Saturday, May 12, 2007

When Curves Collide: Quadratic Systems Explored...

[Another Update: Mutiple solutions to (a) and (b) are now provided in the comments (as of 10:50 PM 5-17-07).]

Target Audience: Algebra 2, Advanced Algebra, and beyond...

The previous post challenged students to consider 'basic' properties of circles and triangles. Now we will look at systems of quadratics - circles and parabolas in particular. The purpose here is to help students go beyond the standard algorithms of solving systems, by analyzing a general type of system using a parameter r. The use of parameters has become the norm on the AP Calculus Exam. Algebra students may benefit from an early introduction.

Consider the system:
x2 + y2 = r2
y = r2 - x2

Here, r denotes a positive constant. Depending on the value of r, this system will have either 2, 3, or 4 solutions!

Simple problem of a parabola intersecting a circle, right? We will assume here that students have already solved specific cases of such systems both graphically and algebraically (substitution, etc.). They have been shown that parabolas and circles may intersect in 0, 1, 2, 3, or 4 points. As a review, begin this investigation by asking pairs of students to sketch (no equations here) graphs depicting each of these cases. That visualization is a powerful context for the algebraic solutions and may motivate the students to consider why varying the parameter r in this lesson leads to different conclusions. This type of analysis goes beyond standard problems and prepares students for the open-ended free-response types of standardized or AP questions they will later encounter in high school or college.


(a) r=1
Solve the following system first graphically, then algebraically:
x2 + y2 = 1
y = 1 - x2
This system demonstrates that a quadratic-quadratic system may have ________ (number) solutions.
For this system, the points of intersection are ___________________________.
For x between -1 and 1, the parabola is (above, below) the circle.

(b) r=2
Solve the following system first graphically, then algebraically:

x2 + y2 = 4
y = 4 - x2

This system demonstrates that a quadratic-quadratic system may have ________ (number) solutions.
For this system, the points of intersection are ___________________________
Restrict the domain to -2 ≤ x ≤ 2. For what values of x is the parabola above the circle in this system? Below the circle?

(c) What set of positive values of r have we not yet considered?
For such values of r, make a conjecture about the number of points of intersection of the system:
x2 + y2 = r2
y = r2 - x2

Now choose a particular value of r in this set, say r = 1/2. Check the validity of your conjecture by solving the system for this particular value both graphically and analytically (algebraically).

(d) Analyze, algebraically, the following system for all positive values of r. Show carefully that your algebraic solution leads to 3 distinct cases for r:
x2 + y2 = r2
y = r2 - x2
For each case, give the solutions (ordered pairs) in terms of the parameter r.
Explain why, for all positive values of r, there will always be at least TWO solutions of this system. That is, the possibility of zero or one solution does not exist for this system...


Anonymous said...

If you keep offering me things a few weeks before they become useful.... This is getting funny. I am actually more likely to give this to one group or one student to explore (we are looking for projects).

I notice that you had a little traffic bump (the good blog kind, not the bad motor vehicle kind)

Dave Marain said...

give me your curriculum for the next 2 weeks and i'll try to plan it out just right for you!

if students are savvy enough to google some math topics they might find these problems and 'rip off' the solutions - probably unlikely, right?

yup, i'm averaging around 100 unique visitors per day now, probably a result of the Carnival, your support, denise's referrals, etc. I also believe that the vast majority of these visits are from teachers or students searching for solutions to actual problems or instructional ideas. As the number of different topics on this blog increase, there's a greater likelihood that a search will match something I have. This is from a casual analysis of referrals over the past couple of weeks.

It's gratifying to see this kind of interest, but I have no idea if more than a handful appreciate what I'm doing or offering. I'm enjoying writing these but it is intensive, since the problems are all original and new, rather than retreads from 20 years ago. I think these kinds of open-ended activities are becoming my identity now.

Anonymous said...

Very nice identity, in that case.

As far as kids googling answers, hell, let them come to me at lunch, I'll give them answers. On a project it's the explanations I want. And yeah, I'll give them the explanations as well, but they need to study and absorb, so they can do the explaining themselves.

(My mean teacher trick.... problem on the board.... kids working... I announce the answer, and ask them to work on an explanation. Whooo, do they hate that!)

I have just stepped into a slow (means some depth) unit on conic sections. That's where your previous sheet fit. We wont' do funny rotations, maybe no eccentricity (they have had no trig). When we are done, that's where this sheet might go. Our last unit will be on exponentials, logs, and their graphs. We will be treating them as functions - no applications at this time.

And full sheets are neat, but even an idea for a slightly off-topic, but still related, exploration, that would be great.

It all comes back. No harm in wandering a bit. I like the connections, both the ones they see now, and the ones they will see later.


Dave Marain said...

Ok, here's a piece of the solution...

(a) r = 1:
We can either substitute y = 1-x^2 into the equation for the circle (most students would probably do this) OR
Solve the circle equation for y:
y = +- √ (1-x^2)
Let a = 1-x^2.
√ a = a leads to
a = 0 or a = 1.
a = 0 implies that x = -1 or x = 1.
a = 1 implies x = 0.
Thus there are THREE points of intersection: (-1,0), (1,0) and (0,1).

We can show algebraically that, for this domain, 1-x^2 is between 0 and 1. Therefore,
√ (1-x^2) > 1-x^2.
This implies that for x between -1 and 1, the parabola is BELOW the circle in the case r = 1.

A similar approach can be used for the other cases.
Solving the system algebraically in the general case, in terms of r, can produce up to 4 solutions. (Details later)
However, x = r and x = -r will ALWAYS be two of these solutions since these correspond to the 'common' x-intercepts of the circle and the parabola.

Dave Marain said...

Update - Solution to (b)
r = 2:
Traditional substitution method:
x^2 + (4-x^2)^2 = 4
x^2 + 16 - 8x^2 + x^4 = 4
x^4 - 7x^2 + 12 = 0
(x^2 - 4)(x^2 - 3) = 0
x = 2, -2, √3, -√3

Non-traditional method (suggested in previous comment):

√(4-x^2) = 4-x^2 iff
4-x^2 = 1 or 4-x^2 = 0
This leads to the same result as above.

Parabola below circle when
√3 < |x| < 2
The parabola intersects or is above the circle for 0 < |x| < √3. Other values are outside of the domain of the circle.
The circle and parabola intersect in 4 points in this case. This is the maximum number of points of a quadratic-quadratic system since the product of the degrees is
2x2 = 4.

Anonymous said...

I worked on the problem doing what is obvious, i.e. what was called the traditional substitution method, and got some extraneous solution. Then my son (a sophomore who knows quite a bit about Algebra 2 already, but is required by the school to take Algebra 2 next year) who is working on the problem at the same time points out that I am doing the wrong substitution. Instead of substituting

y = r^2 – x^2

it is better to substitute

x^2 = r^2 – y

into the circle equation and get

r^2 – y + y^2 = r^2
y^2 – y = 0
y(y-1) = 0
y = 0 or 1
y = 0 implies x = +r or -r
y = 1 implies x = sqrt(r^2-1) or -sqrt(r^2-1)
so the solutions are (+r, 0), (-r, 0), (+sqrt(r^2-1), 1), (-sqrt(r^2-1), 1)
when |r| < 1 there are only the first 2 solutions,
when r = 1, there are 3 solutions (1,0), (-1,0), (0,1)
when |r| > 1, there are 4 solutions as listed above

Dave Marain said...

Kudos to your son! Now that's an efficient solution! It makes beautiful sense to choose the variable that's NOT squared to solve for. This led to a quadratic in y as opposed to a 4th degree equation in x that might have extraneous solutions!

My goal was to provide students with a more in-depth investigation of quadratic-quadratic systems, but he made short work of it! Congratulate him for me.

Anonymous said...

Coming in late here, and I see that bd's son has already beat me to my solution. I started with the "common sense" substitution, got the quadratic-quadratic, and thought, "Yuck! That's no fun." I guess I'm more of a geometry person.

So I went back to the original equations and noticed that they were almost the same:
y^2 = r^2 - x^2
y = r^2 - x^2
So the problem reduced to finding:
y^2 = y