# Homework Help: Graph drawing—Finding the points on a curve that are nearest to the origin

1. Nov 29, 2017

### Karol

1. The problem statement, all variables and given/known data

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

3. 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?

Last edited by a moderator: Nov 29, 2017
2. Nov 29, 2017

### Dick

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?

Last edited: Nov 29, 2017
3. Nov 29, 2017

### 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.

4. Nov 29, 2017

### Ray Vickson

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.

Last edited: Nov 29, 2017
5. Nov 29, 2017

### WWGD

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.

6. Nov 29, 2017

### WWGD

Minimize S?

7. Nov 29, 2017

### Ray Vickson

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

8. Nov 29, 2017

### WWGD

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

9. Nov 29, 2017

### Ray Vickson

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.

10. Nov 30, 2017

### 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

11. Nov 30, 2017

### ehild

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

12. Nov 30, 2017

### 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.

13. Nov 30, 2017

### 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.)

14. Nov 30, 2017

### 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 $~s'_{+y}=0$

15. Nov 30, 2017

### epenguin

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

16. Nov 30, 2017

### 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?
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

17. Dec 1, 2017

### ehild

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

18. Dec 1, 2017

### epenguin

This question

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

Last edited: Dec 1, 2017
19. Dec 1, 2017

### 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:
$$9x^2(5-4x^2)=25-120x^2+64x^4$$
$$45x^2-36x^4=25-120x^2+64x^4$$
$$84x^2=100x^4+25~\rightarrow~100x^4-84x^2+25=0$$
$$x^2=a,~\frac{84+\sqrt{7056-10,000}}{200}$$

20. Dec 1, 2017

### ehild

The second equation is wrong
Wrong. Is 2x5x8=120?

21. Dec 1, 2017

### 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 $~x^2=a$

22. Dec 2, 2017

### ehild

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.

23. Dec 2, 2017

### 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

24. Dec 2, 2017

### ehild

You have to minimize s=x2+y2.
X=1 y=1 gives s=1. What about the other root x2=1/4?

Last edited: Dec 2, 2017
25. Dec 2, 2017

### 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 .

Last edited: Dec 2, 2017