Finding the Closest Point on a Surface to the Origin

[tuablink]

Homework Statement

Find the point on the surface
z=exp(sin(x+y)) that is closest to the origin(0,0,0)

Homework Equations

The Attempt at a Solution

x = 0, y = 0
z = exp(sin(0)) = exp(0) = 1
The point on the surface z=exp(sin(x+y)) that is closest to the origin (0,0,0) is (0,0,1)

 

[tiny-tim]

Welcome to PF!

Hi tuablink! Welcome to PF!

Find the point on the surface
z=exp(sin(x+y)) that is closest to the origin(0,0,0)
…
x = 0, y = 0
z = exp(sin(0)) = exp(0) = 1
The point on the surface z=exp(sin(x+y)) that is closest to the origin (0,0,0) is (0,0,1)

Nooo …

let the distance squared = r² = x² + y² + z².

Is ∂(r²)/∂x = 0 at (0,0,1) ?

 

[tiny-tim]

Why have you started a new thread??

solve('2*x + 2*exp(2*sin(x + y))*cos(x + y)-2*y - 2*exp(2*sin(x + y))*cos(x + y)')

Noooo!

Two equations … solve('2*x + 2*exp(2*sin(x + y))*cos(x + y) = 2*y + 2*exp(2*sin(x + y))*cos(x + y) = 0')

 

[tuablink]

Could you give me hints or guide me?
thank you

 

[tiny-tim]

uhh?

ok …

solve('

2*x + 2*exp(2*sin(x + y))*cos(x + y) =

2*y + 2*exp(2*sin(x + y))*cos(x + y) =

0')

 

[tuablink]

I tried and it gave me an error "Warning: Explicit solution could not be found"

 

[tiny-tim]

oh put the computer down and just look at the two equations.

 

[Dick]

Homework Statement

Find the point on the surface
z=exp(sin(x+y)) that is closest to the origin(0,0,0)

Homework Equations

The Attempt at a Solution

Here is my result:
x = 0, y = 0
z = exp(sin(0)) = exp(0) = 1
The point on the surface z=exp(sin(x+y)) that is closest to the origin (0,0,0) is (0,0,1)

How did you decide x=0, y=0? However you did it, it's not right. What's an expression for the distance from (x,y,z) to the origin?