191k views
4 votes
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
by
9.1k points

1 Answer

7 votes

(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
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories