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Tangent line toa curve that minimizes the area of a triangle

  1. Jan 6, 2008 #1
    Sorry if the title is a bit vague :/

    The Problem: Find the point on the parabola y=1-x^2 at which the tangent line cuts from the first quadrant a triangle with the smallest area.


    Relevant Equations: y = 1-x^2 ; y' = -2x ; A= 1/2bh



    I'm basically stuck near square one, I found this site through a google search of the above problem and found a similar topic here, but it wasn't quite the same (or at least I didn't see how to relate my problem to it) so I'd like to ask someone to help me out here :)

    What I have so far is that the derivative of the tangent line to the curve is -2x. Where should I go from here?
     
  2. jcsd
  3. Jan 6, 2008 #2
    Aha

    Through a closer look at the problem and running through it a lot in my mind and looking again at the other topic I think I have figured it out. Thanks to anyone who was or is working on replying (if any, lol). All I had to do was change A= 1/2 bh to A= 1/2 xy and I know that any point on the parabola is P(p,1-p^2) so I was able to go from here :) Always makes me feel good when I figure things out :D
     
  4. Jan 6, 2008 #3
    Ah! I was trying for ya on/off :p GJ! At least now I've gained another way of approaching random problems.
     
  5. Jan 6, 2008 #4
    hmmm, i may be stuck again, if i find A(p) by setting it equal to 1/2 * (p^2+1)/2p * (p^2+1) i get that A(p) = p^4+2p^2+2/4p and by differentiating this to find its critical numbers to find a min, i get -4p^2+40p+8/4p^4. Now, in order for this to be 0, -4p^2+40p+8 has to be 0. I've tried to use the quadratic formula for this but I keep getting wierd roots that don't even make it 0. Am I doing something wrong? sorry for the early cancellation lol :/ :stumped:

    I'll keep trying but any help is appreciated :)
     
  6. Jan 6, 2008 #5
    I'm not sure what you did, but this is how I'm starting out.

    [tex]A=\frac 1 2 xy[/tex]
    [tex]y=1-x^2[/tex]

    [tex]A(x)=\frac 1 2 x(1-x^2)[/tex]
    [tex]A(x)=\frac 1 2 (x-x^3)[/tex]

    [tex]A'(x)=\frac 1 2 (1-3x^2)[/tex]

    Setting it equal to 0, and then plugging in the values found into my y-eq.
     
  7. Jan 6, 2008 #6
    Wow i completely overlooked that :/ I used a wrong reference point, man i did a lot of extra worthless work :/ at least it was kind of fun and interesting lol Thanks a ton I'll see if I can finish with that, I'm working on a couple other problems now, 3 to be exact and I think I'm doing ok, only 1 is giving me problems right now :/ maybe you can help :)

    it appears simple : find highest and lowest points of curve x^2+xy+y^2=12

    By implicit differentiation i get that y' = 2x+y / 2y+x . Is that right? if so I don't know what i can do to move on with the problem. I suppose I have to make it zero somehow ?? its always the simple concepts that get me...
     
  8. Jan 6, 2008 #7
    Lol, at least you got both of us started on that problem. I was lost on what to do :-]

    I think you forgot the product rule for your middle term, xy. You weren't given any points?

    Well by just plugging in values, the points are P(2,2) b/c that equals 12.
     
    Last edited: Jan 6, 2008
  9. Jan 6, 2008 #8
    no the exact wording is: "Find the highest and lowest points on the curve x^2+xy+y^2=12"

    I did D(x) (x^2 + xy + y^2) = D(x) (12)

    = 2x + y +xy' + 2yy' = 0

    the y + xy' is a result of the product rule (unless i messed up)

    =2yy'+xy' = 2x+y

    =y'(2y+x) = 2x+y

    = y' = 2x+y / 2y+x
     
  10. Jan 6, 2008 #9
    Ah yes you're right, I messed up sorry.
     
  11. Jan 6, 2008 #10
    heh np, i've done worse :P
     
  12. Jan 6, 2008 #11
    Well the way I went about this is, that when y=0 for f', then that tells me that my f is at a max/min, so [tex]y'=\frac{2x+y}{2y+x}[/tex]

    set y=0, that gives me y'=2, which is the same as P(2,2) from which our original equation equals 12.
     
  13. Jan 6, 2008 #12
    ah, didn't think of that. what about the other extrema? set x=0 ? that would give me y=1/2 no?
     
  14. Jan 6, 2008 #13
    I'm not sure about that, I wish I could plug it into my calculator. LOL
     
  15. Jan 6, 2008 #14
    lol same, quick question, from y'=2, how did you get P(2,2) I may just be overlooking that aswell. I thought i got it but now i realized i messed up.
     
  16. Jan 6, 2008 #15
    Ah, I just read my Calculus book and we need to do it over again. It says to set y'=0, and basically just set the numerator equal to 0.

    [tex]y'=\frac{2x+y}{2y+x}[/tex]

    [tex]y'=0=2x+y[/tex]

    [tex]y=-2x[/tex]

    So plug that back into our original equation ...

    [tex]x^2+x(-2x)+(-2x)^2=12[/tex]

    Set it equal to 0 and solve for x. Then plug it back into the numerator of our y'.
     
    Last edited: Jan 6, 2008
  17. Jan 6, 2008 #16
    lol ok, from where? :)
     
  18. Jan 6, 2008 #17
    Ok sorry, refresh!!! I just updated after figuring out what my book did.
     
  19. Jan 6, 2008 #18
    so 3x^2-12=0 ?
     
  20. Jan 6, 2008 #19
    wait wait wait duh, can't i just go x^2=4 and my roots are +/- 2 ??
     
  21. Jan 6, 2008 #20
    Yes, those are the roots and just plug back into the numerator of y'.

    Are there answers and what book is this from?
     
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