Optimisation question #2

[Win_odd Dhamnekar]

Hello Forumites,

2) I want to find the radius r of a circle c whose center is on a fixed circle C of radius R such that the arc length of the part of c within C is a maximum.

 

[mmm4444bot]

As always, please share your work and comments, so that tutors can understand where you're at and what needs to be done next. Thank you! :cool:

 

[Win_odd Dhamnekar]

As always, please share your work and comments, so that tutors can understand where you're at and what needs to be done next. Thank you! :cool:

Hello,
Intuitively speaking, the arc lenth of the part of c within the circle C is maximum only when r=R.

 

[MarkFL]

This is a problem I find very interesting. I was able to solve it, however I had to resort to a numeric root finding technique to get an approximate solution. I will be interested to see if it can be solved exactly.

 

[Win_odd Dhamnekar]

This is a problem I find very interesting. I was able to solve it, however I had to resort to a numeric root finding technique to get an approximate solution. I will be interested to see if it can be solved exactly.

Hello, Which method did you use to find numeric root? Newton's method , secant method, bisection method.

 

[MarkFL]

Hello, Which method did you use to find numeric root? Newton's method , secant method, bisection method.

I used a CAS (W|A), but if I were to do it manually, I would use Newton.

 

[MarkFL]

I am attaching a diagram that should help you get started.



I hope you can see it...it got shrunk considerably for some reason.

 

[Win_odd Dhamnekar]

Optimisation question 2

I am attaching a diagram that should help you get started.

View attachment 9940

I hope you can see it...it got shrunk considerably for some reason.

Hello,
So, how to prove 2*s is maximum? \(\displaystyle 2*s=2*r\theta\). I require to differentiate \(\displaystyle 2*s=2*r*\theta\).w.r.t. r and then make it=0.

 

[MarkFL]

Hello,
So, how to prove 2*s is maximum? \(\displaystyle 2*s=2*r\theta\). I require to differentiate \(\displaystyle 2*s=2*r*\theta\).w.r.t. r and then make it=0.

Because of symmetry, we can simply maximize \(\displaystyle s\). What I did was let the point above the \(\displaystyle x\)-axis where the two circles intersect be \(\displaystyle (x,y)\). We know:

\(\displaystyle s=r\theta\)

Can you write \(\displaystyle r\) and \(\displaystyle \theta\) in terms of \(\displaystyle x\) and \(\displaystyle y\)?

 

[Win_odd Dhamnekar]

Optimisation question 2

Because of symmetry, we can simply maximize \(\displaystyle s\). What I did was let the point above the \(\displaystyle x\)-axis where the two circles intersect be \(\displaystyle (x,y)\). We know:

\(\displaystyle s=r\theta\)

Can you write \(\displaystyle r\) and \(\displaystyle \theta\) in terms of \(\displaystyle x\) and \(\displaystyle y\)?

Hello,
\(\displaystyle r=\sqrt{(R-x)^2+y^2}\) and \(\displaystyle \theta=\arcsin\frac{y}{r}\)

 

[MarkFL]

Hello,
\(\displaystyle r=\sqrt{(R-x)^2+y^2}\) and \(\displaystyle \theta=\arcsin\frac{y}{r}\)

Okay, let's first look at:

\(\displaystyle r=\sqrt{(R-x)^2+y^2}\)

If we expand the radicand, then we have:

\(\displaystyle r=\sqrt{R^2-2Rx+x^2+y^2}\)

Now, we know \(\displaystyle x^2+y^2=R^2\) so we can write:

\(\displaystyle r=\sqrt{R^2-2Rx+R^2}=\sqrt{2R(R-x)}\)

In an effort to use only the variables \(\displaystyle x\) and \(\displaystyle y\), I chose to write:

\(\displaystyle \tan(\theta)=\dfrac{y}{R-x}=\dfrac{\sqrt{R^2-x^2}}{R-x}=\sqrt{\dfrac{R+x}{R-x}}\)

But, what you chose will work too:

\(\displaystyle \sin(\theta)=\dfrac{y}{r}=\dfrac{\sqrt{R^2-x^2}}{\sqrt{(R-x)^2+y^2}}=\dfrac{\sqrt{R^2-x^2}}{\sqrt{2R(R-x)}}=\sqrt{\dfrac{R+x}{2R}}\)

Now we can express \(\displaystyle s\) as a function of just one variable:

\(\displaystyle s(x)=\sqrt{2R(R-x)}\arcsin\left(\sqrt{\dfrac{R+x}{2R}}\right)\)

Can you proceed?

 

[MarkFL]

We can also see from the diagram that:

\(\displaystyle \cos(\theta)=\frac{r}{2R}\)

And so:

\(\displaystyle s(r)=r\arccos\left(\dfrac{r}{2R}\right)\)

This still requires getting a numeric approximation when finding the root of \(\displaystyle s'(r)=0\), but the intermediary computations are simpler.

 

[Win_odd Dhamnekar]

Hint

We can also see from the diagram that:

\(\displaystyle \cos(\theta)=\frac{r}{2R}\)

And so:

\(\displaystyle s(r)=r\arccos\left(\dfrac{r}{2R}\right)\)

This still requires getting a numeric approximation when finding the root of \(\displaystyle s'(r)=0\), but the intermediary computations are simpler.

Hello,
Author has provided a hint: You can answer in terms of R and angle satisfying certain equation.

 

[MarkFL]

This is how I finished the problem:

Now, let's write:

\(\displaystyle s=2R\cdot\dfrac{r}{2R} \arccos\left(\dfrac{r}{2R}\right)\)

Let:

\(\displaystyle u=\dfrac{r}{2R}\)

And we now have the arc-length as a function of one variable:

\(\displaystyle s(u)=2Ru\arccos(u)\) where \(\displaystyle 0\le u\le1\)

So, we set about finding the critical value:

\(\displaystyle s'(u)=2R\left(-\dfrac{u}{\sqrt{1-u^2}}+\arccos(u)\right)=0\)

This implies:

\(\displaystyle \dfrac{u}{\sqrt{1-u^2}}=\arccos(u)\)

This cannot be algebraically solved to get an exact value, so using a numeric root finding method, we obtain:

\(\displaystyle u\approx0.652184623909186793860487473\)

And so the radius in question is:

\(\displaystyle r\approx1.304369247818373587720974946R\)

Now, since:

\(\displaystyle s(0)=s(1)=0\), we know the critical value we found is at a maximum.

Here is a graph showing \(\displaystyle s\) as \(\displaystyle r\) varies from \(\displaystyle 0\) to \(\displaystyle 2R\), with the maximum we found.

 

[Win_odd Dhamnekar]

Final answer

This is how I finished the problem:

Now, let's write:

\(\displaystyle s=2R\cdot\dfrac{r}{2R} \arccos\left(\dfrac{r}{2R}\right)\)

Let:

\(\displaystyle u=\dfrac{r}{2R}\)

And we now have the arc-length as a function of one variable:

\(\displaystyle s(u)=2Ru\arccos(u)\) where \(\displaystyle 0\le u\le1\)

So, we set about finding the critical value:

\(\displaystyle s'(u)=2R\left(-\dfrac{u}{\sqrt{1-u^2}}+\arccos(u)\right)=0\)

This implies:

\(\displaystyle \dfrac{u}{\sqrt{1-u^2}}=\arccos(u)\)

This cannot be algebraically solved to get an exact value, so using a numeric root finding method, we obtain:

\(\displaystyle u\approx0.652184623909186793860487473\)

And so the radius in question is:

\(\displaystyle r\approx1.304369247818373587720974946R\)

Now, since:

\(\displaystyle s(0)=s(1)=0\), we know the critical value we found is at a maximum.

Here is a graph showing \(\displaystyle s\) as \(\displaystyle r\) varies from \(\displaystyle 0\) to \(\displaystyle 2R\), with the maximum we found.

Hello, very correct answer.