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

  1. Nov 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Capture.JPG

    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. jcsd
  3. Nov 29, 2017 #2

    Dick

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    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
  4. Nov 29, 2017 #3
    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.
     
  5. Nov 29, 2017 #4

    Ray Vickson

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    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
  6. Nov 29, 2017 #5

    WWGD

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    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.
     
  7. Nov 29, 2017 #6

    WWGD

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    Minimize S?
     
  8. Nov 29, 2017 #7

    Ray Vickson

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    Minimum also solvable by inspection. Original message edited to include this.
     
  9. Nov 29, 2017 #8

    WWGD

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    I agree, but, aren't we trying to find the minimal distance to the origin?
     
  10. Nov 29, 2017 #9

    Ray Vickson

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    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.
     
  11. Nov 30, 2017 #10
    $$5x^2-6xy+5y^2=4~\rightarrow~5y^2-(6x)y+(5x^2-4)=0$$
    $$y=\frac{6x\pm\sqrt{36x^2-20(5x^2-4)}}{10}$$
    $$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]=-\frac{366}{25}x^2+12x\sqrt{5-4x^2}+20$$
    $$s'_{+y}=-\frac{732}{25}x+12\frac{5-8x^2}{\sqrt{5-4x^2}}$$
    $$s'_{+y}=0:~~\frac{x\sqrt{5-4x^2}}{5-8x^2}=\frac{275}{732}$$
    I am sure i am wrong
     
  12. Nov 30, 2017 #11

    ehild

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    The last row is wrong. You miss a square, and made some other mistakes. The denominator 25 divides the whole expression in the numerator.
    $$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]^2$$
     
  13. Nov 30, 2017 #12

    ehild

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    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.
     
  14. Nov 30, 2017 #13

    StoneTemplePython

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    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.)
     
  15. Nov 30, 2017 #14
    $$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]^2$$
    $$s_{+y}=\frac{6x+\sqrt{36x^2-100x^2+80}}{10}=x^2+\frac{1}{25}(12x\sqrt{5-4x^2}+20-7x^2)$$
    $$s'_{+y}=2x+\frac{1}{25}\left[ 12\left( \sqrt{5-4x^2}-\frac{8x^2}{\sqrt{5-4x^2}} \right)-14x \right]$$
    $$s'_{+y}=\frac{36}{25}x+\frac{12}{25}\cdot\frac{5-12x^2}{\sqrt{5-4x^2}}$$
    That doesn't give an answer for ##~s'_{+y}=0##
     
  16. Nov 30, 2017 #15

    epenguin

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    Er, #2 tells you the thing you have to minimise. Can you see that thing somewhere in the question? :oldwink:
     
  17. Nov 30, 2017 #16
    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
     
  18. Dec 1, 2017 #17

    ehild

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    I got positive under the square root. Show your work.
     
  19. Dec 1, 2017 #18

    epenguin

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    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
  20. Dec 1, 2017 #19
    $$s'_{+y}=\frac{36}{25}x+\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}$$
    $$\frac{36}{25}x=\frac{12}{25}\cdot\frac{5-8x^2}{\sqrt{5-4x^2}}$$
    $$3x=\frac{5-8x^2}{\sqrt{5-4x^2}}~\rightarrow~3x\sqrt{5-4x^2}=5-8x^2$$
    I make a square of each side:
    $$9x^2(5-4x^2)=25-120x^2+64x^4$$
    $$45x^2-36x^4=25-120x^2+64x^4$$
    $$84x^2=100x^4+25~\rightarrow~100x^4-84x^2+25=0$$
    $$x^2=a,~\frac{84+\sqrt{7056-10,000}}{200}$$
     
  21. Dec 1, 2017 #20

    ehild

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