Math Is Fun Forum

The time taken for each pipe to fill the tank will depend on the volume of the tank, call this V and it is a constant for our problem. So our final answer will have V in it.

Let the pipes be called A and B.

Let

time for pipe A to fill the tank

Let

time for pipe B to fill the tank

Let

volumetric flow rate of pipe A in cubic measures per minute

Let

volumetric flow rate of pipe B in cubic measures per minute

So we have the following four equations:

(1)   

(2)   

(3)   

(4)   

Substituting (1) and (2) into (3) we have

(5)   

Rearranging (4) we have

=>

Substituting this value into (5) gives us

=>

=>

=>

=>

Using the quadratic formula gives us

6481 is prime so the root won't simplify. This gives you a value for Ar which you can in turn substitute into (1) to find At and then Bt by (3).

All you need is:

Here's how:

Therefore

Therefore

and

Therefore

Look at the numerator, the 1 - c^2 part is equal to s^2.

In the denominator we have c^2+s^2 which is 1 so we now have:

Absolutely.

Oops! I was suprised that the numbers didn't come out nice but more than that, I was very unhappy with the flight time of over 9 seconds. I've corrected it.

Even the big boys made this mistake - http://edition.cnn.com/TECH/space/9909/ … metric.02/

Thanks,
Mitch.

All straight lines in 2 dimensions can have their equations written in the form:

Call this equation (1)

It doesn't matter if the line is horizontal,vertical or anything inbetween.

We could rearrange equation (1) into this form:

Call this equation (2)

This is now in the form y = mx + c

where m = -a/b and c = -d/b

But this function will not be defined if b = 0. Take the example of the line x = 0. Now the points (0,K) and (0,L) are on this line and equation (1) gives us:

and

giving:

and

But this is only possible if b and d both equal zero - crucially b = 0. Thus equation (2) will be undefined. You cannot use the "y = mx + c" version of the formula to describe vertical lines as a vertical line implies b = 0 in equation (2) and equation (2) is not defined when b = 0. But equation (1) IS defined and for the line x = 0 you just set a = 1 and b = d = 0.

As a final example take the line y = x, this is in the "y = mx + c" form and that's okay as b is not equal to zero. The equation (1) form of this line is:

and so a = 1, b = -1 and c = 0 (Crucially b is NOT equal to zero and so the "y = mx + c" form is defined and so we can use it: y = x).

As a lead to other topics note that:

is the equation of a plane in 3 dimensions.

In general, given n + 1 points there will be a unique nth degree polynomial which passes through those points. So for instance given 2 points (n = 1) where will be a unique 1th degree polynomial (straight line) passing theough those two points.

Specifically for the problem of three points.You know that the parabola must have the form

  y = ax^2 + bx + c

What you need to find out is the values of the "constants" a, b,
and c. Because you don't know them yet, you need to treat a, b, and c
as variables.

What DO you know? You know three points (x,y) on the parabola. Let's
say one of these points is (2,3). Then you can plug these values into
the general equation for the parabola:

  3 = a*2^2 + b*2 + c

  3 = 4a + 2b + c

We have made the "variables" x and y into constants. Now, if you do
the same thing with the other two points on the parabola, you'll have
three linear equations in three unknowns. Solving these will give you the proabola.

Instead of using known points like (2,3) etc. You could use unknown points (x1,y1),(x2,y2) and (x3,y3) and then go through the above process solving for a,b, and c. This would lead you to the big equation given by Patrick.

Cool. Quiet right. I simply read (0,3) and (0,-3) as (3,0) and (-3,0) and carried on regardless!

As an alternative definition. You can understand an ellipse as follows:

Stick two pins in a piece of paper and join them together with a length of string (make sure the string has some slack in it). The using a pen, pull the string taught and then keeping the string taught move the pen wherever it will go (keep that string taught) this will draw an ellipse. The sum of the distances of the pen from each pin is always the same.

http://chickscope.beckman.uiuc.edu/expl … /pins.html

If the two pins are pinned in the same position you get a circle. So a circle is a special case of an ellipse.

In the case of your ellipse, the distance between the pins would be 6 (one above the other as your ellipse lies with its major axis along the y-axis) and the length of the string 10.

We need to get

into the form

Here's how:

Look at the expansion of

which is

That looks a lot like the first two terms of our original equation viz

So we can clearly see that

Similarly we have

So we can re-write our original equation as:

and collect the constants together to give

So finally we have:

So our equation is that of a cricle with centre (-1,1) and radius

Thanks, I was still editing

That'll teach me not to read my preview thoroughly before submitting

The problem lies in the following.

On this first iteration you have 11--------

So you can use any 8 digit binary number after the two 1's. This indeed gives 256 possibilities.

Then we move to 011------- giving us a further 128 possibilities.

But we cannot extrapolate out resoning as next we would have:

0011------ giving us a further 64 possibilities BUT we have missed out the number
1011------

And so on. Here is where the missing possibilities lie.

Ok.

Here's the full answer. I think the question is wrongly worded. There should be no part a) and b). Part b) alone is enough. I would have written it thus:

Let f be the function defined as:

f(x) = { x+2 , x < 2
          { ax^2+bx, x >= 2

Find the unique values of a and b that will make f both continuous and differentiable.

Now, we have to make sure that the y values of the two functions are equal when x = 2 - this will ensure the graph is everywhere continuous. We did this is the previous post and got:

Call this equation (a)

Lastly, we need to make sure that the gradients of the two functions are equal at x = 2 - this will ensure differentiability.

So we need to set the differential of the LHS equal to the differential of the RHS at x = 2 giving:

Call this equation (b)

So we solve (a) and (b) simultaineously to give

and

So that the function required is:

f(x) = { x+2 , x < 2
          { -0.5x^2+3x, x >= 2

To see the function go here:

http://mathdemos.gcsu.edu/mathdemos/pie … ility.html

Scroll down to the "Example 2" applet and then delete the second function listed there and for the first function type in (x<=2)?x+2:-0.5*x^2+3*x and then press the "New Functions" button.

Let's say you have three vectors:

Then the set:

being linearly dependent implies and is implied by

So to give an exmaple of it's use - suppose we wish to determine if the set of vectors:

is linearly dependent or linearly independent:

We calculate:

This non-zero determinant shows that { (1,4,0),(1,4,3),(0,1,-1) } is linearly independent.

1)

is as x->∞ clearly equal to zero.

But if you meant

Then the limit as x->∞ is clearly 1