How Do You Draw and Calculate Unit Vectors for Gradients?

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To calculate the gradient of the field h(x,y) = x^2 y at the point (1,3), the result is nabla h = (6, 1). The gradient vector should be drawn starting from the point (1,3) and pointing towards (6,1). To determine the unit vector for walking downhill at a 45-degree angle, the tangent slope of -1 must be used, leading to the equation v dot (6i + j) = -1. This results in a system of equations for the components of the unit vector. The discussion emphasizes understanding gradients and their geometric representation in relation to directional movement.
hexa
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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
 
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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?
 
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.
 

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