# Homework Help: Find Steepest Climb of this Hyperbolic Paraboloid

1. Oct 11, 2009

### UziStuNNa

Last edited: Oct 12, 2009
2. Oct 11, 2009

### lanedance

if you define F(x,y,z) = x^2-y^2+z

then the paraboloid is defined by the level surface
F(x,y,z) = 0

the gradient direction represents the direction of maximum change, and will by defintion be perpindicular to any level surface.

f(x,y) = z = y^2-x^2

$$\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x})$$

this will represenet the direction of greatest change of z = f(x,y), with x & y and i think it should be easy to relate the slope to the magintude of the gradient

3. Oct 11, 2009

### UziStuNNa

Thank you for the help, but one more question...
Do I need to multiply the gradient of F(x,y) by the unit vector to find the direction of greatest ascent?

4. Oct 11, 2009

### lanedance

no worries, but the question you asked doesn't really make sense, as the gradient is a vector, so what do you mean multiplying a vector by a vector?

the steps i outlined, you will give you the x,y direction, which direction of greatest ascent & the value of that slope in that direction

if you want to find the unit vector representing the direction on surface, use the x,y direction with the slope to find a vector in that direction (in 3D), then normalise by divding by the vector magnitude

note as a check, the vector you find will be perpindicular to the gradient you gave in you first post