Intersection of Two Curves: Do They Meet?

[PensNAS]

Homework Statement

Show that these curves do not intersect.

z=(1/a)(a-y)^2
y^2+z^2=a^2/4

Where a is the radius of the circle and other shape.

Homework Equations

There aren't any.

The Attempt at a Solution

I tried setting them equal to each other but got the equation, which doesn't simplify into anything solve able.

a^2/4-y^2=a^2-4ay+6y^2-4y^3/a+y^4/a^2

 

[Mark44]

Homework Statement

Show that these curves do not intersect.

z=(1/a)(a-y)^2
y^2+z^2=a^2/4

Where a is the radius of the circle and other shape.

Homework Equations

There aren't any.

The Attempt at a Solution

I tried setting them equal to each other but got the equation, which doesn't simplify into anything solve able.

a^2/4-y^2=a^2-4ay+6y^2-4y^3/a+y^4/a^2

I think that the easiest way to do this is to start with a sketch of the two curves. The 2nd equation is of a circle - what is its radius?

What sort of figure does the first equation represent?

 

[PensNAS]

The first equation looks similar to z=1/x. The radius of the circle is a/2. I've already sketched out the equations.

 

[Mark44]

The first equation looks similar to z=1/x.

No, not at all. The 1/a part is just a constant. The variable part, (a - y)², is what's important.

The radius of the circle is a/2. I've already sketched out the equations.

That's correct for the circle's radius.

 

[PensNAS]

It is a decreasing, concave curve is what I should have said. How does one show that there is not a solution for both equations?

 

[Mark44]

It is a decreasing, concave curve is what I should have said.

It is concave up, yes, but it's decreasing on part of its domain and increasing on the other part.

What geometric figure is y = K(a - x)²? (I changed the variables so that you might recognize what this is more easily.)

How does one show that there is not a solution for both equations?

Like I said before, a sketch of both curves is a good start. So far, you have only one of them.

 

[PensNAS]

It is a parabola. The book gives us the graph of both functions, but limits it to the first quadrant. I should have soda that too, my bad.

 

[Mark44]

You could try this: Write a function that represents the distance between an arbitrary point on the circle and one on the left side of the parabola. Find the minimum value of that function. If the minimum value is greater than zero, the two curves don't intersect.

Trying to solve the equation that you found seems like a dead end to me.

 

[PensNAS]

Am I wrong in thinking the formula for the distance between the functions is the top curve minus the bottom curve?

 

[Mark44]

Am I wrong in thinking the formula for the distance between the functions is the top curve minus the bottom curve?

Yes, you're wrong. That would give you the vertical distance between the two curves, which isn't the same as the distance I described.

A slightly different approach would be to find the point on the parabola that's closest to the origin. Then the line segment from the origin to the parabola would intersect the circle at some point, and you could easily find the distance between the circle and the parabola.

I'm sort of making this up as I go along, but I think this is a workable way to go.

 

[Mark44]

Offtopic, but do you happen to be in Pensacola?

 

[Dick]

Yes, you're wrong. That would give you the vertical distance between the two curves, which isn't the same as the distance I described.

A slightly different approach would be to find the point on the parabola that's closest to the origin. Then the line segment from the origin to the parabola would intersect the circle at some point, and you could easily find the distance between the circle and the parabola.

I'm sort of making this up as I go along, but I think this is a workable way to go.

Since the second equation is a circle centered at the origin, the distance from a point on the parabola to the origin is going to give you a fourth order equation, not too much different from what you get trying to intersect them. It might pay to be more clever. If pensNAS has got a good picture, think about drawing a tangent line to the circle that has the circle on one side and the parabola on the other. Then if you write down an equation for the tangent line and show it doesn't intersect the parabola, you would be done.

 

[PensNAS]

Thanks for the help! I actually went and talked to my professor, since this is part of a statics problem and not calculus. He was okay with just plotting a few points around where the curves get closest.

@Mark44, I'm from Pensacola.

 

[Dick]

Thanks for the help! I actually went and talked to my professor, since this is part of a statics problem and not calculus. He was okay with just plotting a few points around where the curves get closest.

@Mark44, I'm from Pensacola.

Ok. For the record, I think it's pretty easy to show that the tangent line to the circle passing through x=a/(2sqrt(2)), y=a/(2sqrt(2)) doesn't intersect the parabola. Since the two curves are on different sides of that line, they can't intersect.