Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

that is as far as i have gotten

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

3(x^2)

Answer 11

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

Answer 12

three x squared

Answer 13

3/16 -4/64 = -1?

Answer 14

what is that?

Answer 15

dunno, just a thought really

Answer 16

i think the point is (2,1,8)

8=3(2^2) -4(1^2)

8 = 12-4

8=8 seems better

Answer 17

how did you get that?

Answer 18

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

Answer 19

cant tell......

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

sure, lets try it

Answer 25

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

Answer 26

yeah

Answer 27

i got the same thing

Answer 28

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

Answer 29

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

Answer 30

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?

Answer 31

got a little hyper on that one lol

Answer 32

haha, so we leave z as a variable?

Answer 33

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

Answer 34

ok that makes sense

Answer 35

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

Answer 36

its a saddle shape graph

Answer 37

z=1/8

Answer 38

the graph looks like we should only have 1 answer

Answer 39

...... you sure?

Answer 40

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

Answer 41

im not sure, i just think it looks like that

Answer 42

checking it out on wolframs graphing thing

Answer 43

i have a different graphing software im using

Answer 44

well lets see yours :)

Answer 45

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

Answer 46

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

Answer 47

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<x,y,z>

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

Answer 48

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:

Answer 49

-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>

Answer 50

the equation of the tangent plane is:

3(x+1/4) +2(y-1/8)+2(z-1/8)=0