(a) Show that a differentiable function f decreases most rapidly at x in the direction opposite the gradient vector, that is, in the direction of −∇f(x). Let θ be the angle between ∇f(x) and unit vector u. Then Du f = |∇f| cos θ _____ . Since the minimum value of cos θ _ is -1 occurring, for 0 ≤ θ < 2π, when θ = , the minimum value of Du f is −|∇f|, occurring when the direction of u is the opposite of _ the direction of ∇f (assuming ∇f is not zero). (b) Use the result of part (a) to find the direction in which the function f(x, y) = x4y − x2y3 decreases fastest at the point (2, −5)._______

[tex]\text{direction opposite the gradient vector, that is, in the direction of}[/tex] [tex]-\bigtriangledown f(x)[/tex][tex]\text{. Let}\ \theta \ \text{be the angle between} \bigtriangledown f(x) \ \text{and unit vector u. Then } D_u f = \mathbf{|\bigtriangledown f| \ cos \ \theta }}[/tex]

[tex]\text{Since the minimum value of} \ \ \mathbf{cos \ \theta} \ \ is \mathbf{-1} \ \text{occuring \ for \ 0} \le \ \theta \ < 2x, \\ \\ when \ \theta = \mathbf{\pi} , \text{the mnimum value of} \ D_uf \ is} -|\bigtriangledown f|, \text{occuring when the direction of u is } \\ \\ \ \mathbf{the \ opposite \ of} \ \text{the direction of } \ \bigtriangledown f (assuming \ \bigtriangledown f\ is \ not \ zero)[/tex]

b) [tex]\text{From part A:}[/tex]

[tex]If \ f(x,y) = x^4y -x^2y^2 \ \ decreases \ fastest \ at \ the \point \ (2,-5)\\ \\ F(x,y) = x^4y -x^2y^3 \\ \\ f_x = \dfrac{df}{dx}= \dfrac{d}{dx}(x^4y-x^2y^3) \\ \\ f_x = \dfrac{df}{dx}= y4x^3 -2y^3x \\ \\ For(2,-5) \\ \\ f_x = (-5)4(2)^3 -2(-5)^3(2) \\ \\ \mathbf{ f_x = 340}[/tex]

[tex]However; f_y = \dfrac{df}{dy} = \dfrac{d}{dy}(x^4y - x^2y^3) \\ \\ f_y = x^4 -3x^2y^2 \\ \\ Now, for (2, -5)\\ \\f_y = (2)^4 -3(2)^2(-5)^2 \\ \\ f_y = -284[/tex]

[tex]So; \bigtriangledown = < 340,-284> \text{this is the direction of fastest decrease}[/tex]