 circles described by $z=\frac12(1+e^{i \theta})$ and $z=\frac12(i+ie^{i \theta})$.

The 2 circles are orthogonal ; the cardioids being the images of these circles by conformal transformation $z \to z^2$ (red circle gives magenta cardioid, cyan circle gives blue cardioid), the orthogonality is preserved.

Solution 2 : Sometimes, one of the best methods for studying a polar curve $r=r(\theta)$ is to come back to its equivalent parametric representation under the form

$$x=r(\theta)\cos \theta, \ \ y=r(\theta)\sin(\theta)\tag{1}$$

In particular, (1) permits to obtain easily the polar angle made by a tangent to a curve $r=r(\theta)$ with the horizontal axis :

$$\frac{dy}{dx} = \frac{dy(\theta)}{d\theta}/\frac{dx(\theta)}{d\theta}= \frac{r'(\theta)\sin \theta+r(\theta)\cos(\theta)}{r'(\theta)\cos \theta-r(\theta)\sin(\theta)}=\frac{r'(\theta)\tan \theta+r(\theta)}{r'(\theta)-r(\theta)\tan(\theta)} .\tag{1}$$

Besides, the condition for orthogonality of two curves at a certain intersection point $I$ is that the product of the slopes of their tangents at this point is $-1$, i.e.,

$$\frac{dy_1}{dx} \frac{dy_2}{dx}=-1$$

I think you have now all the ingredients for being able to terminate.

Answer 3

The angle between tangent line and a ray from pole of a polar curve is $$\tan\psi=\dfrac{r}{r'}$$ then for these curves in every $\theta$ on curves $$\tan\psi_1=\dfrac{r_1}{r'_1}=\dfrac{a(1-\cos\theta)}{a\sin\theta}=\tan\dfrac{\theta}{2}$$ $$\tan\psi_1=\dfrac{r_2}{r'_2}=\dfrac{a(1+\cos\theta)}{-a\sin\theta}=-\cot\dfrac{\theta}{2}$$ therefore $$\tan\psi_1\tan\psi_2=-1$$