Reply to: Geometry problem

[northernladuk]

You lost me right after that...

The rest of it was a bit of a car crash.. So thought you might have understood that

[Paralytic]

Exactly what I said earlier

That's why is was so easy.

[NotAllThere]

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

Exactly what I said earlier

Indeed it was. Doh!

[OwlHoot]

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

[AtW]

Turns out it's quite easy.

You lost me right after that...

[NotAllThere]

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.

[MarillionFan]

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 :

[OwlHoot]

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

[NotAllThere]

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.

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.

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.

[courtg9000]

By getting this forum to help with your kids maths homework. Ofsted has the following to say about your home schooling efforts!

[northernladuk]

[AtW]

Who's got a geometry problem?

[BlasterBates]

It isn't, because in this problem the circle centre is constrained to be on the y-axis, which isn't generally the case for an osculating circle.

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.

[OwlHoot]

The answer to your question is here:

The equation of the tangent to a circle - Equations of circles - Higher - CCEA - GCSE Maths Revision - CCEA - BBC Bitesize

It isn't, because in this problem the circle centre is constrained to be on the y-axis, which isn't generally the case for an osculating circle.

[TestMangler]

Scooter says you should take a short position on that stock immediately !