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

[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.

 

[Karol]

$$y'=\pm\frac{1}{2}(s^2-x^2)^{-1/2}\cdot(-2x)=\pm\frac{x}{\sqrt{s^2-x^2}}$$

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

I insert $~x^2=s^2-y^2~$ and get $~y'=\pm\frac{x}{\vert y \vert}$
Doesn't it equal $~\pm\frac{x}{y}~$?

 

[Mark44]

$$y'=\pm\frac{1}{2}(s^2-x^2)^{-1/2}\cdot(-2x)=\pm\frac{x}{\sqrt{s^2-x^2}}$$

I insert $~x^2=s^2-y^2~$ and get $~y'=\pm\frac{x}{\vert y \vert}$
Doesn't it equal $~\pm\frac{x}{y}~$?

No. $\frac x y \ne \frac x {|y|}$
(I'm ignoring the $\pm$ for the moment, but it doesn't change things.)

 

[ehild]

$$y'=\pm\frac{1}{2}(s^2-x^2)^{-1/2}\cdot(-2x)=\pm\frac{x}{\sqrt{s^2-x^2}}$$

It is wrong, you miss the minus in front of x.

I insert $~x^2=s^2-y^2~$ and get $~y'=\pm\frac{x}{\vert y \vert}$
Doesn't it equal $~\pm\frac{x}{y}~$?

No. It should be $y'=\mp\frac{x}{\vert y \vert}$
Better not to use ± if you do not know what it means. A quantity can not be equal to an other quantity and its negative at the same time, unless it is zero. Think: what is y' if y>0? What is y' it y<0?

 

[Karol]

It is wrong, you miss the minus in front of x.

No. It should be $y'=\mp\frac{x}{\vert y \vert}$
Better not to use ± if you do not know what it means. A quantity can not be equal to an other quantity and its negative at the same time, unless it is zero. Think: what is y' if y>0? What is y' it y<0?

In $~y'=\frac{x}{\vert y \vert}~$ y''s sign depends on x in both cases, when y>0 and y<0.
I wrote ± because of the of the square root of $~x^2+y^2=s^2\rightarrow y=\pm\sqrt{s^2-x^2}~$, and i mean that either $~\sqrt{s^2-x^2}~$ or $~-\sqrt{s^2-x^2}~$ satisfies y².
For a circle, for the same x, y' is + and - but for opposite y's. but we didn't talk about a circle but about the distance function.
Indeed y' can't be +y' and -y' at the same time, so i don't know what to say. it came from the root.

 

[epenguin]

Hopefully we are just into being careful about meanings of terms now.
$\pm$ and $\mp$ come in pairs, meaning that the two things can both be ±, but not independently; $\pm$ means something can be either + or - , but if it is then followed by $\mp$ something then that other thing can also be ± , but the second thing has to be - when the first is + and vice versa. I would have thought it is fairly obvious when you come across it, which is not often. There is an example in #45. In that case as they apply to a positive number (½) It means that the two points in question are in the top left and bottom right quadrants. If instead it was a number you didn’t know the sign ot it would mean the two points are just in opposite quadrants.

I don’t know that then this convention is rigidly extended. That is, if you see two ± ‘s near each other it won’t always be intended that if the first is + then so necessarily is the second; it might be explained or you might be expected to work it out for yourself,

 

[ehild]

For a circle, for the same x, y' is + and - but for opposite y's. but we didn't talk about a circle but about the distance function.
Indeed y' can't be +y' and -y' at the same time, so i don't know what to say. it came from the root.

You do not really know how to use ± correctly.
In Post #99 you wrote $y=\pm\sqrt{s^2-x^2}$, and then you took the derivative of y. But $y=\pm\sqrt{s^2-x^2}$ is not a function. It is two functions: $y_1(x)=\sqrt{s^2-x^2}$ and $y_2(x)= - \sqrt{s^2-x^2}$ .
The derivative of the first one is $$y'_1(x)=\frac{-x}{\sqrt{s^2-x^2}}= \frac{-x}{y_1}$$.
The derivative of the second one is $$y'_2(x)= - \frac{-x}{\sqrt{s^2-x^2}}=\frac{-x}{-\sqrt{s^2-x^2}}= \frac{-x}{y_2}$$.
You see y'=-x/y either you take the + sign or - sign in front of the square root in $y=\pm\sqrt{s^2-x^2}$.

 

[Karol]

I try to do the same with the help of $~\sqrt{y^2}=|y|~$:
$$x^2+y^2=s^2\rightarrow y=|\sqrt{s^2-x^2}|=|(s^2-x^2)^{1/2}|$$
$$y'=\frac{1}{2}|\sqrt{s^2-x^2}^{-1/2}\cdot (-2x)|=\vert \frac{x}{\sqrt{s^2-x^2}} \vert=\vert \frac{x}{y} \vert$$
You see, no - sign before the x/y

 

[Mark44]

I try to do the same with the help of $~\sqrt{y^2}=|y|~$:
$$x^2+y^2=s^2\rightarrow y=|\sqrt{s^2-x^2}|=|(s^2-x^2)^{1/2}|$$

No, this isn't right.
$x^2 + y^2 = s^2 \Rightarrow y = \pm \sqrt{s^2-x^2}$
This means that there are two values of y -- one positive and one negative.
With what you wrote, you get only one value of y. Also, $|\sqrt{s^2 - x^2}|$ is exactly the same thing as $\sqrt{s^2 - x^2}$. The square root function returns a value that is nonnegative, so taking the absolute value of the radical does precisely nothing.

$$y'=\frac{1}{2}|\sqrt{s^2-x^2}^{-1/2}\cdot (-2x)|=\vert \frac{x}{\sqrt{s^2-x^2}} \vert=\vert \frac{x}{y} \vert$$
You see, no - sign before the x/y

 

[ehild]

$$y'=\frac{1}{2}|\sqrt{s^2-x^2}^{-1/2}\cdot (-2x)|=\vert \frac{x}{\sqrt{s^2-x^2}} \vert=\vert \frac{x}{y} \vert$$
You see, no - sign before the x/y

Wrong again.
The derivative of the absolute value of a function is not the same as the absolute value of the derivative.
What is the derivative of f(x)=|x|, for example?

 

[Karol]

If i can't use it, why did you write:

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

$$y=|x|=\left\{ \begin{array}{ll}y=x & x>0 \\ y=-x & x<0 \end{array}\right.$$
$$y'=\left\{ \begin{array}{ll}y'=1 & x>0 \\ y'=-1 & x<0\end{array}\right.$$
But if $~y = \pm \sqrt{s^2-x^2}~$ and not $~| \sqrt{s^2-x^2}|~$ then i can't use absolute value at all

 

[ehild]

If i can't use it, why did you write:

$$y=|x|=\left\{ \begin{array}{ll}y=x & x>0 \\ y=-x & x<0 \end{array}\right.$$
$$y'=\left\{ \begin{array}{ll}y'=1 & x>0 \\ y'=-1 & x<0\end{array}\right.$$
But if $~y = \pm \sqrt{s^2-x^2}~$ and not $~| \sqrt{s^2-x^2}|~$ then i can't use absolute value at all

You see you can not say that $y'= |\frac{d\sqrt{s^2-x^2}}{dx}|$
$~y = \pm \sqrt{s^2-x^2}~$ means that s²=x²+y² is satisfied with both a positive and a negative y for a given x value. Read my Post #111.
$~| \sqrt{s^2-x^2}|~$ is the same as $\sqrt{s^2-x^2}$, read Mark's Post #113.

 

[Karol]

And also:
$$y^2=s^2-x^2~\rightarrow~\sqrt{y^2}=\sqrt{s^2-x^2}~\rightarrow~|y|=\sqrt{s^2-x^2}$$
$$\rightarrow~\left\{ \begin{array}{ll}y_1=\sqrt{s^2-x^2} \\ y_2=-\sqrt{s^2-x^2}\end{array}\right.$$

 

[Mark44]

And also:
$$y^2=s^2-x^2~\rightarrow~\sqrt{y^2}=\sqrt{s^2-x^2}~\rightarrow~|y|=\sqrt{s^2-x^2}$$
$$\rightarrow~\left\{ \begin{array}{ll}y_1=\sqrt{s^2-x^2} \\ y_2=-\sqrt{s^2-x^2}\end{array}\right.$$

Yes, this is correct.

 

[Karol]

The last posts here are unnecessarily complicated. 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.

You used twice in #37 $~\frac{d}{dx} (x^2+y^2) = 0~$, once in obtaining $~\frac{d}{dx}(xy)=0~$ and then at the end when you equaled the two derivatives.
If you used the first time, why didn't i get $~\dfrac {dy}{dx}=-\dfrac {x}{y}~~~~~~~~~~~~~~~(3)~$? indeed i used it at the point of interest

 

[epenguin]

If you used the first time, why didn't i get $~\dfrac {dy}{dx}=-\dfrac {x}{y}~~~~~~~~~~~~~~~(3)~$? indeed i used it at the point of interest

Sorry your meaning is not clear to me, the ‘you’ and ‘I’ not making sense.

This equation is the condition satisfied by a point (on a continuous curve corresponding to some relation between x and y) that is at a minimum or maximum distance from (0, 0) Compared to other points on the given curve I should say (a pedant would probably say compared with neighbouring points).

I was going to say as I have already said more than once that you got this equation yourself in #1 ! But only now I realize there is something wrong in that post, and that is the → !
The → leads to the equation that only makes its first appearance at the end of #47 !
You probably got my eq 3 in the same way I did but got your papers mixed up?
It may be (often is) more efficient rather than trying to see where you went wrong, if hopefully you have now got it moderately clear, write out the argument, which is not that long, starting from zero.
If you can also at some point answer the question of #104 by any method(s) that would be splendid!

 

[Karol]

I know how to obtain $~\dfrac {dy}{dx}=-\dfrac {x}{y}~$, but i ask something else.
You used [r²]'=0 twice by taking into consideration that the distance is part of the original equation. it indeed simplified.
But isn't it too much to use the same consideration, same argument twice in the same process? it is a general question, not related only to this question.
The second time you used [r²]'=0 is at the end of #37 at:

Comparing (6) with (3) therefore, at these points

 

[Karol]

Just curious. How about finding the solution like this:
Find a circle C0 centered at the origin with radius r which intersects the ellipse at precisely one point? The value of r is then the shortest distance?

EDIT: $x^2+y^2=r^2 . 5x^2+5Y^2 -6XY=4 =5r^2-6x(\sqrt{r^2-x^2})=4$. Square both sides and find values of $r$ so that there is just one solution. $5r^2-4 =6x(\sqrt{ r^2-x^2} \rightarrow 25r^4-10r^2+16=36x^2(r^2-x^2) \rightarrow -36x^4 +36x^2r^2 -25r^4 +10r^2 -16 =0$ A bi-quadratic Set $r^2=s$ then we get :

$-36x^4 +36x^2s-25s^2 -10s-16=0. -25s^2-(10 -36x^2)s+ 36x^4-16=0 . s=10-36x^2 \pm \sqrt{}$ . .

$$\frac{10-36x^2\pm\sqrt{(10-36x^2)^2+100(36x^4-16)}}{-50}=\frac{10-36x^2\pm 2\sqrt{3}\sqrt{408x^4-60x^2-125}}{-50}$$

 

[WWGD]

$$\frac{10-36x^2\pm\sqrt{(10-36x^2)^2+100(36x^4-16)}}{-50}=\frac{10-36x^2\pm 2\sqrt{3}\sqrt{408x^4-60x^2-125}}{-50}$$

Sure, thanks, and sorrry for leaving without completing. Doesn't this work?

 

[Karol]

It doesn't work since the expression under the root doesn't give precise value.
$$s=\frac{10-36x^2\pm 2\sqrt{3}\sqrt{408x^4-60x^2-125}}{-50}$$
S is positive and because the denominator is negative (-50) the nominator must be positive.
I don't know to continue from here. under the root i denote $~x^2=a$:
$$408a^2-60a-125=0~\rightarrow~a=\frac{60\pm\sqrt{207,600}}{816}$$
I can't continue

 

[ehild]

Just curious. How about finding the solution like this:
Find a circle C0 centered at the origin with radius r which intersects the ellipse at precisely one point? The value of r is then the shortest distance?

EDIT: $x^2+y^2=r^2 . 5x^2+5Y^2 -6XY=4 =5r^2-6x(\sqrt{r^2-x^2})=4$. Square both sides and find values of $r$ so that there is just one solution. $5r^2-4 =6x(\sqrt{ r^2-x^2} \rightarrow 25r^4-10r^2+16=36x^2(r^2-x^2) \rightarrow -36x^4 +36x^2r^2 -25r^4 +10r^2 -16 =0$
.

The line highlighted by red is wrong. It should be
$5r^2-4 =6x(\sqrt{ r^2-x^2} \rightarrow 25r^4-40r^2+16=36x^2(r^2-x^2)$
But it is better to keep $(5r^2-4)^2$ un-expanded.

 

[Karol]

$$\left[ \frac{5r^2-4}{6} \right]^2=x^2(r^2-x^2)$$
Any value that solves is good, +x and -x are always answers.
I can't solve

 

[Dick]

Why not? Do you want to solve for $x$? It's a quadratic equation in the variable $x^2$.

 

[Karol]

$$x^2+y^2=r^2,~~5x^2+5Y^2 -6xy=4~\rightarrow~5r^2-4 =6x(\sqrt{ r^2-x^2})$$
$$\left[ \frac{5r^2-4}{6} \right]^2=x^2(r^2-x^2)~\rightarrow~\left[ \frac{5r^2-4}{6} \right]^2=r^2a-a^2$$
$$a^2-r^2a+\left[ \frac{5r^2-4}{6} \right]^2=0$$
If the root will be 0 there will be only one solution for a=x²
$$\frac{r^2\pm\sqrt{r^4-4\left[ \frac{5r^2-4}{6} \right]^2}}{2}=\frac{r^2\pm\frac{2\sqrt{2}}{3}\sqrt{-2r^4-5r^2+2}}{2}$$
Under the root:
$$-2r^4-5r^2+2=0~\rightarrow~\frac{5\pm\sqrt{25+16}}{-4}$$
Cannot equal this to 0

 

[ehild]

$$x^2+y^2=r^2,~~5x^2+5Y^2 -6xy=4~\rightarrow~5r^2-4 =6x(\sqrt{ r^2-x^2})$$
$$\left[ \frac{5r^2-4}{6} \right]^2=x^2(r^2-x^2)~\rightarrow~\left[ \frac{5r^2-4}{6} \right]^2=r^2a-a^2$$
$$a^2-r^2a+\left[ \frac{5r^2-4}{6} \right]^2=0$$
If the root will be 0 there will be only one solution for a=x²
$$\frac{r^2\pm\sqrt{r^4-4\left[ \frac{5r^2-4}{6} \right]^2}}{2}=\frac{r^2\pm\frac{2\sqrt{2}}{3}\sqrt{-2r^4-5r^2+2}}{2}$$

Check the signs.

 

[Karol]

$$\frac{r^2\pm\sqrt{r^4-4\left[ \frac{5r^2-4}{6} \right]^2}}{2}=\frac{r^2\pm\frac{2\sqrt{2}}{3}\sqrt{-2r^4+5r^2-2}}{2}$$
Under the root, $~r^2=g~$:
$$-2g^2+5g-2=0~\rightarrow~g_1=2,~g_2=\frac{1}{2}$$
The shortest distance is $~g_2=r^2=\frac{1}{2}$
Back to the original equation $~5(x^2+y^2)-6xy=4~\rightarrow~5r^2-4=6xy~~\rightarrow~xy=-\frac{1}{4}$
I insert $~y=\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]~$:
$$xy=-\frac{1}{4}~\rightarrow~x\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]=-\frac{1}{4}$$
$$~\rightarrow~-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})$$
If i square both sides, i don't get rid of the root

 

[Dick]

$$\frac{r^2\pm\sqrt{r^4-4\left[ \frac{5r^2-4}{6} \right]^2}}{2}=\frac{r^2\pm\frac{2\sqrt{2}}{3}\sqrt{-2r^4+5r^2-2}}{2}$$
Under the root, $~r^2=g~$:
$$-2g^2+5g-2=0~\rightarrow~g_1=2,~g_2=\frac{1}{2}$$
The shortest distance is $~g_2=r^2=\frac{1}{2}$
Back to the original equation $~5(x^2+y^2)-6xy=4~\rightarrow~5r^2-4=6xy~~\rightarrow~xy=-\frac{1}{4}$
I insert $~y=\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]~$:
$$xy=-\frac{1}{4}~\rightarrow~x\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]=-\frac{1}{4}$$
$$~\rightarrow~-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})$$
If i square both sides, i don't get rid of the root

Don't square both sides until you have the square root isolated by itself.

 

[Karol]

$$-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})~\rightarrow~-\frac{5}{24x^3}=\pm\sqrt{5-4x^2}$$
After isolating the root and squaring both sides:
$$2304\cdot x^8-2880\cdot x^6+25=0,~~a=x^2~\rightarrow~2304\cdot a^4-2880\cdot a^3+25=0$$
And now what?

 

[tnich]

$$-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})~\rightarrow~-\frac{5}{24x^3}=\pm\sqrt{5-4x^2}$$
And now what?

This part is not correct.
$$x(3x\pm 2\sqrt{5-4x^2})\neq \pm 6x^3\sqrt{5-4x^2})$$

 

[Karol]

$$-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})~\rightarrow~-\frac{5+12x^2}{8x}=\pm\sqrt{5-4x^2}$$
After squaring:
$$400x^4-200x^2+25=0~\rightarrow~16x^4-8x^2+1=0~\rightarrow~16a^2-8a+1=0$$
$$~\rightarrow~\left[ a-\frac{1}{4} \right]^2=0~\rightarrow~x^2=\frac{1}{4},~x=\pm\frac{1}{2}$$

 

[tnich]

$$~x=\pm\frac{1}{2}$$

You are on the right track. Now all you have to do is figure out the corresponding values of y. Be careful with signs.
Once you have done that, try @ehild's suggestion and convert to polar coordinates. That is a much simpler way to solve this problem.

 

[epenguin]

You got there. Actually you got the answer (r) to the question essentially in #130. However, for completeness it is necessary to know the corresponding x and y. You have got x² and so the two x, and from x² and r² you can now get y². Then you can get y, however you previously saw that not every combination of and x and y that you get is allowable.
If that previous conclusion depended on a calculus argument, and here you want to do an algebraic argument, then you need to complete this.

You have been doing things a long way around in my opinion, but I hope you have learned some things.
For example I hope as soon as you saw

$$-2g^2+5g-2=0$$

in #130, you realized that one solution would be the reciprocal of the other.This happens when coefficients of a polynomial equation equidistant from the ends are equal. Called reciprocal equations their solution can be reduced to solving equations of half the degree plus a quadratic. In your case you only have a quadratic anyway, I don't think there is any saving.

The problem which was previously solved by calculus here you are solving by an algebraic argument. That depended on saying that two roots of an equation have to be equal at the shortest distance. I used the same argument in #45, or at least it could be worded that way. I think it has to be admitted that that was a whole lot faster and easier. Probably on analysis your argument could be shown to be equivalent.

To complete this learning and take on board this other point of view, I again invite you to use the approach of #45 to get the maximum distance. If you do not do this by about Monday I will do it here.

 

[Karol]

You have got x² and so the two x, and from x² and r² you can now get y²

Why do i get y² and not y? the original equation is $~5r^2-6xy=4~$ and from that, by inserting r² and x i can get directly the y and both values of y will be valid, because you said:

however you previously saw that not every combination of and x and y that you get is allowable.

 

[epenguin]

Ah yes if you do it that way it comes OK without ambiguity - I was thinking you might go and use r² and x² .

 

[Karol]

How do you do this kind of hyperlink:

 

[tnich]

How do you do this kind of hyperlink:
View attachment 220470

Just type "@" and the user name.

 

[Karol]

I thank you all for your patience:

Dick
Ray
WWGD
ehild
Phyton
epenguin
Nidum
Chestermiller
Mark44
tnich

It was a long ride

 

[epenguin]

It was a long ride

Though a short distance.
You took some long routes around where you saw some hopefully memorable scenery, like rotating axes, reciprocal equations amongst others.
Every method to find this shortest distance was trying to tell you at the same time the greatest distance, but this was always ignored. In order to fix the method of quadratic inequalities (#45) in your mind so that you might draw on it if you meet a suitable problem, I urged you find the largest distance using it. But as this hasn't been done I can't hold back from it .

Re-write the equation of your curve $5x^2-6xy+5y^2=4$ as
$$~~~~~~~~~~~~ 2(x^2+y^2) =4-3(x-y)^2$$
Then maximum of r^2 is 2 and the maximum of r is √2, which occurs when the square on the right is zero: there $x=y=±1$.

 

[Karol]

Then maximum of r^2 is 2

Why?

 

[Ray Vickson]

Why?

This has been explained to you many, many times already; just look at #4, #12, #38.

 

[Karol]

I didn't know you can mix methods. i wouldn't have gotten to that

 

[epenguin]

Why?

Because any square must be ≥ 0, now can the RHS of the expression in #142 be greater than 4? Similar principle was involved in #45.

 

[Ray Vickson]

I didn't know you can mix methods. i wouldn't have gotten to that

What dicatator told you that you cannot mix methods? Most good problem solvers mix methods all the time; learning to do that effectively should be part of any math education.

 

[Karol]

It was a complex problem for me and i still feel unsure.
I thank you all again
This site and your attitude are invaluable for me

 

[epenguin]

It was far more satisfactory to me to interact with a student who saw the thing through to the end, told us all his results and reasoning including misunderstanding, than one who does not do these things. Or even who tells you 'thanks I have solved it' but you wonder if he has as well as he thinks.