Math Is Fun Forum

Off the top of my head, Goldbach’s conjecture:

http://en.wikipedia.org/wiki/Goldbach's_conjecture

It was just a query about X; I deleted it as Sekky’s post appeared to have answered the question.

This differential equation is of the variables-separable type. You can separate the variables and integrate.

Normally, students are asked to derive the Cartesian equation from the parametric ones, not the other way round.

To get a set of parametric equations from the Cartesian one, you can choose x to be some function of t, and then express y in terms of t. The function of t you choose for x is usually such as to yield simple expressions for x and y. The parametric equations for any curve are not unique.

For example, as parametric equations for the straight line y = mx + c (m ≠ 0), you can simply choose x = t, so y = mt + c. You can also chosse x = t/m, y = t + c as your parametric equations instead. Or you can even choose x = t/m−c/m, y = t. There is no one fixed choice.

Refer to Toast’s diagram. The angle θ in the diagram is π⁄3 radians. Twice that is 2π⁄3 radians. So the length of the arc subtending this angle (one vertical half of the lime curve in Toast’s diagram) is (2π⁄3)r. (Get it? If not, read up on the definition of the radian. )

∴ The perimeter of C (i.e. length of the lime curve in Toast’s diagram) is 2×(2π⁄3)r = (2π⁄3)d = . That’s my answer (same as Toast’s).

1. When you have a line whose equation is y = mx + c,  m is the gradient and c is the y-intercept. For example, for the line given by y = 2x + 3, the gradient is 2 and the y-intercept is 3 (i.e. it passes through the point (0.3)).

If a line passes through the points (a,b) and (c,d), the gradient of the line is (b−d)∕(a−c).

In your example, your line passes through (4.0) and (0,−1). The gradient is therefore (0−(−1))∕(4−0) = 1⁄4.

Since the line passes through (0,−1), the y-intercept is −1.

So the equation of your line is y = {gradient}x + {y-intercept} = (1⁄4)x − 1.

2. If the equation you are given is not in the form y = mx + c, you must write it in that form so the gradient can be read. For example, if you are given x + y = 4 , you must re-write it as y = −x + 4; then you can see that the gradient is −1.

(a) In your example, (i) is already in the form y = mx + c. You can read the gradient straightaway. (ii) is not, so you must re-write it in the form y = mx + c.

(b) If two straight lines do not intersect, that means they are parallel. Parallel lines have the same gradient. Thus, if you have a line with equation y = mx + c and another line with equation y = mx + d, and c ≠ d, then the two lines are parallel; they will never intersect.

I hope you can digest all this. If not, read it through slowly, bit by bit.

I’m not sure but I’ll say …

So, is this how it’s supposed to sound (at crotchet=120)?

http://homepage.ntlworld.com/george.law7/GPN.mid

I made the MIDI file using NoteWorthy Composer, by the way.

It’s a riddle. “Seven ate nine.”

Nope, that’s not it. Look closely at the equations again.

I got it.

Now apply Wilson’s theorem and the result should follow.

If two concentric circles have radii r[sub]1[/sub] and r[sub]2[/sub] (r[sub]1[/sub] < r[sub]2[/sub]), the area of the ring in between (called an annulus) is π(r[sub]2[/sub][sup]2[/sup]−r[sub]1[/sub][sup]2[/sup]).
­
And ganesh, the 17.5 is wrong – it should be 12.5.