Question

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= ?

Answer 1

well, lets take the gradient vector

Answer 2

maybe?

Answer 3

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

Answer 4

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

Answer 5

yes

Answer 6

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

Answer 7

i should prolly include a -1 for the z

Answer 8

df/dz = -1 so its part of the gradient

8x(x-Px) -4y(y-Py) -1(z-Pz)

Answer 9

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

Answer 10

cross multiply?

Answer 11

x=2. y=-2

Answer 12

haha, nm

Answer 13

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

Answer 14

z=4(2)^2−2(-2)^2

z = 16 -8 = 8; P(2,-2,8)

Answer 15

am i right?

Answer 16

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

Answer 17

its possible :)

Answer 18

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

Answer 19

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

Answer 20

my answer was 2,-2,2

Answer 21

okay

Answer 22

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

Answer 23

yes, that looks right

Answer 24

thats the equation of our tangent plane right?

Answer 25

z0 = Pz lol mixing memories lol

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

which was also incorrect

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

z=3x^2-4y^2 n=<3,2,2>

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

Answer 38

that made no sense lol

Answer 39

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),fy(a,b),-1>

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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?

Answer 48

got it lol

Answer 49

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

Answer 50

8x=8; x=1 .... not 2 lol

-4y =8; y=-2

4(1)^2 -2(-2^2) = 4-8 = -4

P(1,2,-4)

Answer 51

lets see if I can type this without a typo....

P(1,-2,-4) yay!!

Answer 52

100%!

Answer 53

you are awesome :)

Answer 54

thnx :)