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Let f(x, y) = xe⁽ˣ²⁻ʸ⁾and P=(4,16). (a) Calculate ||▼fp|| (b) Find the rate of change of f in the direction ▼fp. (C) Find the rate of change of f in the direction of a vector making an angle of 45° with ▼fp

User Ken Benoit
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(a) ||∇f_P|| = √263185, (b) the rate of change of f in the direction ∇f_P is ||∇f_P||^2 = 263185 and (c) the rate of change of f in the direction of a vector making an angle of 45° with ∇f_P is √2/2 * ||∇f_P|| = √2/2 * √263185 = √(2 * 263185)/2.

(a) Calculate ||∇f_P||.

The gradient of f is given by:

∇f = (e^(x^2-y) + 2x^2e^(x^2-y), -xe^(x^2-y))

At the point P=(4,16), we have:

∇f_P = (e^(16-16) + 216^2e^(16-16), -4*e^(16-16)) = (1 + 512, -4) = (513, -4)

Now, we calculate the magnitude of ∇f_P:

||∇f_P|| = √(513^2 + (-4)^2) = √(263169 + 16) = √263185

Therefore, ||∇f_P|| = √263185.

(b) Find the rate of change of f in the direction ∇f_P.

The rate of change of f in the direction of a vector u is given by the directional derivative:

D_uf = ∇f · u

In this case, u = ∇f_P, so we have:

D_(∇f_P)f = ∇f_P · ∇f_P = ||∇f_P||^2

Therefore, the rate of change of f in the direction ∇f_P is ||∇f_P||^2 = 263185.

(c) Find the rate of change of f in the direction of a vector making an angle of 45° with ∇f_P.

Let u be a unit vector making an angle of 45° with ∇f_P. Then the cosine of the angle between u and ∇f_P is cos(45°) = √2/2.

The directional derivative of f in the direction of u is:

D_uf = ∇f_P · u = ||∇f_P|| ||u|| cos(45°) = ||∇f_P|| (√2/2)

Therefore, the rate of change of f in the direction of a vector making an angle of 45° with ∇f_P is √2/2 * ||∇f_P|| = √2/2 * √263185 = √(2 * 263185)/2.

User Brian Mego
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