Find the point on the graph of z=4x^2−2y^2 at which vector n=<8,8,−1> is normal to the tangent plane.
P= ?

- anonymous

- schrodinger

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- amistre64

well, lets take the gradient vector

- amistre64

maybe?

- amistre64

fx(x,y) = 8x
fy(x,y) = -4y

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## More answers

- amistre64

<8x,-4y> is an equation for the normal to every point (x,y) if i recall correctly

- anonymous

yes

- amistre64

8x(x-Px) -4y(y-Py) +0(z-Pz) then... maybe?

- amistre64

i should prolly include a -1 for the z

- amistre64

df/dz = -1 so its part of the gradient
8x(x-Px) -4y(y-Py) -1(z-Pz)

- amistre64

simply put; 8x=8 , and -4y=8

- anonymous

cross multiply?

- amistre64

x=2. y=-2

- anonymous

haha, nm

- amistre64

put those into your z equation to get the point for z

- amistre64

z=4(2)^2−2(-2)^2
z = 16 -8 = 8; P(2,-2,8)

- amistre64

am i right?

- anonymous

i follow your thought process and agree with it but it says that the first and 3rd coordinates are incorrect

- amistre64

its possible :)

- anonymous

haha thats what I said! let me recheck i wrote the correct problem

- amistre64

we could plug in our normal and try to work it backwards...

- anonymous

my answer was 2,-2,2

- anonymous

okay

- amistre64

8(x-Px)+8(y-Py)-1(z-z0) = 0 look good?

- anonymous

yes, that looks right

- amistre64

thats the equation of our tangent plane right?

- amistre64

z0 = Pz lol mixing memories lol

- anonymous

yes my book has:
z=f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)

- amistre64

2,-2,2 you say was the answer?

- anonymous

no, thats what I got when i tried to work the problem

- anonymous

which was also incorrect

- amistre64

ohhh..... do we have an answer to go by :)

- anonymous

unfortunetly, no. I do have an example problem similiar to this which I used as model that has an answer

- amistre64

f(x,y,z) = 4x^2 -2y^2-z = 0 is our original right?

- anonymous

its z=3x^2-4y^2 n=<3,2,2> and the answer was (-1/4,1/8,1/8)

- anonymous

yes z=4x^2-2y^2 which is eqivalent to what you said

- amistre64

and the gradient should give us a parameteric for any normal to any point right?

- amistre64

<8x,-4y,-1> at any given P(x,y,z)

- amistre64

z=3x^2-4y^2 n=<3,2,2>
<6x,-8y,-1>
( px, py, pz)
------------
<3 , 2, 2> if my thougths are right

- amistre64

that made no sense lol

- anonymous

i'm not sure about the gradient, my book doesn't mentioned it. but it says:
"vector i normal to the plane: v=

- amistre64

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

- amistre64

your equation has no a or b variables so im assuming it means x,y

- anonymous

in the example probelm it just converts it
fx(x,y)=6x => fx(a,b)=6a

- amistre64

yeah ....... thats what i was thinking :)

- amistre64

<6x,-8y,-1> as our 'generic' normal

- amistre64

<6x,-8y,-1> = <3,2,2> how?

- amistre64

when x=1/2 and y=-1/4 it appears; but lets see if that makes sense to find z :)

- amistre64

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?

- amistre64

got it lol

- amistre64

-12x=3 ; x = -1/4
16y = 2 ; y = 1/8

- amistre64

8x=8; x=1 .... not 2 lol
-4y =8; y=-2
4(1)^2 -2(-2^2) = 4-8 = -4
P(1,2,-4)

- amistre64

lets see if I can type this without a typo....
P(1,-2,-4) yay!!

- anonymous

100%!

- anonymous

you are awesome :)

- amistre64

thnx :)

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