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.

- anonymous

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- amistre64

so you want the gradiennt to be <3,2,2>

- amistre64

F(x,y,z)=3x^2 -4y^2 -z = 0

- amistre64

dF/dx = 6x
dF/dy = -8y
dF/dz = -1

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- anonymous

that is as far as i have gotten

- amistre64

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

- amistre64

6x = -3/2 ; x = -1/4
-8y = -1 ; y = -1/8
-1 = -1 ; z=-1

- anonymous

ok cool. thanks alot! i had no idea where to start

- amistre64

lets work this back in again to see if it gets us something good :)

- amistre64

-1=3((-1/4)^2) -4((-1/8)^2).. is it 3x^2 or 3(x^2)?

- anonymous

3(x^2)

- amistre64

been loooking at these to long...that the same thing lol

- anonymous

three x squared

- amistre64

3/16 -4/64 = -1?

- anonymous

what is that?

- amistre64

dunno, just a thought really

- amistre64

i think the point is (2,1,8)
8=3(2^2) -4(1^2)
8 = 12-4
8=8 seems better

- anonymous

how did you get that?

- amistre64

i used the point i got to begin with and scaled it back to size again... i gotta see if this is right tho..

- amistre64

cant tell......

- anonymous

ok, so you are sure we want the gradient to be the normal vector?

- amistre64

well, the gradient and the normal are the same thing...

- amistre64

they would have to equal each other.....

- amistre64

suppose we scale the gradient fo match the normal; does that sound like a good thing to do?

- anonymous

sure, lets try it

- amistre64

if we scale it by -2; that will make the -1 = 2

- anonymous

yeah

- anonymous

i got the same thing

- amistre64

but that gets me the same results for x and y; <-1/4,1/8>

- amistre64

lets plug those in then and see what we get for a z i guess...

- amistre64

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?

- amistre64

got a little hyper on that one lol

- anonymous

haha, so we leave z as a variable?

- amistre64

z is dependant on x and y right? and it will be the same ragardless if the combo we use for that

- anonymous

ok that makes sense

- amistre64

what does z look like...it seems hyperbolic to me....

- anonymous

its a saddle shape graph

- amistre64

z=1/8

- anonymous

the graph looks like we should only have 1 answer

- amistre64

...... you sure?

- amistre64

the vector can be positioned anywhere...its not anchored into anthing

- anonymous

im not sure, i just think it looks like that

- amistre64

checking it out on wolframs graphing thing

- anonymous

i have a different graphing software im using

- amistre64

##### 1 Attachment

- amistre64

well lets see yours :)

- anonymous

ok, let me see if i can send it in a word document

- anonymous

##### 1 Attachment

- anonymous

ok, well i need to go. Thanks alot for working with me on this problem!

- amistre64

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

- amistre64

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:

- amistre64

-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>

- amistre64

the equation of the tangent plane is:
3(x+1/4) +2(y-1/8)+2(z-1/8)=0

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