# Find the point on the surface

1. Nov 22, 2009

1. The problem statement, all variables and given/known data
Find the point on the surface
z=exp(sin(x+y)) that is closest to the origin(0,0,0)

2. Relevant equations

3. 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)

2. Nov 22, 2009

### tiny-tim

Welcome to PF!

Nooo

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

Is ∂(r2)/∂x = 0 at (0,0,1) ?

3. Nov 22, 2009

### tiny-tim

Why have you started a new thread??
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')

4. Nov 22, 2009

Could you give me hints or guide me?
thank you

5. Nov 22, 2009

### 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')

6. Nov 22, 2009

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

7. Nov 22, 2009

### tiny-tim

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

8. Nov 22, 2009

### Dick

Re: How can I find the point on the surface?

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?

Last edited by a moderator: Nov 22, 2009