# Local minimum of z=(x^2+y^2)^(1/2)

1. Apr 10, 2012

### hivesaeed4

How do I find the local minimum of z=sqrt(x^2+y^2)
I know its simple, but I'm stuck on it. I've tried using the second derivative but it just goes exponential. Then I tried using the second derivative test but did'nt succeed. And kindly could someone solve it step by step as it makes it much more understandable.

2. Apr 11, 2012

### HallsofIvy

I can't imagine how a second derivative of a square root can be 'exponential'!

You have $z=(x^2+ y^2)^{1/2}$. The partial derivatives are $z_x= (1/2)(x^2+ y^2)^{-1/2}(2x)= x(x^2+y^2)^{-1/2}$ and $z_y= y(x^2+y^2)^{-1/2}$. Those are never 0 so you cannot find max or min that way.

However, you can find max and min "geometrically". In polar coordinates, z= r. In the $r, \theta$ plane that is a line. Rotating around the z-axis gives a cone. It should be easy to see where the minimum value is and that there is no maximum.

3. Apr 11, 2012

### chiro

Hey hivesaeed4.

I would consider finding the minimum of z^2 since z is always positive and z^2 preserves the behaviour of the derivative in comparison to how it behaves with z.

This way you won't get any problems with division by zero and you will still get something that reflects the nature of the function.

4. Apr 11, 2012

### AlephZero

The local minimum is at the (only) point in the (x,y) plane where the derivatives of the function are not even defined. A function can have a local minimum without even being continuous, let alone differentiable (but your function is continuous).

As posts #2 and #3 said, you don't need calculus to solve this problem.