anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
so you want the gradiennt to be <3,2,2>
amistre64
  • amistre64
F(x,y,z)=3x^2 -4y^2 -z = 0
amistre64
  • amistre64
dF/dx = 6x dF/dy = -8y dF/dz = -1

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anonymous
  • anonymous
that is as far as i have gotten
amistre64
  • 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
  • amistre64
6x = -3/2 ; x = -1/4 -8y = -1 ; y = -1/8 -1 = -1 ; z=-1
anonymous
  • anonymous
ok cool. thanks alot! i had no idea where to start
amistre64
  • amistre64
lets work this back in again to see if it gets us something good :)
amistre64
  • amistre64
-1=3((-1/4)^2) -4((-1/8)^2).. is it 3x^2 or 3(x^2)?
anonymous
  • anonymous
3(x^2)
amistre64
  • amistre64
been loooking at these to long...that the same thing lol
anonymous
  • anonymous
three x squared
amistre64
  • amistre64
3/16 -4/64 = -1?
anonymous
  • anonymous
what is that?
amistre64
  • amistre64
dunno, just a thought really
amistre64
  • amistre64
i think the point is (2,1,8) 8=3(2^2) -4(1^2) 8 = 12-4 8=8 seems better
anonymous
  • anonymous
how did you get that?
amistre64
  • 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
  • amistre64
cant tell......
anonymous
  • anonymous
ok, so you are sure we want the gradient to be the normal vector?
amistre64
  • amistre64
well, the gradient and the normal are the same thing...
amistre64
  • amistre64
they would have to equal each other.....
amistre64
  • amistre64
suppose we scale the gradient fo match the normal; does that sound like a good thing to do?
anonymous
  • anonymous
sure, lets try it
amistre64
  • amistre64
if we scale it by -2; that will make the -1 = 2
anonymous
  • anonymous
yeah
anonymous
  • anonymous
i got the same thing
amistre64
  • amistre64
but that gets me the same results for x and y; <-1/4,1/8>
amistre64
  • amistre64
lets plug those in then and see what we get for a z i guess...
amistre64
  • 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
  • amistre64
got a little hyper on that one lol
anonymous
  • anonymous
haha, so we leave z as a variable?
amistre64
  • amistre64
z is dependant on x and y right? and it will be the same ragardless if the combo we use for that
anonymous
  • anonymous
ok that makes sense
amistre64
  • amistre64
what does z look like...it seems hyperbolic to me....
anonymous
  • anonymous
its a saddle shape graph
amistre64
  • amistre64
z=1/8
anonymous
  • anonymous
the graph looks like we should only have 1 answer
amistre64
  • amistre64
...... you sure?
amistre64
  • amistre64
the vector can be positioned anywhere...its not anchored into anthing
anonymous
  • anonymous
im not sure, i just think it looks like that
amistre64
  • amistre64
checking it out on wolframs graphing thing
anonymous
  • anonymous
i have a different graphing software im using
amistre64
  • amistre64
amistre64
  • amistre64
well lets see yours :)
anonymous
  • anonymous
ok, let me see if i can send it in a word document
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
ok, well i need to go. Thanks alot for working with me on this problem!
amistre64
  • 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
  • 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
  • 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
  • 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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