If the numerator, or denominator, goes to zero when x = 1, then (x-1) is a factor.
You can factor it out by dividing (x-1) into the numerator, say, in the style of long-division.
Thanks for the input Trebmling, but I'm wondering if we have been talking at cross purposes.
An intrinsic equation, of the form s=f(psi), passes through the origin. s is the distance along the curve from the origin to a point of interest and psi (or tan(psi)) is the slope of the curve at the...
Thanks. I've downloaded the evaluation copy. But I can't find any reference to "intrinsic" in the help files.. There's nothing in the plot menu about intrinsic equations either.
The intrinsic equaion for the parabola is,
s = a tan(psi) sec(psi) + a log(tan(psi)+sec(psi)).
How would I use...
Anyone know of a program/application/applet that will plot/graph intrinsic equations ?
(preferably free)
I've already had suggestions for KmPLot and Maxima, but neither, it seems, can do the job.
KmPLot is mainly for Linux and I'm on Win XP. (also Vista)
Thanks for the reply, metalInferno.
I don't know what the actual digram may have looked like - I only got the question text - but I imagine that the two options may have looked like the below.
Wedge moving to the left
http://img215.imageshack.us/img215/2241/wedgetotheleftxw3.th.jpg...
Homework Statement
A 45 degree wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge. Find its acceleration (gravity is downwards),Homework Equations
The Attempt at a Solution
I answered the above question on another forum, but got...
OK, got my answer.
My rules don't (always) work.
The function f(x,y)=x^2+y^2+4xy has a TP at (x,y) = (0,0) with fxx = fyy = 2, but fxy = 4 and D = -12 < 0. So f(x,y) has a saddle at the origin despite both fxx and fyy having the same sign.
Oh well, now I know.
You have the force acting down the slope as 380.8 N, with 120 N acting up the slope, from the tow rope.
That leaves you with 260.8 N down the slope moving the wagon - which means the wagon is going to move backwards.
But the wagon is already at the bottom of the slope! It can't move...
If we have a function f(x,y) for which we wish to find the stationary points, then there is one set of rules for classifying those points.
Let there be an extremum of f(x,y) at (x,y) = (a,b), and D = fxx(a,b)*fyy(a,b) - {fxy(a,b)}², then
1) D > 0 and fxx > 0 => min
2) D > 0 and fxx < 0 =>...
I don't see anything wrong with your working, except that R1 and R2 should be the other way arouind. i.e. T = (1/2)w²(m2R2 - m1R1)
1) do you have the value for m1/2 ?
2) do you have the answer ?
If you got the angular frequency OK, then you should be able to get the tension by using one of the earlier equations. You should check your working. If you can't find the error, then post your working and someone will check it out.
The launcher is the entire apparatus for setting the projectile in motion. You should rename that velocity as Vstriker.
You only need the velocity of the striker if it is still going to be moving after impact with the projectile. If the striker were still to be moving after impact, then you...
Why don't you secure the launcher rigidly, so that it doesn't move. Then Vlauncher will be zero and your equation will reduce to,
k*x^2 = Mprojectile*Vprojectile^2
a) OB = OA + AB
b = a + AB
therefore,
AB = b - a, not b + a
OP = OA + ½(AB) = a + ½(b - a) = a + ½b - ½a
OP = ½(b + a)
b) OQ = OA + (2/3)AB (Q is 2/3 of way along AB)
c) the book is wrong, you are right!