# Gradients and unit vectors

1. Apr 22, 2006

### hexa

Does someone want to think along with me?

field h(x,y) = x^2 y

calculate gradient for this field and draw it for point (1,3):

calculate this point: nabla h = (2xy, x^2) = (2*1*3, 1) = (6, 1)

But how do I draw this into the field of x^2y? I know I have to draw a vector. Would this vector go from the middle towards (6, 1)?

Then: Imagine you walk in this field in the point (1,3) and you want to walk downhill at an angle of 45 degrees. Calculate the unit vector (is that the correct term?) that shows which direction to walk.

I know how to calculate the unit vector, but how do I put the 45 degrees into this?

Thanks a lot,
Hexa

2. Apr 24, 2006

### siddharth

At the point (1,3) the value of grad h is (6,1). So the direction of the "arrow" will be along the vector 6 i + j

How is gradients precalculus?

3. Apr 24, 2006

### HallsofIvy

Staff Emeritus
And the tail of the vector should be at the given point (1, 3).

As for the second problem: "downward at an angle of 45 degrees" means that the tangent is -1. If v is a unit vector, the derivative at (x,y) in the direction of v is v dot grad f(x,y). You've already calculated that the gradient of f at (1,3) is 6i+ j. Now you need to find a unit vector v[\b]= ui+vj so that v dot 6i+j= 6u+ v= -1. That gives you two equations for u and v.