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Common tangent problems

  1. Apr 23, 2014 #1
    I don't quite understand the context of common tangent problems.

    This is one of the problems I am trying to solve:

    Prove that there is a line that is a common tangent to the parabolas y = x2 and y2 = x.

    This is how I tried to solve it at least:

    y2 = x --> y = [itex]\pm[/itex][itex]\sqrt{x}[/itex]

    CASE I:
    y = [itex]\sqrt{x}[/itex]

    We have two derivatives:
    One is: 2x
    The other is: [itex]\frac{1}{2}[/itex]*[itex]x^{-1/2}[/itex]

    Set these equal to each other and find that x = ([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex]

    --> The tangent has the slope [itex]\frac{1}{2^{1/3}}[/itex]

    In this case, the tangents have the same slope for the same x-value --> It is NOT a common tangent.

    CASE II:
    y = -[itex]\sqrt{x}[/itex]

    Here we have that the tangent of y = -[itex]\sqrt{x}[/itex] has the negative slope of [itex]\sqrt{x}[/itex] for the same x-value.

    And for y = x2, the tangent has a same-value but negative slope for the negative x-value, that is y'(-x) = -y'(x)

    In our case this gives us that:
    The slope for y = -[itex]\sqrt{x}[/itex] at x = ([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex] is -[itex]\frac{1}{2^{1/3}}[/itex] and the slope for y = x2 at x = -([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex] is -[itex]\frac{1}{2^{1/3}}[/itex].

    Therefore, there is a line that is a common tangent for y = x2 and y2 = x

    Is this, the correct way to solve this problem or would you suggest some other way that is appliable for other problems of this sorts where you have to find common tangents...

    Thanks a whole lot in advance.

    Mr. Fest
    Last edited: Apr 23, 2014
  2. jcsd
  3. Apr 23, 2014 #2

    Simon Bridge

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    To understand the methods, you have to start from the definition of "common tangent" ... how do you tell when the definition has been met?

    It also helps to sketch the curves in question.

    ... they have the same x value - but, presumably, not the same y value?
    This sounds tedious.

    I think an all-in-one approach would be to find the equation of the line that is tangent to curve A at some point P and then find the point of intersection of that line with curve B ... call it Q. If PQ is a tangent to curve B, then you have found your common tangent.

    This approach can be used to construct an equation for the conditions that a particular point P lies on a common tangent.

    In your problems, no context has been supplied so there is nothing to understand about the context.
    This is a common situation in pure mathematics.
    Last edited: Apr 23, 2014
  4. Apr 24, 2014 #3
    Common tangent would be when two functions have the same slope for two different points (two different x-values), correct?

    Exactly, they do not have the same y-value.

    Would you care to elaborate? I think I understand what you mean. Are you telling me that I should for example for function A put the slope as (f(x)-f(a))/(x-a) and the actually calculate the derivative for the other function. Then set (f(x)-f(a))/(x-a) = derivative of the other function? (This presuming that we have gotten info for some specific points...)

    Am I correct in my ideas of the common tangent problem or am I completely off? Would appreciate any further help.
  5. Apr 24, 2014 #4

    Simon Bridge

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    Almost ... recall that a tangent is a line. It is possible for two different tangent lines to have the same slope. You are looking for the tangent line or lines that the two curves have in common: i.e. the tangent off one curve is also a tangent off the other one.

    If point P is on curve f(x) and point Q is on curve g(x) then:
    ... the line through P and Q is a common tangent to f(x) and g(x) iff that line is tangent to f(x) at P and tangent to g(x) at Q.

    You can see it clearly in the case of two circles: draw two circles ... now find all the straight lines that touch both circles at only one point each. There's four. Those lines are common tangents.
  6. Apr 24, 2014 #5
    Now I understand it better, thank you.

    I still find a bit troublesome to actually know how to find these points... because this means that the f'(x) = g'(x) but for different x-values. How can we find these different x-values?

    This might sound stupid, but drawing two circles I only find two lines that touch both circles at only one point each. Namely, one above and one under the circles.
  7. Apr 24, 2014 #6

    Simon Bridge

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    Two different points on a line.
    It helps to sketch the situation so you know whereabouts to expect to find the solutions.
    Note - P and Q can have the same x values if the curves happen to coincide at that point (i.e. P=Q)
    i.e. f(x)=x^2 and g(x)=x^3 have a common tangent at x=0.

    More generally:

    If the circles themselves are not touching or intersecting, then the other two common tangents form an X shape between them ;)
    Last edited: Apr 24, 2014
  8. Apr 24, 2014 #7
    Haha that is very true.

    Thank you very much for your help. Have an exam tomorrow and I'm at work at the moment. But will definitely revise when I get home.
  9. Apr 24, 2014 #8

    Simon Bridge

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    I modified post #6 to link you to a paper explaining how to find common tangents between quadratics.
    The approach should help you figure it out for other functions too - but afaik there is no universal formula for finding them, which is why these puzzles are good tests of your mathematical ability.

    In general you have to be prepared to play around a bit - and sketching the functions out is an essential skill.
  10. Apr 24, 2014 #9
    Thanks a lot!! :)
  11. Apr 24, 2014 #10

    Simon Bridge

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    Re. your original problem.
    Carefully sketch both curves on the same axes - one is a quadratic in x and the other a quadratic in y.
    ... using a ruler, find the common tangent line between the two curves: draw that line on your diagram.
    ... looking at that line - what does it's slope have to be?
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