Math Is Fun Forum

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

I'm catching up fast,  On a weather map the region is the surface of the Earth (part of). Temperature is a scalar as it has just magnitude but not direction. Every point has a temperature so the 'field' of tempeatures matches the real world of the surface.

You can show wind simlarly but, as wind has magnitude and direction, this time it's a vector field.

Bob

Your acceleration calculations look ok. Just one thing bothers me. Does it make a difference if the force is applied for a measured amount of time. I've searched for an answer to this and made no progress.  Why ask this question?  If you apply a force of 200N for, say, 5 seconds then the impulse is 200 x 5 = 1000 N.sec and this gives that amount of momentum to the two objects.

Momentum before = 0

Momentum after = 1000

If I consider myself ( as the pusher) as at rest with the box moving away at velocity v after 5 seconds then

50 x v = 1000 so v = 1000/50 = 20 m/s

Using v = u + at     20 = 0 + 5a     so a = 4m/s^2

Oh joy! That's what you got before. So maybe all's well in the world of Newtonian mechanics.

Do you drive yet? Don't try to experience this 'box' effect by driving a car into a tree. The tree will definitely exert a big force on you!

Bob

Yes as it's 10 to the whole expression.

I'm of an age where there were no calculators to use for calculations. So logs were taught early on as you can make use of this property:

This converts a multiplication into an addition. Similarly divisions and roots.

Bob

Your calc leading to -50 is a good one. 

Yes to the frames question.

Speed is a scalar quantity so I wouldn't expect it to be negative. But velocity is a vector so direction is needed. When you start such a question decide which way is positive and if any calc leads to a negative it means it's going the other way.

eg. A ball is thrown vertically upwards at 27 m/s. Taking g as 10, what is its height after 5 seconds.

I'll take 'up' as positive.

g = -10; t = 5; u = 27

using s = ut + 0.5at^2

height = 27 x 5 - 0.5 x 10 x 5^2 = 135 - 125 = 10m (above the ground)

I'll take 'down' as positive

g = 10; t = 5; u = -27

height = -27 x 5 + 0.5 x 10 x 25 = -135 + 125 = -10m  (As down is positive this ball must be 10 m above the ground.

Suppose u = 7, and I throw the ball up leaning sightly over a cliff.

Take 'up' as positive.

height = 7 x 5 - 0.5 x 10 x 25 = 35 - 125 = -90m   The ball has gone up, come down and carried on downwards below my level down the cliff.

When was the ball again level with my position?

t = ?; u = 7; g = -10; s = 0

0 = 7t - 0.5 x 10 x t^2    =>       7t - 5t^2 = 0       t(7-5t) = 0 So we have two solutions, t = 0 and t = 7/5

We should expect the first as we know the ball was at that height when we threw it and started timing. The second is the one that answers the question.

Bob

The way I was taught to do this is to bring the 'WRT-object' to  rest by applying an equal and opposite speed to it, and apply the same speed to everything else.  As you are stationary that means the WRTs in the parts 3 and 4 are just the given information with no change needed.

For part 1 bring the train to rest by applying a speed of 30 m/s in the opposite direction.  Let's say, to avoid confusion, that the train is heading North, and so is the boarder.  So make the whole world go South at 30 m/s so that the train is brought to rest. The boarder is also given a speed South of 30 m/s so its speed North is now (5 - 30) m/s  ie. 25 m/s South.  If a person is on the train they think they're stationary and the whole world is travelling South at 30 m/s. Except the boarder of course who seems to be travelling South at 25 m/s.

You can try this yourself for part 2. You should get the exact opposite answer to part 1

So add the boarder's speed to the train's to get 30 + 5 m/s

Bob

OK, got it.

Are you familiar with matrix multiplication?  If not then there are two useful pages here:

https://www.mathsisfun.com/algebra/matr … lying.html

https://www.mathsisfun.com/algebra/matr … lator.html

In 2D transformation geometry 2 by 1 vectors (coordinates switched row and column) can be transformed by multiplying by a 2 by 2 matrix.

As any such matrix transforms (0,0) to (0,0) this only works when the transform leaves the origin invariant.  The one for the question does as y=x goes through the origin.

I'll call the matrix

y=x is invariant so (x,x) maps onto (x,x)

This gives us two equations

a + b = 1 so b = 1-a         and c + d = 1 so d = 1-c

So the matrix becomes

Now to fix which stretch by considering a single point under the transformation.

I'll start on the line at (2,2) and go one right and one down to (3,1) and again to (4,0)

That's a good point to consider as there's a zero in the calculation.

It's image is at one right and one down and the same again ie at (6,-2)

So (4,0) maps onto (6,-2)

This leads to a = 1.5 and c = -0.5

So the matrix for this transformation is

So what does this do to our points?

QED

Bob