Sunday, 22 February 2009
Two Nice Questions
I have talked before about Dave Marain's web site at Math Notations. He has a new contest for grade ten and under that has a wide enough span of problems that it even attracks some middle schoolers. If you teach in the grade eight to eleven range, you might want to drop in, and perhaps participate. He covers some nice stuff.
In particular, he posts a Math Problem of the Day (with a solution) and all the ones I have seen have been really interesting challenges. Good for bright kids who need a challenge that textbooks often don't offer. One of the problems from his last contest illustrates a nice property that too few teacherx and students know about quadratics.
The question was:
" The graphs of y = 2x+3 and y = -x2 + bx + c intersect in 2 distinct points P and Q, where P is on the y-axis. Let V denote the vertex of the graph of the parabola.
(a) Determine all values of b for which the points Q and V coincide."
Most students who are at all clever know that the constant term, c, of a quadratic will give the y-intercept, and if you ask them what happens when you change "c", they will tell you it just moves the whole graph up or down without changing the shape. So if the two points intersect on the y-axis, they must have the same constant term.. ie, that the c term of the quadratic must be 3.
When asked what the A term of a quadratic does the usual answer is that it makes the quadratic "narrower." I really don't like this answer for several reasons. I hope students, even those who say the curve is "narrower", realize that it has a domain from negative to positive infinity, and therefore it never gets "narrow". What they mean, I hope, is that it goes up more quickly for larger values of A. and less quickly for A values closer to zero.
When you ask them about B, they are often less certain. They may tell you it moves the vertex (or the graph) left or right, and maybe they can even give a specific distance (someting about h==b/2A) but in fact there is a little more happening than that. In the image above, the red line is the locus being traced by the vertex as the variable b is animated and a is held as -1, while c=3. It looks quite clearly as if it follows a parabolic path with a leading coefficient that is the negative of the original function, and a vertex at the y-intercept. The confirmation of this should not be beyond a good Alg II, pre-calc student.
A couple of days ago Dave also posted a "Problem of the Day" about a regular dodecagon that got me thinking about a problem I haven't worked out yet. A regular Dodecagon for the Greek Impaired student is a twelve sided polygon. Think of a clock face with all the hour marks connected in sequence. What was given in the problem, is that if you draw a chord from any vertex, to the vertex four hours away, the length of this segment squared is the area of the dodecagon. Ok, I didn't know that. The proof was a pretty easy use of pre-calc tools, primarily the law of cosines and areas of isosceles triangles... but ... I wondered, how often does that happen? How often does a chord between two vertices of a regular polygon represent a quadrature of the figure (the square root of its area)... and what if we allowed easy extensions of this...(ie from a vertex to a midpoint of some edge).
Anyway, a big point, Dave's Problems of the Day are pretty good for provoking mathematical thought in students and teachers, so check them out.
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10 comments:
Pat,
Wonderful extension of my original problem and a great graphic. I linked my readers to this post. Thanks again for all of your support.
Dave
"When asked what the A term of a quadratic does the usual answer is that it makes the quadratic "narrower." I really don't like this answer for several reasons... What they mean, I hope, is that it goes up more quickly for larger values of A. and less quickly for A values closer to zero."
I don't think the domain issue is a big deal; an infinite thing can still get narrower.
What I don't like is what you imply but don't quite say in your last sentence here. Changing the A value makes it taller or shorter, not narrower or wider! Doubling A makes it twice as tall.
It *looks* narrower, of course, just like making you twice as tall while keeping you the same width would make you skinnier.
That leads to my other underappreciated fact about parabolas.
It's not narrower or taller! It's the same shape!
All parabolas are similar, so doubling A is equivalent to dilating both axes by a factor of 2. (Or contracting, depending on whether you're changing the axis scale or moving the points.) I think it's AMAZING that this "tall skinny" parabola and this "short fat" parabloa are actually similar figures! I mean, a "tall skinny" isosceles triangle and a "short fat" one are far from similar figures. Even with another shape, like an ellipse or hyperbola, if you double the y-coordinates while leaving x alone, the result is not at all a similar figure to the original. But with a parabola it is!
I think it comes down to b^2 - 4ac = 0, multiply that by whatever you want and it's still 0, but if it's nonzero you will change its value.
I can easily see why this is for our conventional y = x^2, but what about for rotated axes (or dilations by a factor of 2 not along the y-axis or x-axis)? Does it still work? I haven't spent enough time thinking about that yet. Maybe posting this comment will motivate me to find the answer.
Josh,
Thanks, and I would agree that the similarity of all parabolic functions.
But about your statement, ".. what you imply but don't quite say in your last sentence here. Changing the A value makes it taller or shorter, not narrower or wider! Doubling A makes it twice as tall.
"
I didn't say that because I didn't mean that, or intend to imply it. I meant exactly that "that it goes up more quickly for larger values of A"
The "parent" parablola, y=x^2 goes up a distance of 2d-1 units between the value x=d-1 and x=d; and if we change A to some other value (assume positive for easy explanation) then between the same two x values, it will go up exactly A times as much. And for someone with your experience at Calculus, it can be stated in terms of the slope being A times as much at any point...
Perhaps you are right and I shouldn't worry about the use of "narrower" but I still wonder when they say it if the realize, somewhere outside that calculator screen, it still gets wider, and wider , and wider.........
Thanks for the comment, I always learn from you.
Pat
We seem to agree that an infinite thing can get "taller" in some sense of the word, but we disagree about whether it's a good way to use that word with beginning students.
To me, the important point is that the transformation from y = x^2 to y = Ax^2 is a multiplication of the y-coords by A, so it's getting taller, not narrower. I like keeping a focus on the coordinate transformation, and S(x,y) = (x,Ay) makes everything A times taller.
A secondary point is that it's not really changing shape, since all parabolas are similar. For a parabola, the image under S(x,y) = (x/A, y/A) is the same as the image under (x, Ay), which is pretty amazing.
I think the issue that it keeps going forever, so it's not really "taller", would be tertiary if that. I would only bring it up if I saw a student having that misconception, or if I had already discussed things like "there are the same number of even numbers as whole numbers".
So I still worry about the use of "narrower" here, but only because they should use "taller" instead.
Thanks for a great conversation!
Dear Mr.Pat,
We know when a quadratic equation ax^2+bx+c=0 has equal roots, then the root is x12=-b/2a where the value of c not involved on the formula. I need your information, is there another expression of x12 for the case of equal roots that still involves the value of c.
Thanks for your help.
Sincerely Yours,
Denaya Lesa.
I'm not Mr. Pat, but if you have a repeated root, then you know the discriminant is 0. That should give you a clue about how to rewrite -b/(2a) in terms of c and a, or c and b, if you wish.
Denaya,
I am Mr. Pat, and hope Mr. Joshua's clue helped you. If you are not familiar with the term "discriminant" of a quadratic, it is the expression b^2 - 4 a c. This term will be zero in the case that the quadratic has both roots equal.
What he was suggesting, I believe, was that you could solve this for either of a or b, and substitute into the -b/2a to get an expression that did NOT have the one you chose to solve for.
IF that is still not clear, do write again and I will write out the two expressions,
Thanks for writing
Many thanks for your attention @Mr.Joshua and @Mr.Pat.
Of course, Denaya understands the requirement of zero discrminant D=b^2-4ac=0 for the equal roots of ax^2+bx+c=0, Coz this subject has just teached by my math teacher last month. But my purpose here, when the above quadratic equation is written as x^2+D1*x+D0=0, where D1=b/a and D0=c/a, is there analytic formula for the equal roots in expression of both D1 and D0 but it still be equal to -D1/2 or -b/2a.
Apologise sir, as I told on this link
http://metrostateatheists.wordpress.com/2009/02/04/lightgod/, and also on
http://metrostateatheists.wordpress.com/2008/12/16/differential-equations-how-they-relate-to-calculus/,
Denaya is Student of Secondary School at Surabaya of East Jawa, Indonesia.
Oh yeah, I know Mr Pat Ballew from this address :
http://math152.wordpress.com/2009/01/13/interesting-proof-by-contradiction/#comment-255
Again thanks you sir for your explanation. Hopefully you still allow me to ask again if Denaya doesn't understand about basic math.
Happy with you sir,
Denaya Lesa.
Good afternoon Mr.Pat and All,
Whether an idea proposed by my parent mr.Rohedi and mrs.Juliani at this link
http://trevorpythag.wordpress.com/2008/02/24/quadratic-formula/
will be useful for development new science future (NSF)?
Mr.Pat and All,
Whether Denaya's posts are chatagorized as spam like one of visitor's comment at this address:
http://thedaythatidie.wordpress.com/
2009/02/05/oh-my-life/,
Apologise if so. Acctually the purpose of my post here especially to introduce my father Mr.Rohedi to International Science Communities. According me, he needs discussion friends for submitting his theory on math or physics journals. One of abstract of his paper has been posted on the Mathworks (please click my address). But I don't know why the counter of views doesn't work as well, so the number of views doesn't change since the paper posted for the first time.
Okay, see you later sir
My Best Regards,
Denaya Lesa.
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