Math Is Fun Forum

Yes it does.  A question ought to start by defining which direction is positive.  I think we had this in a previous post where an object was being thrown upwards.  You can take up or down as the positive direction but this needs to be consistent throughout the question.  eg. if 'up' is positive then gravity is -g.

Bob

The correct terms for the across and up axes are abscissa and ordinate. Descartes thought up the coordinate system and used x and y, so those have kind of stuck.  The page where I found the definitions of  abscissa and ordinate even used x and y to explain which is which

So it is good practice to always label each axis and it's perfectly ok to use your own symbols where the context suggests it (such as t across for time).

It is also usual to make x the independent variable (ie. the one whose values we can choose) and y the dependent variable (ie. the one whose values depend on a formula involving x).

But usual, not compulsory.

In 3D coordinates x is usually shown going right, y going back into the page, and z going up.

Is it confusing? Yes, probably, but using x for time would be worse I think.  A graph is just a way to make a picture out of a mathematical model and usually helps. Choosing the right variables will help if the grapher says what and why.

Saying at the start what symbols you're going to use is essential.  In logic there are several ways to indicate logic operators such as AND, OR, IMPLIES etc. And it is common to use a dot to replace a multiplication symbol to avoid confusion with a variable x.  In an Argand diagram the up axis labels have an 'i' to indicate an 'imaginary' number, but some folk use 'j'.

Bob

You say 'initial' and 'final' so I'm assuming you're using v = u + at.

I think v and u here are vectors. It's what I was taught .... velocity a vector, speed a scalar. But I tried google. The first 17 hits all had v as a vector. None as a scalar.

Strictly there should be some indication such as an overline or bold print,

but nobody seems to bother.  But I guess that's why Khan thinks scalar.

In practice it'll come out in the calculation, so don't worry too much about the distinction.

Bob

distance and time are defined as fundamental units https://en.wikiversity.org/wiki/Fundamental_units

Speed is a derived unit.

The ancients certainly had the concept. Mariners, for example, would throw a piece of wood overboard to estimate the speed of their ship.  And I'm sure many horses were sold taking account of their speed.

Speed is a rate of change (of distance with respect to time) but it doesn't seem like they had the idea of miles per hour (using their own unit for distance).

So maybe Galileo was the first to do this.

Bob

I can see there's a lion approaching.  Would you just like to run in a random direction or would it be helpful to know which direction is best?

My ship is wanting to intercept another ship on the ocean. It is also moving.  How do I plot a course so that we meet at some place ahead? Shall I just choose a new speed or might it also be useful to consider the direction of travel?

Three ropes are tied together in a knot.  Two others are ready to pull on their ropes and want me to pull also so that, together,  our forces exactly balance.  Do you think it matters which directions we each choose for our pull?

Bob

hi Ganesh,

The link works for me as 'bob' but not as 'alter ego' , a member.  It'll be interesting to find out if it works for ktesla39

Bob

Is their answer good enough?  I'd  say no .... a lot has been left out.  Your proof is much better.

Bob

I've found a way.

step 1. consider a parabola with F = (0,a) and D is y = -a

If (x,y) is the general point on this parabola

Set a = 1/4 and we have y = x^2, so this well known curve is a parabola.

step 2. let a = 1/(4A) => y = Ax^2 is also a parabola.

step 3. Transform this by vector (0,d)  => y = Ax^2 + d is a parabola.

step 4. consider y = ax^2 + bx + c

let the bracketed term be d/a

By the vector transform X = x + b/(2a) this becomes y = AX^2 + d and so is also a parabola.

Thus all quadratics of the form y = ax^2 + bx + c are parabolas.

Bob

The defining property of a parabola is  as follows.

There is a point called the focus.
There is a line called the directrix.
The parabola is the locus of all points such that the distance to the directrix is equal to the distance to the focus

focus (1.5,  2)   directrix  y = 0 (ie the x axis)

Let a point on the parabola have coordinates (x,y)

Equating these:

I shall explore some properties of parabolas based on the defintion.

Left part of diagram.

Let the focus be F.  There must be at least one point on the parabola that is nearest to the directrix.  Let that point be R and let S be the point on the directrix so that FR = RS.  I know that RS must be perpendicular to the directrix (distance implies shortest distance) but I have not assumed that FRS is a straight line

But FR = RS, so the triangle FRS is isosceles. Let T be the midpoint of FS.  Then FT = TS so T is a point on the parabola and it is nearer than R.  Unless R=T in which case FRS is a straight line.

So R is unique and FRS is perpendicular to the directrix.

Second part of diagram.

Let P be a point on the parabola and Q the point on the directrix so that FP = PQ.

Reflect P and Q in the line FRS to give points P' and Q'. 

FP' = P'Q' so P' is also on the parabola.

So FRS is a line of symmetry for the parabola.

Bob

Ok so I didn't read the question.

So now I'm getting

3^(n-1)       (1/2)^(n-1)       (3/4)^(n-1)

When n=1 don't we want just one white triangle?

Bob

hi Phro,

I agree with your answers to a and c but not b.

When we start I'll take it that n = 0; we have a single white triangle side =1

At step 1 the white triangles have side = 1/2 = 1/(2^1)

At step 2 the white triangles have side = 1/4 = 1/(2^2)

.....................

If you spot a sequence and use it to get algebraic answers that seems ok to me. But all that's missing is to prove that the sequences are valid.  You can do that by, for instance, in each step the previous number of whites is made into four new smaller triangles, one of which is blue so 3/4 are still white. So each step increases the number of whites by a multiplier of times 3/4

Bob