Question

Prove that every tangent plane to the surface z= xe^(x/y) passes through the origin.

Answer 1

The equation of the tangent plan at any point (a,b) is given by
z= f(a,b) +Gradient(f,(a,b)) dot (x-a,y-b)

Answer 2

First you find the gradient vector at (a,b), you find
\[
\left\{\frac{a e^{a/b}}{b}+e^{a/b},-\frac{a^2 e^{a/b}}{b^2}\right\}
\]

Answer 3

Apply the above formula to find
\[
z=-\frac{a^2 e^{a/b} (y-b)}{b^2}+\left(\frac{a e^{a/b}}{b}+e^{a/b}\right)
(x-a)+a e^{a/b}
\]

Answer 4

If you put x=0 and y=0 above, you find that z=0, meaning that the the plan passes through the origin

Answer 5

you sure?

the normal vector to that surface is

\(\nabla z = \vec n = <e^{\frac{x}{y}}(1+ \frac{x}{y}), - e^{\frac{x}{y}} (\frac{x}{y})^2>\)

\(= e^{\frac{x}{y}} <(1+ \frac{x}{y}), - (\frac{x}{y})^2>\)

the equation of any plane is

\((\vec r - \vec r_o) \bullet \vec n = 0\)

if the origin is on every tangent plane, it follows that for every plane:

\((\vec r - \vec 0) \bullet \vec n = 0\) or \( \vec r \bullet \vec n = 0\)

yet

\(<x, y> \bullet \vec n \)

\( = <x, y> \bullet e^{\frac{x}{y}} <(1+ \frac{x}{y}), - (\frac{x}{y})^2> \ne 0\)

Answer 6

@518nad

Answer 7

what is z(0,0)

Answer 8

oh u can think about it like this actually we have a parametric equation like this
R(x,y)=<x,y,f(x,y)>
R_x cross R_y = normal vector at (x,y,z) = <a,b,c>
plane equation
ax+by+cz=D
z=f(x,y)
use the 2 equatiosn to get expression for D
after subbing this stuff in, you will be left with a plane equation just as a function of x,y,z and if x,y,z=0 = 0 is true then you are done

Answer 9

LOL!!!

Answer 10

hey u stop laughing at me

Answer 11

If have answer already please close:) Thanks!

Answer 12

Just laughing at meself, mate! I didn't write the closed surface first :-(

So one more time:
\(S = x e^{x/y} - z = 0\)


the normal vector to that surface is

\(\vec n = \nabla S = <e^{\frac{x}{y}}(1+ \frac{x}{y}), - e^{\frac{x}{y}} (\frac{x}{y})^2, -1>\)


the equation of any plane is

\((\vec r - \vec r_o) \bullet \vec n = 0\)

if the origin is on every tangent plane, it follows that for every plane:

\((\vec r - \vec 0) \bullet \vec n = 0\) or \( \vec r \bullet \vec n = 0\)

And

\(<x, y, z> \bullet \vec n \)

\( = <x, y, z> \bullet <e^{\frac{x}{y}}(1+ \frac{x}{y}), - e^{\frac{x}{y}} (\frac{x}{y})^2, -1> \color{red}{ = 0}\)