Equation of the tangent on the curve

[imana41]

please help me about this
get Equation of the tangent plane on the surface(x^2-2y^2+2z=4) on p0(x0,y0,z0) and then find a point that tangent plane on the curve on this point is Horizontal ?

i get the Equation of the tangent on the surface(x^2-2y^2+2z=4) on p0(x0,y0,z0)
[URL]http://latex.codecogs.com/gif.latex?\bigtriangledown%20f=(2x_{0},-4y_{0},2)[/URL]

and the P equation is [URL]http://latex.codecogs.com/gif.latex?2x_{0}x-4y_{0}y+2z=2z_{0}+2x_{0}^2-4y_{0}^2[/URL]

but how can i find a point that tangent plane on the surface on this point is Horizontal ??
plaese help and sorry for my bad english

 

[Kostas_P]

Horizontal probably means parallel to the xy-plane therefore a constant z-plane. Your tangent plane equation will be a constant z plane if no x and y appear in the equation.

 

[imana41]

your purpose is ?[URL]http://latex.codecogs.com/gif.latex?\frac{\bigtriangledown%20f}{\left%20|%20\bigtriangledown%20f%20\right%20|}=(0,0,1)[/URL]

or not . pleaes explain more

 

[Kostas_P]

Yes exactly.

 

[imana41]

thanks plaese help about this
find a point on surface z=xy+1 that Nearest point to (0,0,0) ?

 

[HallsofIvy]

Well, why can't you start it out? I presume you know how to write the distance from (x,y,z) to (0,0,0). You need to minimize that with the condition z= xy+ 1. Do you know how to use the "lagrange multiplier" method? If not you could try to use that equation to eliminate z from the distance formula and minimize as a function of two variables. By the way, since the derivative of $\sqrt{f(x)}$ is $f'(x)/(2\sqrt{f(x)})$, which is 0 if and only if f'(x) is 0, it is sufficient, and easier, to minimize the distance squared rather than the distance function.

 

[imana41]

distance from (x,y,z) to (0,0,0) is [URL]http://latex.codecogs.com/gif.latex?\sqrt{(x-0)^2+(y-0)^2+(z-0)^2}[/URL] your purpose is [URL]http://latex.codecogs.com/gif.latex?d=\sqrt{x^2+y^2+x^2y^2+2xy+1}[/URL] and then d'=0 but ratio to x and y then x=0 and y=0 and z=1 and the point =(0,0,1)

 

[imana41]

Well, why can't you start it out? I presume you know how to write the distance from (x,y,z) to (0,0,0). You need to minimize that with the condition z= xy+ 1. Do you know how to use the "lagrange multiplier" method? If not you could try to use that equation to eliminate z from the distance formula and minimize as a function of two variables. By the way, since the derivative of $\sqrt{f(x)}$ is $f'(x)/(2\sqrt{f(x)})$, which is 0 if and only if f'(x) is 0, it is sufficient, and easier, to minimize the distance squared rather than the distance function.

please help me more

 

[HallsofIvy]

The distance, squared, from (x,y,z) to (0, 0, 0) is $x^2+ y^2+ z^2$.

If you are on the surface z= xy+ 1, then the distance, squared, is $x^2+ y^2+ (xy+ 1)^2$.

Minimize that by setting the partial derivatives equal to 0.

(Mimizing f^2 is the same as minimizing f, as long as f is non-negative (and distance is always non-negative) because the derivative of f^2 is 2f f' . That will be 0 only if f= 0 or f'= 0. Obviously if f= 0 that is the lowest possible value of f^2.)

 

[imana41]

The distance, squared, from (x,y,z) to (0, 0, 0) is $x^2+ y^2+ z^2$.

If you are on the surface z= xy+ 1, then the distance, squared, is $x^2+ y^2+ (xy+ 1)^2$.

Minimize that by setting the partial derivatives equal to 0.

(Mimizing f^2 is the same as minimizing f, as long as f is non-negative (and distance is always non-negative) because the derivative of f^2 is 2f f' . That will be 0 only if f= 0 or f'= 0. Obviously if f= 0 that is the lowest possible value of f^2.)

then the nearest point is (0,0,1)

 

[HallsofIvy]

Yes, that is correct.