Graph drawing—Finding the points on a curve that are nearest to the origin

[epenguin]

Can you please direct me to the post number you explained that, epenguin? there are already many posts and i intend to go through all of them

The post where I previously directed you to the post, as well as to a still simpler method, barely more than one line of calculation is #53

 

[ehild]

$$s'=2x+\frac{1}{25}\left[ 12 \left( \sqrt{5-4x^2}-\frac{4x^2}{\sqrt{5-4x^2}} \right)-14x \right]=\frac{60-96x^2-300x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}$$

It is wrong.

 

[Chestermiller]

Has anyone tried to do this problem yet using the method recommended by @ehild in post #12?

 

[Ray Vickson]

Has anyone tried to do this problem yet using the method recommended by @ehild in post #12?

Yes, I did that in post #38.

 

[Chestermiller]

Yes, I did that in post #38.

Oops. Missed that. Sorry.

 

[Karol]

$$s'=2x+\frac{1}{25}\left[ 12 \left( \sqrt{5-4x^2}-\frac{4x^2}{\sqrt{5-4x^2}} \right)-14x \right]=\frac{60-96x^2-36x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}$$
I won't continue since i am not sure it's correct. the root isn't precise here either.
Is the mistake in the derivation?

 

[ehild]

$$s'=2x+\frac{1}{25}\left[ 12 \left( \sqrt{5-4x^2}-\frac{4x^2}{\sqrt{5-4x^2}} \right)-14x \right]=\frac{60-96x^2-36x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}$$
I won't continue since i am not sure it's correct. the root isn't precise here either.
Is the mistake in the derivation?

It is correct so far. And the root is precise.
Ops. @Dick s correct, there is a sign error.

 

[Dick]

$$s'=2x+\frac{1}{25}\left[ 12 \left( \sqrt{5-4x^2}-\frac{4x^2}{\sqrt{5-4x^2}} \right)-14x \right]=\frac{60-96x^2-36x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}$$
I won't continue since i am not sure it's correct. the root isn't precise here either.
Is the mistake in the derivation?

Not correct. It should be: $$\frac{60-96x^2+36x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}$$
Note the change of sign on the '36'.

 

[Karol]

Not correct. It should be: $$\frac{60-96x^2+36x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}$$
Note the change of sign on the '36'.

Yes, i wrote that the root was not precise but i had in mind this equation, with the $~+36x\sqrt{5-4x^2}~$
I found the mistake:
$$s'=\frac{60-96x^2+36x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}=\frac{12\cdot(5-8x^2+3x\sqrt{5-4x^2})}{25\sqrt{5-4x^2}}$$
$$s'=o~\rightarrow~-100x^4+125x^2-25=0~\rightarrow~-100a^2+125a-25=0$$
$$a_1=1,~a_2=\frac{1}{4}~\rightarrow~x_{1,2}=\pm1,~x_{3,4}=\pm\frac{1}{2}$$
Now i am trying to understand:

The calculation of $\frac{d}{dx} (5 x^2 + 5y^2 - 6 xy)$ is carried through in full without exploiting the fact that $\frac{d}{dx} (x^2+y^2) = 0$ (at the point of interest) so you only need to find the condition for $\frac{d}{dx}(xy)=0$ (which is essentially what I have done in #37). Since #15 I have been trying to prod you to recognise and exploit the fact that the distance function $(x^2+y^2)$ Is also part of the equation of the curve.

Inpost #37:

Along the curve defined by (1)
$$\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~~~~(4)$$

I know what is $~y'=dy/dx~$ when i am given explicitly $~y=f(x)~$, for example $~y=2x^2~$. then y' is the tangent to the curve y=2x².
But in $~5x^2-6xy+5y^2=4~$ x and y are mixed, so what does $~dy/dx~$ mean? and where is the right side, the =4, in $~\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~$?

 

[epenguin]

Well I think this

This question can be done either by little thought and hard calculations, or hard thought and easy calculations.

has been vindicated and this

? Without any words I think you won’t yourself know what you’ve done or why when you come back to it in six months.
.

in days and hours rather than months, since earlier you were almost at the answer but seemed to forget.

Yes, i wrote that the root was not precise but i had in mind this equation, with the $~+36x\sqrt{5-4x^2}~$
I found the mistake:
$$s'=\frac{60-96x^2+36x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}=\frac{12\cdot(5-8x^2+3x\sqrt{5-4x^2})}{25\sqrt{5-4x^2}}$$
$$s'=o~\rightarrow~-100x^4+125x^2-25=0~\rightarrow~-100a^2+125a-25=0$$
$$a_1=1,~a_2=\frac{1}{4}~\rightarrow~x_{1,2}=\pm1,~x_{3,4}=\pm\frac{1}{2}$$
Now i am trying to understand:

Inpost #37:

I know what is $~y'=dy/dx~$ when i am given explicitly $~y=f(x)~$, for example $~y=2x^2~$. then y' is the tangent to the curve y=2x².
But in $~5x^2-6xy+5y^2=4~$ x and y are mixed, so what does $~dy/dx~$ mean? and where is the right side, the =4, in $~\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~$?

You also seemed to show that you did know what dy/dx is and how to obtain it even when you do not have an explicit formula of form y = f(x) ( in #1 !). The problem presents us with not that form but an ‘implicit’ form generally symbolised f(x, y) = 0, or f(x, y) = C, C a constant. It still describes a two dimensional curve, and dy/dx means the same thing in either form.

You are trying to solve the problem by reorganising the implicit form into an explicit one, which may seem to you safer. It is a perfectly valid mathematical motivation to try and show what can be done in the confines of a given method or approach even when that is not the most convenient. As long as you do recognise that it is not convenient. Also you would soon run into problems where this approach becomes prohibitively complicated, or totally impossible; yours works only because you can solve relevant equations - other examples will give equations you cannot solve, but you would still be able to differentiate the implicit functions.

For the last question you realize it makes no difference if I include a -4 in the bracket or not? It’s a matter of preference and for preference I would normally have put the whole equation ... = 0 and then I might have included it. But the question was originally framed ... = 4.
I could then have derived the quoted equation from d(4)/dx = 0 .

 

[Karol]

For the last question you realize it makes no difference if I include a -4 in the bracket or not? It’s a matter of preference and for preference I would normally have put the whole equation ... = 0 and then I might have included it. But the question was originally framed ... = 4.
I could then have derived the quoted equation from d(4)/dx = 0

I still don't understand, the english is also a bit difficult for me.
I have to maintain correctness and derive both sides of $~5x^2-6xy+5y^2=4~$
What you did in $~\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~$ is derive one side.
And besides, $~\dfrac {dy}{dx}~$ is suitable for y=f(x), not f(x,y)=C, correct?
And besides, also, when i derived both sides i got the =0, and from that i got $~y'=\frac{x}{y}~$, but you made $~\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)=0~$ in order to find the extremum.

 

[ehild]

I still don't understand, the english is also a bit difficult for me.
I have to maintain correctness and derive both sides of $~5x^2-6xy+5y^2=4~$

To have the derivative of y, you can differentiate F(x,y) = const with respect to x, applying chain rule.

What you did in $~\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~$ is derive one side.

It was a mistake, @epenguin meant dF(x,y)/dx with $F(x,y)=x^{2}+y^{2}-\frac {6}{5}xy$

 

[Karol]

To have the derivative of y, you can differentiate F(x,y) = const with respect to x, applying chain rule.

I know, that's how i got y'=-x/y

It was a mistake, @epenguin meant dF(x,y)/dx with $F(x,y)=x^{2}+y^{2}-\frac {6}{5}xy$

Do you mean $~\frac{dF(x,y)}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~$?
Still it is one side only derivative. how do i get from this to =0 in order to find the extremum?

 

[epenguin]

I still don't understand, the english is also a bit difficult for me.
I have to maintain correctness and derive both sides of $~5x^2-6xy+5y^2=4~$
What you did in $~\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~$ is derive one side.
And besides, $~\dfrac {dy}{dx}~$ is suitable for y=f(x), not f(x,y)=C, correct?
And besides, also, when i derived both sides i got the =0, and from that i got $~y'=\frac{x}{y}~$, but you made $~\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)=0~$ in order to find the extremum.

Well no, sorry, I think these points have already been answered in the very previous post #80, which of them hasn’t?
For your last sentence: the given equation has $(x^2+y^2 -\frac{6}{5}xy)$ as constant ($\frac{4}{5}$, but for any other number it would be the same) and therefore its d/dx equals 0 at all points on the curve. So a way of saying what we are trying to do is find a point where the curve slope given by that is equal to that of an origin-centred circle through the same point (your $~y'=-\frac{x}{y}~$) i.e. cotangency – see #29. In the calculus method when we made these two slopes equal at the point we got the answer.

We are in our fifth page of calculations for a problem which was solved by one line of algebra and then it all fell out (#45) or three or four lines of calculus (#37), and you were there yourself after a calculation just a bit more complicated than necessary but not a huge problem, except for missing a concluding line in #52. I don’t know if this is a problem for you, but the impression you gave me is that you just write out formulae and calculations but not the thread of argument so you sometimes lose sight of it and hence e.g. of distinction between equations that are true always, ones that are true at every point of a given curve, and ones that are true just at particular points.

 

[ehild]

I know, that's how i got y'=-x/y

Do you mean $~\frac{dF(x,y)}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~$?
Still it is one side only derivative. how do i get from this to =0 in order to find the extremum?

As F(x,y)=const, dF(x,y)/dx = 0. From r² being extremum, you got that y' = -x/y. Combine that with dF/dx=0..

 

[epenguin]

As F(x,y)=const, dF(x,y)/dx = 0. From r² being extremum, you got that y' = -x/y. Combine that with dF/dx=0..

This was done in #52. (Even if not in quite the most expeditious way possible see #53, 37.) Not sure why we are still talking about it.

 

[Karol]

the given equation has $(x^2+y^2 -\frac{6}{5}xy)$ as constant ($\frac{4}{5}$, but for any other number it would be the same) and therefore its d/dx equals 0 at all points on the curve. So a way of saying what we are trying to do is find a point where the curve slope given by that is equal to that of an origin-centred circle through the same point (your $~y'=-\frac{x}{y}~$) i.e. cotangency – see #29. In the calculus method when we made these two slopes equal at the point we got the answer.

How can dF(x,y)/dx be =0 on all the curve? i know that it should be 0 since the right side is constant, but the tangent to an ellipse changes!
Isn't dF(x,y)/dx the tangent to the ellipse?

Here the radius of the circle is the required shortest distance. What Is the equation of a circle? How does this fit with the equation of the curve? What conditions have to be satisfied to get a situation like diagram?

Equation of a circle whose center is at (h,k) and with radius r:
$$(x-h)^2-(y-k)^2=r^2$$
For origin centered: $~x^2-y^2=r^2~$. i derive both sides and get $~y'=\frac{x}{y}~$.

From r2 being extremum, you got that y' = -x/y. Combine that with dF/dx=0

I didn't get y' = -x/y by use of any r. it just came from deriving both sides of $~x^2+y^{2}-\frac {6}{5}xy=\frac{4}{5}~$
At the extremum point the tangents of the ellipse and circle are equal, i try to equal the circle's tangent $~y'=\frac{x}{y}~$ and that of the ellipse $~y'=-\frac{x}{y}~$ and get nothing:
$$\frac{x}{y}=-\frac{x}{y}~\rightarrow~\frac{x}{y}=0$$
The r doesn't appear and with these two y' equations i can seek only where the tangents equal. later i will find r, so i think

 

[ehild]

How can dF(x,y)/dx be =0 on all the curve? i know that it should be 0 since the right side is constant, but the tangent to an ellipse changes!
Isn't dF(x,y)/dx the tangent to the ellipse?

No, it is the tangent of the function F(x, y(x)), a constant function. The tangent of the ellipse is dy/dx.

Equation of a circle whose center is at (h,k) and with radius r:
$$(x-h)^2-(y-k)^2=r^2$$

It is not the equation of a circle.

 

[epenguin]

How can dF(x,y)/dx be =0 on all the curve? i know that it should be 0 since the right side is constant,RIGHT but the tangent to an ellipse changes!
Isn't dF(x,y)/dx the tangent to the ellipse? NO.

Equation of a circle whose center is at (h,k) and with radius r:
$$(x-h)^2-(y-k)^2=r^2$$
For origin centered: $~x^2-y^2=r^2~$. I DONT KNOW WHY YOU HAVE STARTED CHANGING EQUATIONS FOR A CIRCLE AND FOR DISTANCE CHANGING A + INTO A -
i derive both sides and get $~y'=\frac{x}{y}~$. BUT IT HAS MADE YOU GET WRONG WHAT YOU GOT RIGHT IN #1

I didn't get y' = -x/y by use of any r. it just came from deriving both sides of $~x^2+y^{2}-\frac {6}{5}xy=\frac{4}{5}~$
At the extremum point the tangents of the ellipse and circle are equal, i try to equal the circle's tangent $~y'=\frac{x}{y}~$ and that of the ellipse $~y'=-\frac{x}{y}~$ and get nothing: I THINK YOU ALREADY GOT THE RIGHT EQUATIIONS #52 AND SEEM TO HAVE FORGOTTEN
$$\frac{x}{y}=-\frac{x}{y}~\rightarrow~\frac{x}{y}=0$$
The r doesn't appear and with these two y' equations i can seek only where thhe tangents equal. later i will find r, so i think

 

[Karol]

No, it is the tangent of the function F(x, y(x)), a constant function

Where can i learn this subject, what's it's name?
What does it mean a tangent to a function? i visualize the tangent as it's name, tangent to a curve
$$(x-h)^2+(y-k)^2=r^2~\rightarrow~y'=-\frac{x}{y}$$
The tangent to the ellipse function (?) is also $~y'=-\frac{x}{y}~$, how can it be, they are different graphs?
Ehild said that dF(x,y)/dx is the tangent to the function, not the elipse, so how can i interpret $~y'=-\frac{x}{y}~$ and use it?

 

[epenguin]

Where can i learn this subject, what's it's name?
What does it mean a tangent to a function? i visualize the tangent as it's name, tangent to a curve
$$(x-h)^2+(y-k)^2=r^2~\rightarrow~y'=-\frac{x}{y}$$
The tangent to the ellipse function (?) is also $~y'=-\frac{x}{y}~$, how can it be, they are different graphs?
Ehild said that dF(x,y)/dx is the tangent to the function, not the elipse, so how can i interpret $~y'=-\frac{x}{y}~$ and use it?

The tangent to the ellipse is not that. The slope of the tangent to this ellipse is given in #47, 50. (To be precise, you sometimes say ‘tangent’ when you mean slope of the tangent but we knew what you meant.)

I think the name of the subject is ‘implicit differentiation’, or ‘differentiation of implicit functions’. But I think you do already know how to do implicit differentiation – you did it to obtain the slope of the tangent to the circle! Ray Vickson did it for the not much more complicated case of this ellipse in #47, 50 which I guess you understood.

So it looks to me more like a question of getting clarity about what has been done, from a few key posts on this thread which I already mentioned, and eventually there are many textbooks with calculus and geometry together, treating conic sections In particular with examples.

 

[ehild]

What does it mean a tangent to a function? i visualize the tangent as it's name, tangent to a curve
$$(x-h)^2+(y-k)^2=r^2~\rightarrow~y'=-\frac{x}{y}$$
The tangent to the ellipse function (?) is also $~y'=-\frac{x}{y}~$, how can it be, they are different graphs?
Ehild said that dF(x,y)/dx is the tangent to the function, not the elipse, so how can i interpret $~y'=-\frac{x}{y}~$ and use it?

Remember the original problem.

This equation defines an ellipse. At the same time, it defines two y(x) function, you derived in Post #61,
$$y=\frac{3x\pm 2\sqrt{5-4x^2}}{5}$$ (*)
The the graphs of these functions constitutes the ellipse.
* is the explicit definition of y(x). 5x^ˇ2-6xy+5y²=4 is the implicit definition of y(x). (**)
You have to find a point on the curve which is closest to the origin. That means x²+y²=s is minimum for that point. s has its minimum value if its derivative with respect x is zero.at that point. Remember that point is a point of the curve.
s has its extrems where if dy/dx = -x/y.
You know how x and y are related. Substitute y(x) and dy/dx from formula (*) and its derivative into dy/dx = -x/y. You get an equation for x, that you need to solve.
But you do not need the explicit form of y(x). With implicit differentiation, you get y' = dy/dx from the original formula (*).
When you differentiate 5x²-6xy+5y²=4 you get 10(x+ xy' ) - (6y+x y') = 0 .
Now you have three equation for the x, y coordinates of the point of the curve nearest to the origin:
5x²-6xy+5y²=4
y' = -x/y and
10(x+ y y' ) - 6(y+x y') = 0 .
Solve.

Note I did not speak about a circle.

 

[Karol]

Tangent to the ellipse:
$$y' = \frac{5x-3y}{3x-5y}$$
Equals the tangent to the circle:
$$\frac{5x-3y}{3x-5y}=-\frac{x}{y}~\rightarrow~y^2=x^2$$
For $y=\pm x~\rightarrow~x_{1,2}=\pm 1~$ and it's wrong

The solution is very simple if you write the equation of the curve in polar coordinates r, θ. r is the distance from the origin, so you need to find the minimum r (or r^2) in terms of the angle θ

$$x=r\cdot \cos\theta,~y=r\cdot \sin\theta$$
$$x^2+y^2-\frac{6}{5}xy=\frac{4}{5}~\rightarrow~r^2(\sin^2\theta+\cos^2\theta)-\frac{6}{5}r^2\cdot\sin\theta\cdot\cos\theta=\frac{4}{5}$$
$$r^2=\frac{4}{5-6\cdot\sin\theta\cdot\cos\theta}$$
$$[r^2]'=4\frac{6(\cos^2\theta-\sin^2\theta)}{(5-6\cdot\sin\theta\cdot\cos\theta)^2}$$
$$[r^2]'=0~\rightarrow~\sin^2\theta=\cos^2\theta~\rightarrow~\theta=\pm 45^0$$
For $\displaystyle ~\theta=-45^0~\rightarrow~r=\frac{1}{\sqrt{2}}$
Correct

 

[ehild]

Tangent to the ellipse:
$$y' = \frac{5x-3y}{3x-5y}$$
Equals the tangent to the circle:
$$\frac{5x-3y}{3x-5y}=-\frac{x}{y}~\rightarrow~y^2=x^2$$
For $y=\pm x~\rightarrow~x_{1,2}=\pm 1~$ and it's wrong

x=±1 if y=x
What is x if y = -x ?

 

[epenguin]

Tangent to the ellipse:
$$y' = \frac{5x-3y}{3x-5y}$$
Equals the tangent to the circle:
$$\frac{5x-3y}{3x-5y}=-\frac{x}{y}~\rightarrow~y^2=x^2$$
For $y=\pm x~\rightarrow~x_{1,2}=\pm 1~$ and it's wrong

Wrong only in a manner of speaking. Does satisfy the mathematical conditions imposed. However those conditions will give you the maximum as well the minimum distance from centre.

See last line of #50.

 

[Karol]

My mistake, again:
$$\frac{5x-3y}{3x-5y}=-\frac{x}{y}~\rightarrow~y^2=x^2$$
For $~y=x:~x_{1,2}=\pm 1~$.
For $~y=-x:~x_{1,2}=\pm \frac{1}{2}~$.
The second is the minimum

 

[Karol]

You have to find a point on the curve which is closest to the origin. That means x2+y2=s is minimum for that point. s has its minimum value if its derivative with respect x is zero.at that point. Remember that point is a point of the curve.
s has its extrems where if dy/dx = -x/y.

Is dy/dx=-x/y the tangent at the extremum point only or is it a tangent to all point on the distance formula $~x^2+y^2=s~$?
Because in the explicit form y=f(x) i derive: y'=f'(x) and then equal to zero f'(x)=0 in order to find the extremum.
With the implicit form i don't equal to zero, although i know that when deriving both sides of $~x^2+y^2=s~$ indeed the right side is 0. is this the cause?

 

[ehild]

Is dy/dx=-x/y the tangent at the extremum point only or is it a tangent to all point on the distance formula $~x^2+y^2=s~$?
Because in the explicit form y=f(x) i derive: y'=f'(x) and then equal to zero f'(x)=0 in order to find the extremum.
With the implicit form i don't equal to zero, although i know that when deriving both sides of $~x^2+y^2=s~$ indeed the right side is 0. is this the cause?

dy/dx is not tangent to anything. A straight line can be the tangent of a curve. dy/dx is the slope of the tangent line.
The distance of a point from the origin is sqrt(x²+y²) For a given F(x,y)=0, the distance is extremum if ds/dx =0, that is, dy/dx = -x/y.
Around a circle, the tangent lines have the slope dy/dx=-x/y
In the explicit form of the curve, y(x), the derivative y' is not zero.

 

[Karol]

With the equation of a circle $~x^2+y^2=s^2~$ ,in the explicit form,i get a slightly different derivative:
$$x^2+y^2=s^2\rightarrow y=\pm\sqrt{s^2-x^2}=\pm(s^2-x^2)^{1/2}$$
$$y'=\pm\frac{1}{2}(s^2-x^2)^{-1/2}\cdot(-2)=\pm\frac{x}{\sqrt{s^2-x^2}}$$
I insert $~x^2=s^2-y^2~$ and get: $~y'=\pm\frac{x}{y}~$.

 

[Mark44]

With the equation of a circle $~x^2+y^2=s^2~$ ,in the explicit form,i get a slightly different derivative:
$$x^2+y^2=s^2\rightarrow y=\pm\sqrt{s^2-x^2}=\pm(s^2-x^2)^{1/2}$$
$$y'=\pm\frac{1}{2}(s^2-x^2)^{-1/2}\cdot(-2)=\pm\frac{x}{\sqrt{s^2-x^2}}$$

In the middle expression above, you have omitted a factor of x in the numerator.
It should be $\pm\frac{1}{2}(s^2-x^2)^{-1/2}\cdot(-2x)$

I insert $~x^2=s^2-y^2~$ and get: $~y'=\pm\frac{x}{y}~$.

 

[ehild]

$$y'=\pm\frac{1}{2}(s^2-x^2)^{-1/2}\cdot(-2)=\pm\frac{x}{\sqrt{s^2-x^2}}$$
I insert $~x^2=s^2-y^2~$ and get: $~y'=\pm\frac{x}{y}~$.

You are wrong again.
$\sqrt {y^2}= |y|$ instead of y.

 

[Mark44]

I'm amazed that this thread has gone on for over 100 posts over a month and a half...

 

[ehild]

I'm amazed that this thread has gone on for over 100 posts over a month and a half...

It is like some sport, finding the errors of Karol.

 

[epenguin]

Well maybe we shouldn’t be too harsh about what are just slips or not quite right terminology when we know what he means - would that more students here actually finished a problem and gave an answer instead of just disappearing.

He appears to be on top of the calculus method now – having on the way been misled I guess by superficial resemblances to more familiar things into supposing e.g. dy/dx = 0 or dF/dx = 0 would give solutions to this problem as many students would.

I think it would be useful and instructive and would invite him, before ending to try to get the maximum distance of the curve from the centre (which he almost already has by calculus) by the approach of #45 which is a very short calculation.

 

[Mark44]

Well maybe we shouldn’t be too harsh about what are just slips or not quite right terminology when we know what he means - would that more students here actually finished a problem and gave an answer instead of just disappearing.

I have to admire his determination -- that's definitely a plus.