Find slope of multivariable function

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To find the steepest points on the hill described by the function f(x,y) = 3/(1+x²+y²), the gradient ∇f(x,y) must be calculated, yielding the partial derivatives d/dx(f(x,y)) = -6x/(1+x²+y²)² and d/dy(f(x,y)) = -6y/(1+x²+y²)². The maximum slope occurs where the magnitude of the gradient ||∇f(x,y)|| is maximized. To find these extremal points, one can differentiate the expression for ||∇f(x,y)|| with respect to x and y. Simplifying the process by maximizing the square of the gradient instead of the square root can also be beneficial.
dolle39
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Homework Statement


A hill is described with the following function:

f(x,y) = 3/(1+x2 +y2)

Where f(x,y) is the height. Find the points where the hill is steepest!

Homework Equations


∇f(x,y) = d/dx(f(x,y))i + d/dy(f(x,y))j

The Attempt at a Solution


d/dx(f(x,y)) = -6x/(1+x2+y2)2
d/dy(f(x,y)) = -6y/(1+x2+y2)2

Know as far as I understand the maximum slope should occur when:

||∇f(x,y)|| has it max. But how should I find that? And am I even going in the right direction?
 
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Yes, that should work. You can write out the expression for ||∇f(x,y)|| and differentiate wrt to x, y, to find out where its extremals are. You can make life a bit easier by noting that you can discard the square root operation, since a positive value is maximised when its square is maximised.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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