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anonymous
 5 years ago
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
 5 years ago
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.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so you want the gradiennt to be <3,2,2>

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0F(x,y,z)=3x^2 4y^2 z = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dF/dx = 6x dF/dy = 8y dF/dz = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is as far as i have gotten

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.06x = 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
 5 years ago
Best ResponseYou've already chosen the best response.06x = 3/2 ; x = 1/4 8y = 1 ; y = 1/8 1 = 1 ; z=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok cool. thanks alot! i had no idea where to start

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets work this back in again to see if it gets us something good :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01=3((1/4)^2) 4((1/8)^2).. is it 3x^2 or 3(x^2)?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0been loooking at these to long...that the same thing lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dunno, just a thought really

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i think the point is (2,1,8) 8=3(2^2) 4(1^2) 8 = 124 8=8 seems better

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get that?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i used the point i got to begin with and scaled it back to size again... i gotta see if this is right tho..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so you are sure we want the gradient to be the normal vector?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, the gradient and the normal are the same thing...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0they would have to equal each other.....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0suppose we scale the gradient fo match the normal; does that sound like a good thing to do?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if we scale it by 2; that will make the 1 = 2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but that gets me the same results for x and y; <1/4,1/8>

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets plug those in then and see what we get for a z i guess...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0got a little hyper on that one lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, so we leave z as a variable?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0z is dependant on x and y right? and it will be the same ragardless if the combo we use for that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what does z look like...it seems hyperbolic to me....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its a saddle shape graph

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the graph looks like we should only have 1 answer

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the vector can be positioned anywhere...its not anchored into anthing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not sure, i just think it looks like that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0checking it out on wolframs graphing thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have a different graphing software im using

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well lets see yours :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, let me see if i can send it in a word document

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, well i need to go. Thanks alot for working with me on this problem!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0heres what i have deduced: The gradient is a replica of the normal; for example: y=2x+8; 0 = 2xy+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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0z = 3x^2 4y^2; 0=3x^24y^2z 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
 5 years ago
Best ResponseYou've already chosen the best response.01/2<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
 5 years ago
Best ResponseYou've already chosen the best response.0the equation of the tangent plane is: 3(x+1/4) +2(y1/8)+2(z1/8)=0
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