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anonymous
 5 years ago
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
 5 years ago
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= ?

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2well, lets take the gradient vector

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2fx(x,y) = 8x fy(x,y) = 4y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2<8x,4y> is an equation for the normal to every point (x,y) if i recall correctly

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.28x(xPx) 4y(yPy) +0(zPz) then... maybe?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2i should prolly include a 1 for the z

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2df/dz = 1 so its part of the gradient 8x(xPx) 4y(yPy) 1(zPz)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2simply put; 8x=8 , and 4y=8

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2put those into your z equation to get the point for z

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2z=4(2)^2−2(2)^2 z = 16 8 = 8; P(2,2,8)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i follow your thought process and agree with it but it says that the first and 3rd coordinates are incorrect

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha thats what I said! let me recheck i wrote the correct problem

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2we could plug in our normal and try to work it backwards...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.28(xPx)+8(yPy)1(zz0) = 0 look good?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, that looks right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2thats the equation of our tangent plane right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2z0 = Pz lol mixing memories lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes my book has: z=f(a,b)+fx(a,b)(xa)+fy(a,b)(yb)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.22,2,2 you say was the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, thats what I got when i tried to work the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which was also incorrect

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2ohhh..... do we have an answer to go by :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0unfortunetly, no. I do have an example problem similiar to this which I used as model that has an answer

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2f(x,y,z) = 4x^2 2y^2z = 0 is our original right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its z=3x^24y^2 n=<3,2,2> and the answer was (1/4,1/8,1/8)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes z=4x^22y^2 which is eqivalent to what you said

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2and the gradient should give us a parameteric for any normal to any point right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2<8x,4y,1> at any given P(x,y,z)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2z=3x^24y^2 n=<3,2,2> <6x,8y,1> ( px, py, pz)  <3 , 2, 2> if my thougths are right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2that made no sense lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'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>

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2fx(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
 5 years ago
Best ResponseYou've already chosen the best response.2your equation has no a or b variables so im assuming it means x,y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in the example probelm it just converts it fx(x,y)=6x => fx(a,b)=6a

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2yeah ....... thats what i was thinking :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2<6x,8y,1> as our 'generic' normal

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2<6x,8y,1> = <3,2,2> how?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2when x=1/2 and y=1/4 it appears; but lets see if that makes sense to find z :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2z = 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
 5 years ago
Best ResponseYou've already chosen the best response.212x=3 ; x = 1/4 16y = 2 ; y = 1/8

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.28x=8; x=1 .... not 2 lol 4y =8; y=2 4(1)^2 2(2^2) = 48 = 4 P(1,2,4)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2lets see if I can type this without a typo.... P(1,2,4) yay!!
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