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Math Help - Serious problem with common tangent

  1. #1
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    Serious problem with common tangent

    I would like to form common tangent equation for the functions f(x) and g(x) on the attachement. First of all I tried to form tangent equation for simple functions as y=x^2+2x+2 and y=-x^2-6x-14 but didn't manage. I am confused by now and any help would be appreciated.
    Maybe it would be enough to show the way to get tangent equation for y=x^2+2x+2 and y=-x^2-6x-14 by explaining step by step (not only operations)

    Thanks in advance.
    Attached Thumbnails Attached Thumbnails Serious problem with common tangent-puutuja.jpg  
    Last edited by totalnewbie; August 17th 2005 at 06:34 AM.
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  2. #2
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    The tangent to y = f(x) at the point (x0,y0) is the line through (x0,y0) with slope f'(x0): so y - f(x0) = f'(x0)(x-x0). So the tangent to y = log(x) is y-log(x0) = (1/x0)(x-x0). Similarly, the tangent to y = f(x) at the point (x1,y1) is y - f(x1) = f'(x1)(x-x1). So the tangent to y = x^2/2e is y - x1^2/2e = (x1/e)(x-x1).

    You want these to be the same line. So you need the slopes and intercepts to be the same: that is, x1^2/2e = 1/x0 (slope) and log(x0)-1 = x1^2/2e - x1^2/e. You have to solve these to find x0 and x1: that gives you the equation for the common tangent.
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  3. #3
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    I think this is right.
    1, y=x^2 + 2x + 2
    2, y=-x^2 - 6x - 14

    OK suppose that the point of tangency of 1, is located at x=a. We can now calculate the y coordinate as a^2 + 2a + 2

    The gradient function (y') of 1 is y' = 2x + 2. At x=a this becomes y'= 2a + 2

    The equation of any line with gradient m and passing through (x1, y1) is

    y - y1 = m(x-x1)
    or * y=mx -mx1 + y1
    At the point x=a the gradient m = 2a+2 x1 = a and y1 = a^2 + 2a +2
    Plugging this into * gives
    y=(2a+2)x - (2a+2)(a) + a^2 + 2a + 2
    simplifying
    y=(2a + 2)x -2a^2 - 2a + a^2 + 2a + 2
    y=(2a + 2)x -a^2 + 2

    Assume that the point of tangency on 2 is at x=b where y = -b^2 - 6b - 14
    The gradient function of 2 is y' = -2b - b. Like before we can use y - y1 = m(x - x1)

    y= (-2b - 6)x - (-2b - 6)(b) + (-b^2 - 6b - 14)
    y= (-2b - 6)x +2b^2 + 6b - b^2 - 6b - 14
    ** y= (-2b - 6)x + b^2 - 14

    You should realise that * and ** are equations of the same line and so we can equate coefficients.

    2a + 2 = -2b - 6
    or a = -b - 4

    - a^2 + 2 = b^2 - 14
    - a^2 + 16 = b^2

    Substituting
    -(-b - 4)(-b - 4) + 16 = b^2
    -(b^2 + 8b + 16) + 16 = b^2
    0=2b^2 +8b
    0=b^2 + 4b
    b = 0 or b = -4
    a= -4 or a=0

    It's now possible to sub in these values for x

    -b^2 - 6b - 14

    (0,-14) and (-4, -6)

    a^2+2a+2
    (-4, 10) and (0, 2)

    I feel sure I've made an error but I haven't time to sort it.
    Hope it helps.
    Last edited by Cold; August 18th 2005 at 12:12 AM.
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  4. #4
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    There's a further technique that works when one of the equations is of the form y = quadratic in x (which it is here of course). Let y - f(x1) = f'(x1)(x-x1) be the equation for the tangent to the first curve: write this as y=mx+c. The line y = mx+c is tangent to the curve y = g(x) = quadratic in x if and only if the equation mx + c = g(x), which is a quadratic in x, has a repeated root: that is, iff its discriminant (b^2-4ac) is zero.

    Example: y= f(x) = x^2 + 2x + 2 and y= g(x) = -x^2 - 6x - 14.
    The tangent y-y1 = f'(x1)(x-x1): y -(x1^2 + 2x1 + 2) = (2x1+2)(x-x1). Substitute y = (2x1+2)x + (x1^2+2x1+2) - (2x1+2)x1 = (2x1+2)x - x1^2 + 2. Substitute into second equation: (2x1+2)x - x1^2 + 2 = -x^2 - 6x - 14. As a quadratic in x this is x^2 + x(2x1+2+6) - x1^2+2 + 14. The discriminant is (2x1+8)^2 - 4(16-x1^2) = 4x1^2 + 32x1 + 64 - 64 + 4x1^2 = 8x1^2 + 32x1 and this = 0 when x1 = 0 or -4.
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  5. #5
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    Cold, I analyzed your solution but it's someway strange.
    I tried another way and according to the math program it is right.
    So,
    y=x^2+2x+2
    y=-x^2-6x-14
    The tangent we are going to find must be the same line which touch both parabola at different point.
    The First touching point is (x1;y1)
    The Second touching point is (x2;y2)
    In order to not make equations hard to understand, I do substitution:
    a=X1
    b=x2
    Because slope of the line is the same, I can equate the first derivatives of the functions, which means:
    2a+2=-2b-6
    2a+2b=-8
    a+b=-4
    b=-4-a
    Now I can declare y1 and y2 by using a
    y1=a^2+2a+2
    y2=-b^2-6b-14=-(-a-4)^2-6(-a-4)-14)=-a^2-2a-6
    Now I can form tangent of the first parabola, which means:
    y-y1=k(x-x1) where k is slope
    y-(a^2+2a+2)=(2a+2)(x-a)
    y-a^2-2a-2=(2a+2)x-2a^2-2a
    y=(2a+2)x-a^2+2
    As far as here, it was easy way. The following seems to me little bit wierd but nevertheless it brought me to the finish. Maybe some of you can explain it.
    I considered the second touching point (x2;y2) and did the following substitution:
    y2=(2a+2)x2-a^2+2
    Now I substituted x2 and y1 by using a to get an equation to find the value of a:
    -a^2-2a-6=(2a+2)(-a-4)-a^2+2
    2a^2+8a=0
    a=0
    a=-4
    It appears that there are two common tangent:
    IF a=0
    y1=0^2+2*0+2=2
    b=0-4=-4
    y2=-(-4^2)-6*(-4)-14=-6
    According to the formula x-x1/x2-x1=y-y1/y2-y1 I can substitute the values and get the equation y=2x+2
    IF a=-4
    y1=(-4)^2+2*(-4)+2=10
    b=-(-4)-4=0
    y2=-0^2-6*0-14=-14
    According to the formula x-x1/x2-x1=y-y1/y2-y1 I can substitute the values and get the equation y=-6x-14

    And FINALLY I add an attachment to show that the equations of the line are true.
    Attached Thumbnails Attached Thumbnails Serious problem with common tangent-jama.jpg  
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  6. #6
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    What program did you use to do the graph?
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  7. #7
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    Quote Originally Posted by Dustin
    What program did you use to do the graph?
    I think that you have no use if I say to you the program's name because it's name is Function (application file is function.exe and I dont even recall where I donwloaded it).
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  8. #8
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    Oh, thanks anyway.
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