Let f(x,y)=5x^3−4xy^2.
(A) Find the direction of f when f is increasing fastest at point (-2,-3).
(B) What is the rate of increase in that direction?
(A) Find the direction of f when f is increasing fastest at point (-2,-3).
(B) What is the rate of increase in that direction?
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(A)
The (vector) gradient ∇f(x, y) points in the direction of greatest increase of f(x, y).
(∇f)(x,y) = (15x² - 4y²)i + (-8xy)j
(∇f)(-2, -3) = (15(-2)² - 4(-3)²)i + ((-8)(-2)(-3))j
 = 24i - 48j
I don't know what you mean by "the direction of f ". f is a (scalar) function, not a vector; it doesn't have a direction. If you're asking for a unit vector in the direction of greatest increase of f, this is (∇f) / |∇f| .
(B)
Speaking of which, the rate of increase in that direction is |(∇f)(-2, -3)| = √(24² + (-48)²) = √2880 = 24√5 ≈ 53.67.
So that unit vector I mentioned in (A) is (24i - 48j)/(24√5) = (√5/5)i - (2√5/5)j
The (vector) gradient ∇f(x, y) points in the direction of greatest increase of f(x, y).
(∇f)(x,y) = (15x² - 4y²)i + (-8xy)j
(∇f)(-2, -3) = (15(-2)² - 4(-3)²)i + ((-8)(-2)(-3))j
 = 24i - 48j
I don't know what you mean by "the direction of f ". f is a (scalar) function, not a vector; it doesn't have a direction. If you're asking for a unit vector in the direction of greatest increase of f, this is (∇f) / |∇f| .
(B)
Speaking of which, the rate of increase in that direction is |(∇f)(-2, -3)| = √(24² + (-48)²) = √2880 = 24√5 ≈ 53.67.
So that unit vector I mentioned in (A) is (24i - 48j)/(24√5) = (√5/5)i - (2√5/5)j