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My grand children just finished calculus of functions of one real variable. They can take derivatives. The tangent at a point on a curve is defined intuitively as the limiting position of secant line through the point. I am supposed to introduce functions of two variables without dot-cross products. Tangents to curves in spaces and differentiability are not in the syllabus. Question: With my hands tied, how can I convince my kids, intuitively psychologically and emotionally but need not be logically mathematically, that the plane z=f(a,b)+D_x f(a,b)(x-a)+D_y f(a,b)(y-b) is usually frequently TANGENT to the NICE surface z=f(x,y) at (a,b)? What does it mean by a tangent plane? It appears that I have only one option, that is
What is your alternative solutions to this teaching situation? Thank you in advance. Twma 07:06, 5 February 2007 (UTC)
Linear subspaces, flats were not available to me for explanation. Exactly two slices as you described were at my disposal. I did that and followed up with ice cream, candy and cookies. My grandchildren were happy. At the end in a subtle way, I pleased them to be good drivers (get good marks) without knowing anything about petrol chemistry (mathematical theory). Am I an acceptable grand father? Thanks. Twma 00:50, 7 February 2007 (UTC)
if X = , then X = ? In other words, do I multiply numbers where the exponents(?) are equal? Is that against any mathematical laws? ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:43, 5 February 2007 (UTC)
am x bm = (ab)m is true.
and the square root is 162x^2 - you half the exponent - (taking logs, dividing by two - then exponentiating again..)
square root (nx) = nx/2
In fact using the first rule you supplied (which is correct) you can get
(nx/2)2 = nx/2 x nx/2 = nx/2+x/2 = nx
so:
square root (nx) = square root ((nx/2)2) = nx/2
(I get to say QED now) 87.102.8.103 15:45, 5 February 2007 (UTC) Hope that helps. 87.102.8.103 15:34, 5 February 2007 (UTC)
K Thanks very much. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:23, 6 February 2007 (UTC)
What precisely are bernoulli's curves of quickest descent?? —Preceding unsigned comment added by 59.183.58.204 ( talk • contribs)
I have been told that when we markup costs for our work instead of just for example muliplying the cost times 1.20 for a 20% makup we should divide by 0.80. Which is correct?
This isn't a homework question, but it's based on something our teacher explained in class. I now have some idea why he didn't continue any farther. It's a geometric coloring problem. Take a circle of radius r, a point on that circle, and two colors. Every point on the circle must be a different color from any point on the circle that's unit distance away from it. Is it, or is it not, possible to color this circle? In some cases, where r=1 for instance, if you start coloring clockwise from a point, you return to your starting point within a finite number of steps. If this covers an odd number of points, the coloring is impossible, otherwise it's easy. What about when r is irrational, though? If you have to cover an infinite number of points before you return to the starting point, do you wind up with a contradiction, or not? Or is it like quantum mechanics, where you get both? I guess what I'm asking is, is infinity even or odd? Black Carrot 20:15, 5 February 2007 (UTC)
No, I mean an irrational radius, if I worked the formula for chord through correctly. The problem says "unit distance", not "unit distance along the circle". I think this is slightly different from Thomson's lamp and the Grand Hotel, and you're right, the divisibility of infinity may not be related either. I can't imagine, though, what "yes" even means, let alone how you got it as an answer. Could you give more detail? Black Carrot 22:46, 5 February 2007 (UTC)
You're right, I copied the formula wrong. So, supposing it's irrational. I can pretty much understand the group action page, it seems to mean "shuffle everything in the set around." Naturally, we can say more specifically that the action consists of rotating the circle by a particular angle, and mapping everything from before to after. We chose irrational angles specifically to make the orbit infinite, so naturally the orbit is infinite. No problems so far. That's about where you lose me, though. How is it "countably" infinite? I could have sworn it would hit every point on the circle if we chose the angle/circumference relation properly, which would make it "uncountably" infinite, right? A complete line segment (and therefore the circumference of a circle) contains uncountably many points, corresponding to the real numbers on the interval [0,1). Then, you mention the axiom of choice, defined in the article as
It says nothing about what that choice would entail, and reading related articles, I don't think it's supposed to. This can be applied to the current situation, as far as I can tell, by saying "if I'm trying to color this set of points, then at each and every point, there exists at least one color I can specifically choose." I kind of already knew that. So, anyway, we move on with the proof. We pick an orbit (by my reckoning, the orbit, but whatever) and pick a point in it. We then move along the orbit, coloring each point, alternating (in this case) red and black. ...Then it stops. How has that gotten me farther along? As far as I can tell, you're interpreting the axiom of choice as saying
...which is ridiculous. But hopefully I misunderstood. Black Carrot 01:46, 6 February 2007 (UTC)
I appreciate the numbering, that's a lot clearer. And I'm sorry, that makes sense as an application of the axiom of choice. I don't see that fixes the proof, but at least that wasn't the problem. Now, though I still don't quite follow how there are uncountably many countably-sized orbits, let's take that as assumed.
How do you never reach a conflict? Sure, it would take infinitely many steps to get back to your exact starting point, that's why I chose that angle, but what's wrong with that? Black Carrot 05:30, 7 February 2007 (UTC)
And in what may be a related question, does induction extend to uncountably long sequences? The article used to say it doesn't, but now it doesn't mention it, and I never found out why or even whether that's true. Black Carrot 05:30, 7 February 2007 (UTC)
Mathematics desk | ||
---|---|---|
< February 4 | << Jan | February | Mar >> | February 6 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
My grand children just finished calculus of functions of one real variable. They can take derivatives. The tangent at a point on a curve is defined intuitively as the limiting position of secant line through the point. I am supposed to introduce functions of two variables without dot-cross products. Tangents to curves in spaces and differentiability are not in the syllabus. Question: With my hands tied, how can I convince my kids, intuitively psychologically and emotionally but need not be logically mathematically, that the plane z=f(a,b)+D_x f(a,b)(x-a)+D_y f(a,b)(y-b) is usually frequently TANGENT to the NICE surface z=f(x,y) at (a,b)? What does it mean by a tangent plane? It appears that I have only one option, that is
What is your alternative solutions to this teaching situation? Thank you in advance. Twma 07:06, 5 February 2007 (UTC)
Linear subspaces, flats were not available to me for explanation. Exactly two slices as you described were at my disposal. I did that and followed up with ice cream, candy and cookies. My grandchildren were happy. At the end in a subtle way, I pleased them to be good drivers (get good marks) without knowing anything about petrol chemistry (mathematical theory). Am I an acceptable grand father? Thanks. Twma 00:50, 7 February 2007 (UTC)
if X = , then X = ? In other words, do I multiply numbers where the exponents(?) are equal? Is that against any mathematical laws? ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:43, 5 February 2007 (UTC)
am x bm = (ab)m is true.
and the square root is 162x^2 - you half the exponent - (taking logs, dividing by two - then exponentiating again..)
square root (nx) = nx/2
In fact using the first rule you supplied (which is correct) you can get
(nx/2)2 = nx/2 x nx/2 = nx/2+x/2 = nx
so:
square root (nx) = square root ((nx/2)2) = nx/2
(I get to say QED now) 87.102.8.103 15:45, 5 February 2007 (UTC) Hope that helps. 87.102.8.103 15:34, 5 February 2007 (UTC)
K Thanks very much. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:23, 6 February 2007 (UTC)
What precisely are bernoulli's curves of quickest descent?? —Preceding unsigned comment added by 59.183.58.204 ( talk • contribs)
I have been told that when we markup costs for our work instead of just for example muliplying the cost times 1.20 for a 20% makup we should divide by 0.80. Which is correct?
This isn't a homework question, but it's based on something our teacher explained in class. I now have some idea why he didn't continue any farther. It's a geometric coloring problem. Take a circle of radius r, a point on that circle, and two colors. Every point on the circle must be a different color from any point on the circle that's unit distance away from it. Is it, or is it not, possible to color this circle? In some cases, where r=1 for instance, if you start coloring clockwise from a point, you return to your starting point within a finite number of steps. If this covers an odd number of points, the coloring is impossible, otherwise it's easy. What about when r is irrational, though? If you have to cover an infinite number of points before you return to the starting point, do you wind up with a contradiction, or not? Or is it like quantum mechanics, where you get both? I guess what I'm asking is, is infinity even or odd? Black Carrot 20:15, 5 February 2007 (UTC)
No, I mean an irrational radius, if I worked the formula for chord through correctly. The problem says "unit distance", not "unit distance along the circle". I think this is slightly different from Thomson's lamp and the Grand Hotel, and you're right, the divisibility of infinity may not be related either. I can't imagine, though, what "yes" even means, let alone how you got it as an answer. Could you give more detail? Black Carrot 22:46, 5 February 2007 (UTC)
You're right, I copied the formula wrong. So, supposing it's irrational. I can pretty much understand the group action page, it seems to mean "shuffle everything in the set around." Naturally, we can say more specifically that the action consists of rotating the circle by a particular angle, and mapping everything from before to after. We chose irrational angles specifically to make the orbit infinite, so naturally the orbit is infinite. No problems so far. That's about where you lose me, though. How is it "countably" infinite? I could have sworn it would hit every point on the circle if we chose the angle/circumference relation properly, which would make it "uncountably" infinite, right? A complete line segment (and therefore the circumference of a circle) contains uncountably many points, corresponding to the real numbers on the interval [0,1). Then, you mention the axiom of choice, defined in the article as
It says nothing about what that choice would entail, and reading related articles, I don't think it's supposed to. This can be applied to the current situation, as far as I can tell, by saying "if I'm trying to color this set of points, then at each and every point, there exists at least one color I can specifically choose." I kind of already knew that. So, anyway, we move on with the proof. We pick an orbit (by my reckoning, the orbit, but whatever) and pick a point in it. We then move along the orbit, coloring each point, alternating (in this case) red and black. ...Then it stops. How has that gotten me farther along? As far as I can tell, you're interpreting the axiom of choice as saying
...which is ridiculous. But hopefully I misunderstood. Black Carrot 01:46, 6 February 2007 (UTC)
I appreciate the numbering, that's a lot clearer. And I'm sorry, that makes sense as an application of the axiom of choice. I don't see that fixes the proof, but at least that wasn't the problem. Now, though I still don't quite follow how there are uncountably many countably-sized orbits, let's take that as assumed.
How do you never reach a conflict? Sure, it would take infinitely many steps to get back to your exact starting point, that's why I chose that angle, but what's wrong with that? Black Carrot 05:30, 7 February 2007 (UTC)
And in what may be a related question, does induction extend to uncountably long sequences? The article used to say it doesn't, but now it doesn't mention it, and I never found out why or even whether that's true. Black Carrot 05:30, 7 February 2007 (UTC)