Tough Problem

[Moose352]

The line y=mx+b intersects the parabola y=x^2 at two points A and B. Find the point P on arc AOB that maximizes the area of triangle APB.

I want to see how you guys solve this problem. I wrote an equation for shortest distance between a point on the parabola and the line, then tried to maximize it. The problem is, I end up trying to find the zeros of a third degree polynomial. I may be screwing up the simple stuff, but I don't know.

[pnaj]

Probably the thing to do is to derive an expression for the area of the triangle in terms of x, and then maximise that.

[HallsofIvy]

First clarify: does "arc AOB" mean the circular arc (most common meaning of arc) or the portion of the parabola from A to B?

[pnaj]

That's a bit picky, isn't it?

[AKG]

x_A = \frac{m + \sqrt{m^2 + 4b}}{2}

x_B = \frac{m - \sqrt{m^2 + 4b}}{2}

A(x_A, x_A ^2)\ \ B(x_B, x_B ^2)\ \ P(x_P, x_P ^2)

Area = \frac{1}{2} |\vec{AB} \times \vec{AP}|*

Express all this in terms of m, b, and x_P and then differentiate area with respect to x_P.

Doing this, and setting the derivative to zero, I get:

x_P = \frac{x_A + x_B}{2} = \frac{m}{2}

Therefore:

P(\frac{m}{2}, \frac{m^2}{4})

*EDITED: "det(...)" to "|...|"

[Moose352]

Thanks AKG. But, how did you end up with the formula for the area? I thought the magnitude of the cross product is equal to the area of the triangle * 2. I don't understand how you have the determinant.

By the way, I figured out that I was making a stupid mistake differentiating, and I managed to find the answer that way too. But your way is a lot nicer.

And halls, AOB i guess means the portion of the parabola.

Moose

[AKG]

Hmm, how did you end up with the formula for the area? Thanks again. I figured out that I was making a stupid mistake differentiating, and I managed to find the answer that way too. But your way is a lot nicer.

And halls, AOB i guess means the portion of the parabola.

Moose:) I imagine the problem would get very ugly if you didn't have that formula.

I can't remember the derivation, but the determinant of the cross product of two vectors gives the area of the parallelogram formed by those two vectors (of course, those vectors would be 2 of the 4 sides, but you should get the idea). The area of the triangle would just be half the area of that parallelogram. I'm at work right now, so I can't really try to figure out a derivation, but if you can't find one online I'll try to provide one (although I'm sure someone here will do that before I get a chance).

EDIT: Ignore most of the stuff above. I was a bit rusty with this stuff, so I looked it up on mathworld where their notation confused me, so what I wrote didn't make perfect sense. It's not the determinant of the cross product, but the magnitude of it which gives you the area of the parallelogram. If you have vectors \vec{u} = (u_x, u_y) and \vec{v} = (v_x, v_y) then the area of the parallelogram is:

A = |\vec{u} \times \vec{v}| = \det \left[\begin{array}{cc}u_x&u_y\\v_x&v_y\end{array}\right]

Now, how do we prove this?

Well, |\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin \theta, where \theta is the angle between the vectors. Now, it's not hard to see that if the absolute value of the cross product is what's given on the right, that we have the area of the parallelogram. If |\vec{u}| represents the base of the parallelogram, we can see with some trigonometry that |\vec{v}|\sin \theta is the height/altitude of the parallelogram, and clearly the product of these two values is the area. Now, how do we know that the equation at the start of this paragraph is true? Well, I suppose it would be messy, but you'd just have to show that:

u_xv_y - v_xu_y = |\vec{u}||\vec{v}|\sin \theta

You can use the fact that:

\sin \theta = \sqrt{1 - \cos ^2 \theta} = \sqrt{1 - \left (\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \right )}

[NateTG]

:) I imagine the problem would get very ugly if you didn't have that formula.

I can't remember the derivation, but the determinant of the cross product of two vectors gives the area of the parallelogram formed by those two vectors (of course, those vectors would be 2 of the 4 sides, but you should get the idea). The area of the triangle would just be half the area of that parallelogram. I'm at work right now, so I can't really try to figure out a derivation, but if you can't find one online I'll try to provide one (although I'm sure someone here will do that before I get a chance).

The trick is that the cross product can also be written as a determinant, even if I don't understant the notation that AKG used (I thought the cross product was a vector...)

[Moose352]

I understand it now! Thanks AKG.

[AKG]

As I thought, the proof is "ugly". Well, not really, it's just rather involved, you can't be expected to just see it. Anyway, check out this link (http://www.mcraeclan.com/MathHelp/GeometryVectorCrossProduct.htm) and look at the headings, "Magnitude of a x b" and the following heading "Area of a parallelogram".

[Gza]

I imagine the problem would get very ugly if you didn't have that formula.

How's this for ugly: I did the problem with Heron's formula for the area of a triangle. I got the answer, but it took a good page of algebra. (Don't worry, I won't even think about posting it after seeing AKG's elegant solution.)

[uart]

That's a bit picky, isn't it?

pnaj, Asking him to clarify the question is not being "picky". On the contrary, it's essentual. On reading the question I immediately wondered the same thing, was the "arc" an arc (three points define a unique cirular arc) or was the "arc" really the parabola. How can you do a problem that is not properly defined? I can't imagine how you could see that as being "picky".

[pnaj]

It was just a comment, that's all.

I suppose that, technically, an arc refers to part of a circle, but I see the word 'arc' used in context with curves other than a circle all the time.

And I just didn't see the possible circle in this problem.

[HallsofIvy]

It was just a comment, that's all.

I suppose that, technically, an arc refers to part of a circle, but I see the word 'arc' used in context with curves other than a circle all the time.

And I just didn't see the possible circle in this problem.

No, it wasn't a comment it was a question!

(NOW, I'm being picky!)

The only difference is that you assumed one possible meaning and I asked. I don't consider that "picky", just cautious.

[pnaj]

HallsofIvy,

Apologies for my question.

pnaj