At 5/29/07 03:39 PM, Glaiel-Gamer wrote:
At 5/29/07 09:22 AM, dELtaluca wrote:
1. (Not sure, perhaps one of the further modules)
f'(x,y) in terms of x: 2x-4y
in terms of y: -4x+3y^2+4
if by 'f'(x,y) in terms of X' you mean, partial derivitave of f with respect to 'X' then ill agree :P get your terminology correct please.
(4/3, 2/3, z) and (4, 2, z) are the critical points (stationary).
there is no 'z' so ill ignore your next comment, but the stationary points are correct yes.
ANSWER + 'correct' method
using 'p' for the partial derivitive symbol:
pf/px = 2x - 4y, pf/py = -4x + 3y^2 + 4, rearrange pf/px to give x = 2y, substitute into pf/py
3y^2 - 8y + 4 = 0
(3y - 2)(y- 2) = 0, y = 2/3, 2 (use pf/px ->)
stationary points: (4/3,2/3) (4,2)
p^2f/px^2 = 2, p^2f/py^2 = 6y, p^2f/pxpy = p^2f/pypx = -4, forming the hessian matrix:
H =
| 2 , -4 |
| -4 , 6y |
with determinant |H| = 12y - 16, neither stationary points are degenerate, continue to classify
| H - hI | =
|2 - h , -4 |
| -4, 6y- h| = (2-h)(6y-h)-16
for y = 2, we have:
h^2- 14h +8 = 0
(h-7)^2 = 41
h = 7 pm sqrt{41}, both values of h are positive: therefore point (4,2) is a local minimum
for y = 2/3 we have:
h^2 - 6y - 6 = 0
(h-3)^2 = 15
h = 3 pm sqrt{15}, one negative, one positive: therefore point(4/3,2/3) is a saddle point
2.
tan-1(sqrt(e^2x-1))
ok, not the answer i had in mind, but a quick test (comparison to my own answer by trying values of x) they are equivalent functions, although i'd have no idea about how to convert one to the other, so ill give you half marks for using a calculator :P
Ok, so I worked backwards after finding the solution on the calculator. But seriously, how the hell would anyone figure out to do that on their own?
like this:
first substitution:
e^x = u, dx/du = 1/u
int 1/sqrt(e^{2x}-1}) dx = int 1/sqrt(u^2 - 1)*1/u du
second substitution:
u = secv, du/dv = secvtanv
int 1/sqrt(u^-1)*1/u du = int 1/tanv*1/secv*tanvsecv dv = int 1 dv = v + k
int 1/sqrt(u^2 - 1)*1/u du = sec^{-1}v + k
int 1/sqrt(e^{2x} - 1) dx = sec^{-1}(e^x) + k
like that ;)
3.
2x+x(dy/dx)+y+2y(dy/dx)=0
correct (...wait for it)
(0, sqrt(3)) and (0, -sqrt(3))
EASY!
completely wrong
2x+x(dy/dx)+y+2y(dy/dx)=0, dy/dx = (-2x - y)/(x + 2y) <=> x + 2y =/= 0
let dy/dx = 0, 2x + y = 0: substitute into original equation: y = -2x
x^2 - 2x^2 + 4x^2 = 3
3x^2 = 3, x = pm 1, y = mp -2
both stationary points satisfy x + 2y =/= 0: stationary points are:
(1, -2) (-1, 2)
4.
sin(2y)/4+y/2 = sinx-xcosx+C
correct, although it would've been nicer if you had actually shown your working out for the two integrals: