maths problem

y=ax²-2x+1
question: for which "a" does the tangent of point P(1;?) go through point Q(0;0)?

Since this is in fact my maths homework, it would be nice if someone posted an answer quickly... one that I can understand 😀

Angie 😀

You wish RHP would do your homework, don't ya? That ain't going to happen, but here are some hints:

- evaluate the derivative of y in general
- evaluate the derivative at the point where x=1
- evaluate the value of y at the point where x=1
- remember that the derivative of this curve is also it's slope at that point
- do you know any other formulas for slopes? use one to find the slope from point (0,0) to (1, y(1)) and equate it with the derivative

From there on in, it's fairly straightforward.

Originally posted by angie88
P(1;?)

Tell me what this expression means and I will consider helping you through this problem.

oh angie angie...you dissapoint me lol 😉

i would help but im crap at maths 🙁

fred

Originally posted by DoctorScribbles
Tell me what this expression means and I will consider helping you through this problem.

it means that the point P has the values x=1 and y=???, i.e. you need to find that out.
thx,
Angie 😀

Originally posted by angie88
it means that the point P has the values x=1 and y=???, i.e. you need to find that out.
thx,
Angie 😀

Very good. Can you express the y coordinate of P in a more precise way than ???, perhaps in terms of a?

Originally posted by DoctorScribbles
Very good. Can you express the y coordinate of P in a more precise way than ???, perhaps in terms of a?

yeah... P(1;(a-1))
that was about as far as I got 🙁

Originally posted by angie88
yeah... P(1;(a-1))
that was about as far as I got 🙁

Correct. Now, tell me what tangent means, and if you can:

1. Give a general expression for the line that is tangent to ax^2 - 2x + 1, for any given point (x, ax^2 - 2x + 1) on the curve, in terms of a and x. If you can't give a complete expression for this line, can you at least give an expression for its slope?

2. Give an expression for the particular line that is tangent to ax^2 - 2x + 1 at the point (1, a-1) in terms of a. If you can't give a complete expression for this line, can you at least give an expression for its slope?

Whenever I do some maths, I start off with a little mental exercise called Picking My Battles. By this I mean that I look at the problem and say ''Is it sexy? Is it a special case of something simple? If so, is the most obvious method pointlessly long?''

In your case, Dr. S. is helping you with a general method to mess about with all sorts of things to do with tangents, and he's doing a good job of it. However, depending on why you're taking this maths class, developing your mathematical intuition might be better than learning a lot of general methods.

If a is not 0, what do we call the curve defined by that equation? Picture one of these curves, placed wherever you want it.

Originally posted by angie88
y=ax²-2x+1
question: for which "a" does the tangent of point P(1;?) go through point Q(0;0)?

Since this is in fact my maths homework, it would be nice if someone posted an answer quickly... one that I can understand 😀

Angie 😀

The ? is something you can calculate; The point P(1;?) has 1 for x, wich means you fill in x in your equation.

What is the tangent? And how do you calculate the tangent to a point of a given function?

What does it mean for a line to go through (0;0) ?

Combine those two things and you'll find a value for a, wich you can fill into your equation to actually get a number for ?, instead of some expresion with a in it.

Originally posted by angie88
yeah... P(1;(a-1))
that was about as far as I got 🙁

?

i got x=1, y=2a-2

thank you for all your help my friends 😛
In fact, I figured it out this morning (yes, I did it myself!) 😀

the answer is a=1.

Once again thanks,

Angie 😀

Nicely done. I suppose you tried a = 1, noticed it gives (x-1)^2, and saw that that gives the (horizontal) tangent you need, no differentiating required, because you Picked Your Battles 😉?

Originally posted by royalchicken
I suppose you tried a = 1, noticed it gives (x-1)^2, and saw that that gives the (horizontal) tangent you need, no differentiating required, because you Picked Your Battles 😉?

I hope that:

1. That was not the method used, or
2. If not (1), then she realizes that that method fails in general due to the instances of polynomials in which a=1 is only one solution among many, and
3. If (2) and not (1), then she acknowledges that she just got lucky this time.

Math is a war, not a battle!

Originally posted by DoctorScribbles
I hope that:

1. That was not the method used, or
2. If not (1), then she realizes that that method fails in general due to the instances of polynomials in which a=1 is only one solution among many, and
3. If (2) and not (1), then she acknowledges that she just got lucky this time.

Math is a war, not a battle!

In such a trivial instance, it doesn't really matter. In general, general methods are useful, and even essential, to know, but it's more important to know the details of what goes into the method (the facts about what a tangent is, how it relates to the derivative, etc.). Thus in the War, we choose from our mathematical knowledge what we need to knkow to construct a method. In an easy battle, we choose as few facts as possible and construct a simple, ad hoc, possibly elegant method. In a hard battle, we construct a powerful, general, complicated method. Part of the reason for this is aesthetics, part is the fact that sometimes, if we want to be efficient, well-developed intution helps.