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Geometry problem

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    #11
    Originally posted by OwlHoot View Post
    How is the curve represented?

    * Explicitly, via an equation of the form y = f(x), e.g. y = e^x - 5

    * Implicitly, via f(x, y) = 0, e.g. x^3 y^2 + cos(x) log(y) = 0, from where one cannot necessarily derive an explicit finite expression for either x or y

    * Parametrically, via x = u(t), y = v(t), e.g. x = e^(-2t), y = e^(-3t)
    It doesn't matter. All that matters is the tangent at the chosen point.

    Originally posted by courtg9000 View Post
    By getting this forum to help with your kids maths homework. Ofsted has the following to say about your home schooling efforts!
    Very good. It's for an undergraduate science degree, and the professor was quite happy to accept a trial and error approach. However, my daughter is interested in how to calculate.

    Originally posted by BlasterBates View Post
    Yes but you simply set y=0 in whatever equation you figure out. You have the tangent, now you find the circle with y=0.

    If you didn't have the circle constrained you would have an infinite number of solutions.
    Looks promising.
    Down with racism. Long live miscegenation!

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      #12
      Originally posted by NotAllThere View Post

      It doesn't matter. All that matters is the tangent at the chosen point. ..
      As the circle centre is constrained to be on a line (the Y axis) then what you want is the normal to the curve at that point, i.e. the line perpendicular to the tangent at the point.

      Then the centre of the circle will be where this normal and the line/axis intersect, and the radius is the distance between that point of intersection and the point on the curve.

      (If the normal to the curve at the point is parallel to the line/axis then the radius will be infinite.)

      If the direction cosines of the tangent are a & b, then those of the normal are -b & a
      Work in the public sector? Read the IR35 FAQ here

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        #13
        Originally posted by NotAllThere View Post
        Given a continuous, differentiable curve, find the mid point and radius of a circle that bisects the y-axis and is tangential to the curve at a specific point on the curve.

        It looks something like the blue circle in this diagram. These are Mohr's circles.



        We know the point where the circle touches the curve, we know the gradient at that point, we know that the circle must be perpendicular to the x axis at y = 0. I think that's sufficient information. But how to calculate it? Geometrically or, more interesting, analytically?
        Are you trying to use geometry to justify the EU's raid on our vaccines by comparing their curves to our?!?!?

        I think you'll find this sums it up better?

        Define Ursula von der Leyen
        Answer :
        What happens in General, stays in General.
        You know what they say about assumptions!

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          #14
          Turns out it's quite easy. We know the point of intersection and the tangent there. So figure out the (unique) line perpendicular to it. The length of the line to the x axis is the radius, and the point of intersection with the x axis is the centre of the circle.
          Down with racism. Long live miscegenation!

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            #15
            Originally posted by NotAllThere View Post
            Turns out it's quite easy.
            You lost me right after that...

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              #16
              Originally posted by NotAllThere View Post

              Turns out it's quite easy. We know the point of intersection and the tangent there. So figure out the (unique) line perpendicular to it. The length of the line to the x axis is the radius, and the point of intersection with the x axis is the centre of the circle.
              Exactly what I said earlier
              Work in the public sector? Read the IR35 FAQ here

              Comment


                #17
                Originally posted by OwlHoot View Post
                As the circle centre is constrained to be on a line (the Y axis) then what you want is the normal to the curve at that point, i.e. the line perpendicular to the tangent at the point.

                Then the centre of the circle will be where this normal and the line/axis intersect, and the radius is the distance between that point of intersection and the point on the curve.

                (If the normal to the curve at the point is parallel to the line/axis then the radius will be infinite.)

                If the direction cosines of the tangent are a & b, then those of the normal are -b & a
                Originally posted by OwlHoot View Post
                Exactly what I said earlier
                Indeed it was. Doh!
                Down with racism. Long live miscegenation!

                Comment


                  #18
                  Originally posted by OwlHoot View Post
                  Exactly what I said earlier
                  That's why is was so easy.

                  Comment


                    #19
                    Originally posted by AtW View Post
                    You lost me right after that...
                    The rest of it was a bit of a car crash.. So thought you might have understood that
                    'CUK forum personality of 2011 - Winner - Yes really!!!!

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