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Two curves with two shared tangents

  1. Feb 27, 2006 #1
    I'm stuck on this problem. I am hoping someone can walk me through it or get me past my choking point. The problem states:

    Two equations have two shared tangent lines between them. Find the equations of these tangent lines analytically.

    [tex]g(x)=x^2[/tex]
    [tex]f(x)=-x^2+6x-5[/tex]

    The first step I took was to define what is needed for two curves to share a single tangent line:

    -[tex]g'(x_0)=f'(x_1)[/tex] slopes at two different x-values must be the same
    -the tangent line to [tex](x_0,y_0)[/tex] must pass through the point [tex](x_1,y_1)[/tex] on the other curve with the same slope

    I found the general equation for the tangent of a line [tex]f(x)[/tex] at [tex]x[/tex] to be [tex]f'(x)k-xf'(x)+f(x)[/tex] where [tex]k[/tex] is to be substituted for [tex]x[/tex] in the final equation.

    I did this with both of the functions in the problem, and set them equal. My final result for this was [tex]4xk-2x^2=6k-5[/tex] Assuming you leave k in there and just try to get the sides looking equal, I got nowhere. After substituting x for k, there is no solution.
     
    Last edited: Feb 27, 2006
  2. jcsd
  3. Feb 27, 2006 #2

    0rthodontist

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    It is easier to approach it from the perspective of the number of solutions of g(x) - L(x) = 0 and L(x) - f(x) = 0. If both equations have one solution (which since they are both quadratic means that b^2 - 4ac = 0) then L(x) is tangent to both curves. There are two solutions for m and b of L(x) = mx + b.
     
    Last edited: Feb 27, 2006
  4. Feb 27, 2006 #3
    Could you go through that approach? I sort of understand your logic, and sort of don't. I don't understand what L(x) is doing when those two equations can be simplified to g(x)-f(x)=0.
     
  5. Feb 27, 2006 #4

    0rthodontist

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    L(x) is the equation of the line, L(x) = mx + b for m and b that you are trying to determine.

    For example this is what you do for the first equation, g(x) - L(x) = 0
    x^2-(mx + b) = 0
    x^2 - mx - b = 0
    This equation has a unique solution iff m^2 + 4b (the expression under the radical sign in the quadratic formula) = 0. So we have m^2 + 4b = 0, and we can work with L(x) - f(x) = 0 in a similar manner to get another equation involving m and b.

    The idea is that if g(x) is tangent to L(x), then g(x) - L(x) has exactly one root (only true because g(x) is quadratic, as it is here). Drawing a picture can make that clear.

    You can't simply combine the equations because you are not actually stating that either g(x) - L(x) = 0 or L(x) - f(x) = 0 is true. You are just setting them equal to 0 because you need to find how many roots each has.
     
    Last edited: Feb 27, 2006
  6. Feb 28, 2006 #5

    VietDao29

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    What's that for? One can differentiate f(x) with respect to x to get f'(x) = 2x, then use the tangent line equation at (x0, y0):
    y - y0 = f'(x0)(x - x0), to arrive at the equation:
    y = 2x0(x - x0) + y0 = 2x0x - 2x02 + x02 = 2x0x - x02.
    -----------------------
    You can also approach it from this way:
    Say P(x0, y0), P is on the graph of f(x), and Q(x1, y1), Q is on the graph of g(x) share the same tangent line.
    So it means that:
    [tex]\left\{ \begin{array}{l} f'(x_0) = g'(x_1) \\ \frac{y_0 - y_1}{x_0 - x_1} = f'(x_0) \end{array} \right.[/tex]
    Can you see why? Can you go from here? :)
     
  7. Feb 28, 2006 #6
    No, I don't think I can. But that seems to make a lot more sense than the L(x) stuff.

    I still don't see how to solve either of those methods. The first method you described is essentially what I have been trying for. The tangent equation for f(x) (the one you described was actually g(x)) ends up being [tex]y-y_1=f'(x_1)(x-x_1)[/tex] or [tex]y=(-2x_1+6)(x-x_1)+y_1[/tex] Can you check that?

    Even once that's done, I still don't understand how to solve for distinct tangent lines.
     
    Last edited: Feb 28, 2006
  8. Mar 1, 2006 #7

    VietDao29

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    It can be simplified to:
    [tex]y = (-2x_1 + 6) (x - x_1) + y_1 = -2x_1 x + 2x_1 ^ 2 + 6x - 6x_1 - x_1 ^ 2 + 6x_1 - 5 = (-2x_1 + 6) x + x_1 ^ 2 - 5[/tex]
    By the way, the equation f'(x0) = g'(x1) means that the tangent line of f(x) at x0, and the tangent line of g(x) at x1 are parallel. Do you know why???
    The next equation:
    [tex]\frac{y_1 - y_0}{x_1 - x_0} = f'(x_0)[/tex] is to assure that Q is on the tangent line of f(x) at x0, and P is on the tangent line of g(x) at x1. Or in other words, PQ is the tangent line of f(x) and g(x) at x0, and x1 respectively.
    I'll start of for you:
    f'(x0) = g'(x1)
    <=> -2x0 + 6 = 2x1
    <=> x0 = 3 - x1
    Now this means that the tangent line of g(x) at x1 is parallel to the tangent line of f(x) at 3 - x1. Do you see why?
    Now you can substitute that to the equation:
    [tex]\frac{y_1 - y_0}{x_1 - x_0} = f'(x_0)[/tex], and solve for x1. What's y1 in terms of x1, what's y0 in terms of x0? Note that P, and Q are on the graph of f(x), and g(x) respectively. Can you go from here? :)
     
    Last edited: Mar 1, 2006
  9. Mar 1, 2006 #8

    0rthodontist

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    Didn't I just explain it to KingNothing? It is a faster and simpler technique in this situation than using calculus. Again, the idea is, the following two statements are equivalent when f(x) is quadratic and L(x) is a line:
    "L(x) is tangent to f(x)"
    "f(x) - L(x) has exactly one root"
     
    Last edited: Mar 1, 2006
  10. Mar 1, 2006 #9

    VietDao29

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    So I'll get m2 = -4b for L(x). Now I define K(x) in the same manner that's the tangent line to f(x), K(x) = lx + u. Then I'll get:
    (6 + l)2 = 20 - 4u. Then how can I go from there?
    Am I missing something?
     
  11. Mar 1, 2006 #10

    0rthodontist

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    It has to be the same line that is tangent to both quadratics. L(x) = mx+b for both quadratics. The other equation you get is 16-12m+m^2-4b = 0
     
  12. Mar 1, 2006 #11

    VietDao29

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    Yes, it's simplier than my way. :rolleyes: Thanks for pointing that out. :)
    KingNothing, now you have 2 ways to tackle this problem. Both are good, I think. Can you go from here? :)
     
    Last edited: Mar 1, 2006
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