Find the point(s) on the grapf of z=3(x^2) -4(y^2) at which the vector <3,2,2> is normal to the tangent plane.

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Find the point(s) on the grapf of z=3(x^2) -4(y^2) at which the vector <3,2,2> is normal to the tangent plane.

Mathematics
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so you want the gradiennt to be <3,2,2>
F(x,y,z)=3x^2 -4y^2 -z = 0
dF/dx = 6x dF/dy = -8y dF/dz = -1

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that is as far as i have gotten
6x = 3 -8y = 2 -1 = 2 ....gonna have to scale it i spose. <3,2,2>(-1/2) = <-3/2, -1,-1> for this to work
6x = -3/2 ; x = -1/4 -8y = -1 ; y = -1/8 -1 = -1 ; z=-1
ok cool. thanks alot! i had no idea where to start
lets work this back in again to see if it gets us something good :)
-1=3((-1/4)^2) -4((-1/8)^2).. is it 3x^2 or 3(x^2)?
3(x^2)
been loooking at these to long...that the same thing lol
three x squared
3/16 -4/64 = -1?
what is that?
dunno, just a thought really
i think the point is (2,1,8) 8=3(2^2) -4(1^2) 8 = 12-4 8=8 seems better
how did you get that?
i used the point i got to begin with and scaled it back to size again... i gotta see if this is right tho..
cant tell......
ok, so you are sure we want the gradient to be the normal vector?
well, the gradient and the normal are the same thing...
they would have to equal each other.....
suppose we scale the gradient fo match the normal; does that sound like a good thing to do?
sure, lets try it
if we scale it by -2; that will make the -1 = 2
yeah
i got the same thing
but that gets me the same results for x and y; <-1/4,1/8>
lets plug those in then and see what we get for a z i guess...
i betcha thats our answers then..... (-1/4,-1/8,z) (-1/4,-1/8,z) (-1/4,1/8,z) (1/4,-1/8,z) (1/4,1/8,z)...whatcha think?
got a little hyper on that one lol
haha, so we leave z as a variable?
z is dependant on x and y right? and it will be the same ragardless if the combo we use for that
ok that makes sense
what does z look like...it seems hyperbolic to me....
its a saddle shape graph
z=1/8
the graph looks like we should only have 1 answer
...... you sure?
the vector can be positioned anywhere...its not anchored into anthing
im not sure, i just think it looks like that
checking it out on wolframs graphing thing
i have a different graphing software im using
well lets see yours :)
ok, let me see if i can send it in a word document
1 Attachment
ok, well i need to go. Thanks alot for working with me on this problem!
heres what i have deduced: The gradient is a replica of the normal; for example: y=2x+8; 0 = 2x-y+8 the gradient is <2,-1,0>; we can find the normal by picking any point on the line and using it against the gradient like this: g(Px,Py,Pz) = n g=2,-1,0 P=4,16,0 --------- n=2,-1,0 simply becasue there is no place to use the point in 'g'. Hence the gradient and the normal are scalar vectors of one another
z = 3x^2 -4y^2; 0=3x^2-4y^2-z g(x,y,z)=6x,-8y,-1 P=(x0,y0,z0) 6(x0),-8(y0),-1 is parallel to <3,2,2>; so lets scale this normal vector to suit our needs:
-1/2<3,2,2> = <-3/2,-1,-1> 6x,-8y,-1 x0, y0, z0 ---------- -3/2,-1,-1 x0=-1/4 y0= 1/8 z0 = 3(-1/4)^2 -4(1/8)^2 z0 = 3/16 - 4/64 = 1/8 P(-1/4,1/8,1/8) is a point that is on the tangent plane; with the normal vector <3,2,2>
the equation of the tangent plane is: 3(x+1/4) +2(y-1/8)+2(z-1/8)=0

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