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well, lets take the gradient vector
fx(x,y) = 8x fy(x,y) = -4y
<8x,-4y> is an equation for the normal to every point (x,y) if i recall correctly
8x(x-Px) -4y(y-Py) +0(z-Pz) then... maybe?
i should prolly include a -1 for the z
df/dz = -1 so its part of the gradient 8x(x-Px) -4y(y-Py) -1(z-Pz)
simply put; 8x=8 , and -4y=8
put those into your z equation to get the point for z
z=4(2)^2−2(-2)^2 z = 16 -8 = 8; P(2,-2,8)
am i right?
i follow your thought process and agree with it but it says that the first and 3rd coordinates are incorrect
its possible :)
haha thats what I said! let me recheck i wrote the correct problem
we could plug in our normal and try to work it backwards...
my answer was 2,-2,2
8(x-Px)+8(y-Py)-1(z-z0) = 0 look good?
yes, that looks right
thats the equation of our tangent plane right?
z0 = Pz lol mixing memories lol
yes my book has: z=f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)
2,-2,2 you say was the answer?
no, thats what I got when i tried to work the problem
which was also incorrect
ohhh..... do we have an answer to go by :)
unfortunetly, no. I do have an example problem similiar to this which I used as model that has an answer
f(x,y,z) = 4x^2 -2y^2-z = 0 is our original right?
its z=3x^2-4y^2 n=<3,2,2> and the answer was (-1/4,1/8,1/8)
yes z=4x^2-2y^2 which is eqivalent to what you said
and the gradient should give us a parameteric for any normal to any point right?
<8x,-4y,-1> at any given P(x,y,z)
z=3x^2-4y^2 n=<3,2,2> <6x,-8y,-1> ( px, py, pz) ------------ <3 , 2, 2> if my thougths are right
that made no sense lol
i'm not sure about the gradient, my book doesn't mentioned it. but it says: "vector i normal to the plane: v=
fx(a,b) is df/dx which is the gradient part... the a,b seems odd to me tho; but they seem to be generic fillins for the inputs that are usually x,y
your equation has no a or b variables so im assuming it means x,y
in the example probelm it just converts it fx(x,y)=6x => fx(a,b)=6a
yeah ....... thats what i was thinking :)
<6x,-8y,-1> as our 'generic' normal
<6x,-8y,-1> = <3,2,2> how?
when x=1/2 and y=-1/4 it appears; but lets see if that makes sense to find z :)
z = 1/2 when we input those; and we are looking to match: (-1/4,1/8,1/8) (1/2,-1/4,1/2) i wonder if we scale the gradient.... -2<6x,-8y,-1> = <-12x,16y,2> perhaps?
got it lol
-12x=3 ; x = -1/4 16y = 2 ; y = 1/8
8x=8; x=1 .... not 2 lol -4y =8; y=-2 4(1)^2 -2(-2^2) = 4-8 = -4 P(1,2,-4)
lets see if I can type this without a typo.... P(1,-2,-4) yay!!
you are awesome :)