Find minimum x value from a polar function

Question

I am mainly examining limacon functions. For the equation r= b + a*cos(theta), it is easy to find the minimum radius, but I want to find the most negative value (between a given range). Take function r=1.5 + 1.25*cos(theta) for 0<=theta<=180:

We know there is a minimum radius at 180, but I am interested in what looks to be two minimum x-values. How can I find these values?

I tried parameterizing with what we know:

r=1.5 + 1.25*cos(theta)

x=r*cos(theta) => x=(1.5 + 1.25*cos(theta))*cos(theta) (substitution)

I then looked at the minimum of this function, but it is not the value I am expecting. Any input is appreciated.

Answer 1

From $r = A + B \cos \theta$, multiply by $r$, yielding $r^2 = A r + B x$ when we make your substitution $r \cos \theta \rightarrow x$. Solve for $x$, yielding $x = \frac{r^2 - A r}{B} = \frac{r(r-A)}{B}$. This last is minimized when $r = A/2$ (by symmetry in the numerator). Substituting this in, we find the minimum $x$ value is $-\frac{A^2}{4B}$.

Note: It could be maximized if we only look at symmetry. Since the coefficient of $r^2$ is positive, we know it's a minimum.

For your example, $A = 1.5, B = 1.25$ and $-\frac{1.5^2}{4(1.25)} = -0.45$.

Answer 2

If $$x(t)=\cos (t) \left(\frac{5 \cos (t)}{4}+\frac{3}{2}\right)$$ differentiation leads to $$\frac{dx(t)}{dt}=-\frac{1}{2} \sin (t) (5 \cos (t)+3)$$ the solutions of which being $t=0$ and $t=\pm \cos ^{-1}\left(-\frac{3}{5}\right)$. For this last value, you should find $x=-\frac{9}{20}$

Answer 3

The derivative of $$y=1.5\cos(x)+1.25\cos^2(x)$$ is $$y'=-1.5\sin(x)-2.5\cos(x)\sin(x)=-0.5\sin(x)(3+5\cos(x))$$ One zero of the derivative is found when $x=0$ but it corresponds to the maxima. The second zero occurs, when $$\cos(x)=-\frac{3}{5}$$ so now plug it in $$y_{min}=-\frac{9}{10}+\frac{9}{20}=-\frac{9}{20}$$

Answer 4

Try to think of everything in terms of the $(x,y)$ co-ordinates.

First, let's convert $r=1.5+1.25\cos(\theta)$ into Cartesian ($(x,y)$) co-ordinates.

Multiplying through by $r$ yields: $$\underbrace{r^2}_{x^2+y^2}=1.5\underbrace{r}_{\sqrt{x^2+y^2}}+1.25\underbrace{r\cos(\theta)}_{x}.$$

Simplifying, we get: $$(*) x^2+y^2=1.5\sqrt{x^2+y^2}+1.25x $$

Now, $(*)$ gives an implicit relationship between $x$ and $y$.

Differentiate $(*)$ (implicitly) wrt $x$ to give: $$\frac{dy}{dx}=\frac{\frac{1.5x}{\sqrt{x^2+y^2}}-2x+1.25}{2y-\frac{1.5y}{\sqrt{x^2+y^2}}}.$$

The points you're referring to are the points of infinite gradient (when $\frac{dx}{dy}=0$). i.e. when $$(**) 2y-\frac{1.5y}{\sqrt{x^2+y^2}}=0 .$$

Now, $(**)$ gives a relationship between $y$ and $x$.

Now, solve $(*)$ and $(**)$ simultaneously!