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

[Karol]

Homework Statement

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling
When f'(x)>0 ? the function rises

The Attempt at a Solution

Deriving relative to x:
$$10x-6(yy'+x)+10yy'=0~\rightarrow~y'=-\frac{x}{y}$$
What do i do with that?

 

[Dick]

Homework Statement

View attachment 215800

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling
When f'(x)>0 ? the function rises

The Attempt at a Solution

Deriving relative to x:
$$10x-6(yy'+x)+10yy'=0~\rightarrow~y'=-\frac{x}{y}$$
What do i do with that?

You don't do anything with that. The quantity you want to minimize is distance, or $x^2+y^2$. The curve is a 'constraint'. Do you know the Lagrange multiplier method?

 

[Karol]

I don't know the Lagrange method and i am sure it's not to be used, since this chapter is the third in the book, and the book started from "scratch", it taught systematically, and it didn't teach it.
$$s=x^2+y^2~\rightarrow~s'=2x+2yy'$$
I insert $~y'=-\frac{x}{y}~$: s'=0
So the distance is a constant and the equation is a circle.
$$5x^2-6xy+5y^2=4~\rightarrow~x^2+y^2=\frac{4+6xy}{5}$$
I don't know to get the radius and the center, but the circle's center isn't at the origin, so i need to find the line from the center to the origin and then intersect it with the circle to find the point closest.

 

[Ray Vickson]

I don't know the Lagrange method and i am sure it's not to be used, since this chapter is the third in the book, and the book started from "scratch", it taught systematically, and it didn't teach it.
$$s=x^2+y^2~\rightarrow~s'=2x+2yy'$$
I insert $~y'=-\frac{x}{y}~$: s'=0
So the distance is a constant and the equation is a circle.
$$5x^2-6xy+5y^2=4~\rightarrow~x^2+y^2=\frac{4+6xy}{5}$$
I don't know to get the radius and the center, but the circle's center isn't at the origin, so i need to find the line from the center to the origin and then intersect it with the circle to find the point closest.

The curve is not a circle; it is a tilted ellipse.

You can proceed in one of two ways.
(1) Solve the curve equation for y in terms of x; that has two roots, one for the upper part of the curve, and one for the lower part. For each part you can substitute in your formula for y = y(x), to get a squared distance $s = x^2 + y^2$ that is a function of $x$ alone, then use standard calculus methods to find maxima and minima for each of the two parts.
(2) Alternatively, you can figure out how to rotate to a new (orthogonal) coordinate system $(X,Y)$, so that when you substitute in $x = a_1 X + a_2 Y, y=b_1 X + b_2 Y$ you get a new curve $F(X,Y) = 4$ that has no cross-product terms $XY$ in it. For an orthogonal transformation we have $x^2+y^2 = X^2 + Y^2$, so the problem is to maximize/minimize $S = X^2 + Y^2$ on the curve $F(X,Y) = 4$. It is solvable by inspection, needing no calculus at all.

 

[WWGD]

I don't know the Lagrange method and i am sure it's not to be used, since this chapter is the third in the book, and the book started from "scratch", it taught systematically, and it didn't teach it.
$$s=x^2+y^2~\rightarrow~s'=2x+2yy'$$
I insert $~y'=-\frac{x}{y}~$: s'=0
So the distance is a constant and the equation is a circle.
$$5x^2-6xy+5y^2=4~\rightarrow~x^2+y^2=\frac{4+6xy}{5}$$
I don't know to get the radius and the center, but the circle's center isn't at the origin, so i need to find the line from the center to the origin and then intersect it with the circle to find the point closest.

The curve would be a circle only if $(6)xy$ is constant , so that $x^2+y^2$ is itself constant. You don't have any condition guaranteeing this.

 

[WWGD]

The curve is not a circle; it is a tilted ellipse.

, so the problem is to maximize $S = X^2 + Y^2$ on the curve $F(X,Y) = 4$. It is solvable by inspection, needing no calculus at all.

Minimize S?

 

[Ray Vickson]

Minimize S?

Minimum also solvable by inspection. Original message edited to include this.

 

[WWGD]

Minimum also solvable by inspection. Original message edited to include this.

I agree, but, aren't we trying to find the minimal distance to the origin?

 

[Ray Vickson]

I agree, but, aren't we trying to find the minimal distance to the origin?

Minimizing/maximizing $S = x^2 + y^2$ is equivalent to minimizing/maximizing $\sqrt{S} = \sqrt{x^2 + y^2} = \text{distance to origin}.$ However, the squared distance problem tends to be a bit easier, analytically at least.

 

[Karol]

$$5x^2-6xy+5y^2=4~\rightarrow~5y^2-(6x)y+(5x^2-4)=0$$
$$y=\frac{6x\pm\sqrt{36x^2-20(5x^2-4)}}{10}$$
$$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]=-\frac{366}{25}x^2+12x\sqrt{5-4x^2}+20$$
$$s'_{+y}=-\frac{732}{25}x+12\frac{5-8x^2}{\sqrt{5-4x^2}}$$
$$s'_{+y}=0:~~\frac{x\sqrt{5-4x^2}}{5-8x^2}=\frac{275}{732}$$
I am sure i am wrong

 

[ehild]

$$5x^2-6xy+5y^2=4~\rightarrow~5y^2-(6x)y+(5x^2-4)=0$$
$$y=\frac{6x\pm\sqrt{36x^2-20(5x^2-4)}}{10}$$
$$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]=-\frac{366}{25}x^2+12x\sqrt{5-4x^2}+20$$

I am sure i am wrong

The last row is wrong. You miss a square, and made some other mistakes. The denominator 25 divides the whole expression in the numerator.
$$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]^2$$

 

[ehild]

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.

 

[StoneTemplePython]

One more approach:

Draw the picture.

apply $GM \leq AM$

$xy = (x^2 y^2 )^{\frac{1}{2}} \leq \gamma (x^2 + y^2)$

where $\gamma$ is some positive scalar we don't need to worry about.

Via two different domain restrictions, you get conditions for maximum distance $r^2$ and minimum distance $r^2$. (I.e. there are 4 quadrants in the graph, but you only need to worry about 2 of them, as symmetry then covers the other two.)

 

[Karol]

$$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]^2$$
$$s_{+y}=\frac{6x+\sqrt{36x^2-100x^2+80}}{10}=x^2+\frac{1}{25}(12x\sqrt{5-4x^2}+20-7x^2)$$
$$s'_{+y}=2x+\frac{1}{25}\left[ 12\left( \sqrt{5-4x^2}-\frac{8x^2}{\sqrt{5-4x^2}} \right)-14x \right]$$
$$s'_{+y}=\frac{36}{25}x+\frac{12}{25}\cdot\frac{5-12x^2}{\sqrt{5-4x^2}}$$
That doesn't give an answer for $~s'_{+y}=0$

 

[epenguin]

Er, #2 tells you the thing you have to minimise. Can you see that thing somewhere in the question?

 

[Karol]

I minimize the distance $~s=x^2+y^2$. in which question do you mean i don't see the distance? in my answer in #15?
The original question sure asks about the distance.
I correct a little mistake:
$$s'_{+y}=\frac{36}{25}x+\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}$$
But still it gives negative under the root, when i try to solve a quadratic equation

 

[ehild]

I minimize the distance $~s=x^2+y^2$. in which question do you mean i don't see the distance? in my answer in #15?
The original question sure asks about the distance.
I correct a little mistake:
$$s'_{+y}=\frac{36}{25}x+\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}$$
But still it gives negative under the root, when i try to solve a quadratic equation

I got positive under the square root. Show your work.

 

[epenguin]

I minimize the distance $~s=x^2+y^2$. in which question do you mean

This question

Homework Statement

View attachment 215800

The equation that appears in the first line of the question #1 I don't seem to be able to reproduce here.

 

[Karol]

$$s'_{+y}=\frac{36}{25}x+\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}$$
$$\frac{36}{25}x=\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}$$
$$3x=\frac{5-8x^2}{\sqrt{5-4x^2}}~\rightarrow~3x\sqrt{5-4x^2}=5-8x^2$$
I make a square of each side:
$$9x^2(5-4x^2)=25-120x^2+64x^4$$
$$45x^2-36x^4=25-120x^2+64x^4$$
$$84x^2=100x^4+25~\rightarrow~100x^4-84x^2+25=0$$
$$x^2=a,~\frac{84+\sqrt{7056-10,000}}{200}$$

 

[ehild]

$$s'_{+y}=\frac{36}{25}x+\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}=0$$
$$\frac{36}{25}x=\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}$$

The second equation is wrong

$$3x=\frac{5-8x^2}{\sqrt{5-4x^2}}~\rightarrow~3x\sqrt{5-4x^2}=5-8x^2$$
I make a square of each side:
$$9x^2(5-4x^2)=25-120x^2+64x^4$$
$$45x^2-36x^4=25-120x^2+64x^4$$

Wrong. Is 2x5x8=120?

 

[Karol]

$$\frac{36}{25}x=\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}~\rightarrow~-\frac{36}{25}x=\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}$$
$$-3x=\frac{5-8x^2}{\sqrt{5-4x^2}}~\rightarrow~-3x\sqrt{5-4x^2}=5-8x^2$$
$$9x^2(5-4x^2)=25-80x^2+64x^4~\rightarrow~100x^4-125x^2+25=0$$
$$x^2=a,~~\frac{125+\sqrt{15,625-10,000}}{200}=\frac{125+75}{200}=1~~\rightarrow~x=1$$
$$5y^2-6y+1=0,~~y_1=1,~~y_2=\frac{1}{2}$$
These x and y are for the upper part of the ellipse, how will i decide whether it's y₁ or y₂? and also i took the positive x, but $~x^2=a$

 

[ehild]

$$9x^2(5-4x^2)=25-80x^2+64x^4~\rightarrow~100x^4-125x^2+25=0$$

The equation is equivalent to $$4x^4-5x^2+1=0$$, and there are two roots for x². Find the corresponding y-s and check, if they are really roots of the original equation. Find the x,y pairs which result minimum s.

 

[Karol]

The pairs (1,1) and (-1,-1) give true for the original equation.
$y'=-\frac{x}{y}~$ and $~y'(1,1)=0$, deflection point. $~y'(-1,-1)=2~\rightarrow~$ minimum

 

[ehild]

The pairs (1,1) and (-1,-1) give true for the original equation.
$y'=-\frac{x}{y}~$ and $~y'(1,1)=0$, deflection point. $~y'(-1,-1)=2~\rightarrow~$ minimum

I don't follow you.
You have to minimize s=x²+y².
X=1 y=1 gives s=1. What about the other root x²=1/4?

 

[epenguin]

I expect you will get there in the end. But I do not think that your way of doing it is what the questioners want.
Sometimes these training questions require indeed a lot of slogging, but more often they are not but rather they want you to find the best way.
This question can be done either by little thought and hard calculations, or hard thought and easy calculations.
Even Outside textbook problems scientists and mathematicians look for simplifications of the problems.

I have managed to do this, that did take me some time - there are some traps or sticking places, and I am not yet 100 % satisfied with my formulation, bit I've got the answer. (May be related to same ideas as #12, but I do it staying in the original rectangular Cartesian coordinates.)

Useless to give it to you if you don't even take the first step, which is to answer my question #15 - for this also look back at your #3 .

 

[Karol]

$$s=x^2+y^2=x^2+[y(x)]^2~\rightarrow~4x^4-5x^2+1=0~\rightarrow~x_{1,2}=\pm 1$$
$$x_1=1~\mbox{in}~5x^2-6xy+5y^2=4:~y_{1,2}=1,\frac{1}{2}$$
$$(1,1)~\mbox{in}~5x^2-6xy+5y^2=4:~~\mbox{True}$$
$$\left( 1,\frac{1}{2} \right)~\mbox{in}~5x^2-6xy+5y^2=4:~\mbox{False}$$
$$x_2=-1~\mbox{in}~5x^2-6xy+5y^2=4:~y_{1,2}=-1,-\frac{1}{2}$$
$$(-1,-1)~\mbox{in}~5x^2-6xy+5y^2=4:~\mbox{True}$$
$$\left(1,\frac{1}{2} \right)~\mbox{in}~5x^2-6xy+5y^2=4:~\mbox{False}$$
So there are only two couples: (1,1) and (-1,-1) that make the original equation true.
I can't talk only about x2=1/4 because there should be couples, points.

 

[ehild]

So there are only two couples: (1,1) and (-1,-1) that make the original equation true.
I can't talk only about x2=1/4 because there should be couples, points.

You certainly meant x²=1/4, x=±1/2, other solutions of the equation $100x^4-125x^2+25=0$, with the possible y values. You should check which couples mean the points closest to the origin.

 

[ehild]

This is the graph of the curve 5x²-6xy-5y²=4. Are the points (1,1) and (-1,-1) really closest to the origin?

 

[epenguin]

Well I thought all along it would have been helpful to you to draw a picture. You can’t do this

- without numbers that come from a solution, But you can sketch something like this.
This would also help you see when you have some mistakes.
You have done a lot of work but I just can’t tell how close you are, because you produce just a series of equations with no explanation of you are doing, no narrative.

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?

Answering these questions can lead you to the answer with far less calculation than what you have been doing.

 

[epenguin]

You are probably quite close to answer. This is what your solution gives.

Almost certain you have had an equality of squares and have made the choice of square roots between + and - that gives you a maximum instead of minimum. Easy mistake to make.

 

[Karol]

(2) Alternatively, you can figure out how to rotate to a new (orthogonal) coordinate system $(X,Y)$, so that when you substitute in $x = a_1 X + a_2 Y, y=b_1 X + b_2 Y$ you get a new curve $F(X,Y) = 4$ that has no cross-product terms $XY$ in it. For an orthogonal transformation we have $x^2+y^2 = X^2 + Y^2$, so the problem is to maximize/minimize $S = X^2 + Y^2$ on the curve $F(X,Y) = 4$. It is solvable by inspection, needing no calculus at all.

Rotated coordinate sysytem:
$$X=x(\cos\alpha) + y(\sin\alpha)=ax+\sqrt{1-a^2}\cdot y$$
$$Y=x(-\sin\alpha) + y(\cos\alpha)=-\sqrt{1-a^2}\cdot x + ay$$
Because a horizontal ellipse is $~k_1 x^2+k_2 y^2=1~$:
$$5x^2-6xy+5y^2=4~\rightarrow~\frac{5}{4}x^2-\frac{3}{2}xy+\frac{5}{4}y^2=1$$
$$\frac{5}{4}x^2-\frac{3}{2}xy+\frac{5}{4}y^2=\frac{5}{4}[ax+\sqrt{1-a^2}\cdot y]+\frac{5}{4}[\sqrt{1-a^2}\cdot x+ay]$$
This is the equation in order to find the $~a=\cos\alpha~$, but the a "disappeared":
$$\frac{5}{4}x^2-\frac{3}{2}xy+\frac{5}{4}y^2=\frac{5}{4}(x^2+y^2)$$

 

[Nidum]

Isn't this problem easily solvable by considering a radial line with one end centred on the origin and the other end intercepting the problem curve . Conceptually sweep the line through a range of angles and calculate the length of the line between origin and intercept . Determine angles where line length is a minimum .

Edit : Essentially as proposed by @ehild in post #12 .

 

[epenguin]

It seemed to me that you were nearly there before, and it was just a question of taking an alternative root, after all you got a maximum and getting the minimum is essentially part of the same calculation. So I wonder why you don’t complete that.

There is an alternative quite easy calculation that I keep hinting at as well.

 

[Karol]

It seemed to me that you were nearly there before, and it was just a question of taking an alternative root, after all you got a maximum and getting the minimum is essentially part of the same calculation. So I wonder why you don’t complete that.

I didn't complete since i discovered the mistake in the calculation and found that (0.5,0.5) solves also, and it's easy to compare the distance to the origin of the 2 solutions (1,1) and (0.5,0.5).
Now i want to solve in different methods and want to find the angle with which to rotate the axes like Ray said

 

[Ray Vickson]

I didn't complete since i discovered the mistake in the calculation and found that (0.5,0.5) solves also, and it's easy to compare the distance to the origin of the 2 solutions (1,1) and (0.5,0.5).
Now i want to solve in different methods and want to find the angle with which to rotate the axes like Ray said

Just substitute $x = X \cos \theta - Y \sin \theta$ and $y = X \sin \theta + Y \cos \theta$ into your function $f(x,y) = 5(x^2+y^2) -6 xy$, to get a function $F(X,Y)$. Determine $\theta$ by requiring that there be no product term $X Y$ in the final $F$.