Where's the teachable moment in this math problem?
October 14, 2018 12:24 PM Subscribe
I am a long-term substitute teacher filling in for someone on maternity leave in a precalculus classroom. Last week I gave the first test of the marking period, which included a question on finding the intersections of a line with a circle. One of the students had an interesting alternate answer.
For FERPA reasons, I'm not including the student's actual work, but the I've re-scanned the original question and my own handwriting of the usual route to the solution at this link. You basically use the linear equation to solve for y in terms of x (or you can do x in terms of y), and substitute that solution back into the circular equation to turn it into a one-variable quadratic that can be solved with several methods.
But another student came to me with a question: By multiplying both sides of the linear equation, it's possible to get that equal to the same constant as the circular equation. Setting them equal to each other doesn't solve the question, but can be re-written as a new circular equation. (My re-written version of the student's work is the top part of this second image. .
I also noticed later that it's possible, using some tricky "completing the square", to set them equal to each other and turn both sides into a difference of squares, getting you binomials and smaller numbers that resolve to the same new circular equation (that's the bottom of the second image).
So my question is, what does any of this mean? The intersection of the original line and circle are the point of the exercise, is there anything teachable about these alternate methods? The new circle equation would have the same intersections with the original line, but since there are only two points, I'm pretty sure there's an infinite number of circles that could do the same, no?
If there's something useful about the alternate methods that might help as we continue to discuss systems of equations, I'd love to hear it. Thanks!
For FERPA reasons, I'm not including the student's actual work, but the I've re-scanned the original question and my own handwriting of the usual route to the solution at this link. You basically use the linear equation to solve for y in terms of x (or you can do x in terms of y), and substitute that solution back into the circular equation to turn it into a one-variable quadratic that can be solved with several methods.
But another student came to me with a question: By multiplying both sides of the linear equation, it's possible to get that equal to the same constant as the circular equation. Setting them equal to each other doesn't solve the question, but can be re-written as a new circular equation. (My re-written version of the student's work is the top part of this second image. .
I also noticed later that it's possible, using some tricky "completing the square", to set them equal to each other and turn both sides into a difference of squares, getting you binomials and smaller numbers that resolve to the same new circular equation (that's the bottom of the second image).
So my question is, what does any of this mean? The intersection of the original line and circle are the point of the exercise, is there anything teachable about these alternate methods? The new circle equation would have the same intersections with the original line, but since there are only two points, I'm pretty sure there's an infinite number of circles that could do the same, no?
If there's something useful about the alternate methods that might help as we continue to discuss systems of equations, I'd love to hear it. Thanks!
Best answer: When your student set the two equations equal to each other and tossed out the 18, they threw away some information. That 18 could have been a 12 or a 20 or a ...
The resulting circle is the range of intersection points of these circles/lines of different sizes that would all have led to the same one equation.
Visually:
Here's your original circle/line and circle.
Here's another circle/line/circle but I have multiplied the 18 thrown away by a common factor. The third circle stays the same but still intersects the two points.
posted by vacapinta at 12:44 PM on October 14, 2018 [9 favorites]
The resulting circle is the range of intersection points of these circles/lines of different sizes that would all have led to the same one equation.
Visually:
Here's your original circle/line and circle.
Here's another circle/line/circle but I have multiplied the 18 thrown away by a common factor. The third circle stays the same but still intersects the two points.
posted by vacapinta at 12:44 PM on October 14, 2018 [9 favorites]
Best answer: Vacapinta has it. You start out with a system of two equations with two unknowns. Setting them both equal to the same value and then therefore each other isn't *wrong*, its just not helpful because now you have a single equation with two unknowns. You've lost some information.
Substitution works because it lets you reduce your system to a single (quadratic) equation with a single unknown, find the solutions to that, and then back substitute to find the values of the other variable in your system.
posted by Reverend John at 1:29 PM on October 14, 2018 [2 favorites]
Substitution works because it lets you reduce your system to a single (quadratic) equation with a single unknown, find the solutions to that, and then back substitute to find the values of the other variable in your system.
posted by Reverend John at 1:29 PM on October 14, 2018 [2 favorites]
It's not quite what you have, but Lagrange multipliers are a nearby idea that might be interesting to your student.
posted by solitary dancer at 1:30 PM on October 14, 2018
posted by solitary dancer at 1:30 PM on October 14, 2018
Response by poster: Vacapinta, can you clarify what you’re saying? The points on the new circle are a collection of all the possible intersection points of the original circle and line if they had been made equal to any arbitrary constant?
posted by The Pluto Gangsta at 3:19 PM on October 14, 2018
posted by The Pluto Gangsta at 3:19 PM on October 14, 2018
Best answer: Not vacapinta, but yes. I made you a little Desmos graph to illustrate.
https://www.desmos.com/calculator/vgzx2cbcez
If you set the slider to 18, you have your original system, illustrated by the green circle and purple line. The black slider "forgets" the constant. If you play with the slider, you can see the intersections between the original circle and line stay on the black circle.
posted by ktkt at 4:14 PM on October 14, 2018 [6 favorites]
https://www.desmos.com/calculator/vgzx2cbcez
If you set the slider to 18, you have your original system, illustrated by the green circle and purple line. The black slider "forgets" the constant. If you play with the slider, you can see the intersections between the original circle and line stay on the black circle.
posted by ktkt at 4:14 PM on October 14, 2018 [6 favorites]
Also: not all constants will yield solutions -- if k<0, the green circle disappears entirely, and if k is too large, the line and circle will no longer intersect.
posted by ktkt at 4:16 PM on October 14, 2018
posted by ktkt at 4:16 PM on October 14, 2018
I see nothing wrong with the alternate solution. I think the teachable concept is that many problems have multiple paths to a solution. This is true for most major math theorems. The proofs taught in schools now are usually not the original proofs.
posted by SemiSalt at 5:50 PM on October 14, 2018
posted by SemiSalt at 5:50 PM on October 14, 2018
I think the teachable moment might be to give your student some extra ( ie fun) problems to solve that have semi-easily-reachable ‘alternative’ processes to get there. I can’t point to any resources for this. . .but man, if I was that kid I would feel pretty good knowing that a teacher saw my creative thinking and encouraged me to follow up with it.
posted by ananci at 8:22 PM on October 14, 2018 [2 favorites]
posted by ananci at 8:22 PM on October 14, 2018 [2 favorites]
What your student did is really a very nice solution to a different but related question!
The process they laid out will always lead to a quadratic equation in x and y of the form
x^2 + y^2 + ax + by = 0
which, as you say, can be seen to be the equation of a circle by completing the square.
But it's not just any circle, because any equation of the above form is satisfied when (x,y) = (0,0).
In other words, given the equation of the circle and the line, you have found the (unique!) circle passing through the origin, P, and Q, where P and Q are the two points of intersection you were originally looking for. Another way to say this is that you've computed the circumcircle of triangle OPQ.
posted by escabeche at 8:34 PM on October 14, 2018 [7 favorites]
The process they laid out will always lead to a quadratic equation in x and y of the form
x^2 + y^2 + ax + by = 0
which, as you say, can be seen to be the equation of a circle by completing the square.
But it's not just any circle, because any equation of the above form is satisfied when (x,y) = (0,0).
In other words, given the equation of the circle and the line, you have found the (unique!) circle passing through the origin, P, and Q, where P and Q are the two points of intersection you were originally looking for. Another way to say this is that you've computed the circumcircle of triangle OPQ.
posted by escabeche at 8:34 PM on October 14, 2018 [7 favorites]
I don't know a thing about math, but I do know my math loving daughter who consistently finds the right answer the "wrong" way has adored the math teachers who gave her positive feedback for her alternate methods. Of course she's also had to get the feedback that making up her own system is not always the wisest approach, but teachers who have noticed and valued her style have really helped keep her love and passion for math alive.
posted by latkes at 8:56 PM on October 14, 2018
posted by latkes at 8:56 PM on October 14, 2018
Response by poster: The resulting circle is the range of intersection points of these circles/lines of different sizes that would all have led to the same one equation.
That definitely is something teachable, thanks vacapinta for the insight, ktkt for the Desmos link and everyone who contributed!
posted by The Pluto Gangsta at 7:02 AM on October 15, 2018
That definitely is something teachable, thanks vacapinta for the insight, ktkt for the Desmos link and everyone who contributed!
posted by The Pluto Gangsta at 7:02 AM on October 15, 2018
Also: not all constants will yield solutions -- if k<0, the green circle disappears entirely, and if k is too large, the line and circle will no longer intersect.
We can find the value at which k becomes too large fairly easily:
Call the circle your student found C. Consider the set of equations -12x+6y=k; re-written in slope-intercept form, this is y=2x+k/6, so this represents a set of parallel lines with slope 2.
For some values of k, the line intersects C at two points, at others, not at all; and for two values of k, the line is tangent to C and meets it at exactly one point. Trivially, one of these values must be k=0; as ktkt points out, the original circle disappears entirely for k<0.
Since these lines are parallel, the other point must be the opposite point on C. We know the center of C is (-6,3), so the opposite point is (-12,6). Solve for k at this point and you get k=180.
Therefore, the set of equations x2+y2=k, -12x+6y=k [or, if you like, -4x+2y-k/3=0] has two solutions for 0<k<180, one solution at k=0 or k=180, and no (real) solutions for k<0 or k>180.
posted by DevilsAdvocate at 12:56 PM on October 16, 2018 [1 favorite]
We can find the value at which k becomes too large fairly easily:
Call the circle your student found C. Consider the set of equations -12x+6y=k; re-written in slope-intercept form, this is y=2x+k/6, so this represents a set of parallel lines with slope 2.
For some values of k, the line intersects C at two points, at others, not at all; and for two values of k, the line is tangent to C and meets it at exactly one point. Trivially, one of these values must be k=0; as ktkt points out, the original circle disappears entirely for k<0.
Since these lines are parallel, the other point must be the opposite point on C. We know the center of C is (-6,3), so the opposite point is (-12,6). Solve for k at this point and you get k=180.
Therefore, the set of equations x2+y2=k, -12x+6y=k [or, if you like, -4x+2y-k/3=0] has two solutions for 0<k<180, one solution at k=0 or k=180, and no (real) solutions for k<0 or k>180.
posted by DevilsAdvocate at 12:56 PM on October 16, 2018 [1 favorite]
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As another example, if I take two linear equations like x+y=8 and 5x-y=4, then I can use the same technique to get y=3x, which is another line through the point of intersection. But not much else?
posted by MillyMath at 12:44 PM on October 14, 2018