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Suppose f(x,y)=xy, P=(−4,−4) and v=2i+3j. A. Find the gradient of f. ∇f= i+ j Note: Your answers should be expressions of x and y; e.g. "3x - 4y" B. Find the gradient of f at the point P. (∇f)(P)= i+ j Note: Your answers should be numbers C. Find the directional derivative of f at P in the direction of v. Duf= Note: Your answer should be a number D. Find the maximum rate of change of f at P. Note: Your answer should be a number E. Find the (unit) direction vector in which the maximum rate of change occurs at P. u= i+ j Note: Your answers should be numbers

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Answer:

a) The gradient of a function is the vector of partial derivatives. Then

[tex]\nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})=(y,x)=y\hat{i} + x\hat{j}[/tex]

b) It's enough evaluate P in the gradient.

[tex]\nabla f(P)=(-4,-4)=-4\hat{i} - 4 \hat{j}[/tex]

c) The directional derivative of f at P in direction of V is the dot produtc of [tex]\nabla f(P)[/tex] and v.

[tex]\nabla f(P) v=(-4,-4)\left[\begin{array}{ccc}2\\3\end{array}\right] =(-4)2+(-4)3=-20[/tex]

d) The maximum rate of change of f at P is the magnitude of the gradient vector at P.

[tex]||\nabla f(P)||=\sqrt{(-4)^2+(-4)^2}=\sqrt{32}=4\sqrt{2}[/tex]

e) The maximum rate of change occurs in the direction of the gradient. Then

[tex]v=\frac{1}{4\sqrt{2}}(-4,-4)=(\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}})= \frac{-1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}[/tex]

is the direction vector in which the maximum rate of change occurs at P.

Answers:

  • Gradient of f:    [tex]\nabla f = y\hat{i} + x\hat{j}[/tex]
  • Gradient of f at point p: [tex]\nabla f = -4\hat{i} -4\hat{j}[/tex]
  • Directional derivative of f and P in direction of v: [tex]\nabla f(P)v = -20\\[/tex]
  • The maximum rate of change of f at P:  [tex]| \nabla f(P)| = 4\sqrt{2}[/tex]
  • The (unit) direction vector in which the maximum rate of change occurs at P is:  [tex]v = -\dfrac{1}{\sqrt{2}}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{j}[/tex]

Step by step solutions:

Given that:

  • [tex]f(x,y) = xy[/tex]
  • [tex]P = (-4,4)\\[/tex]
  • [tex]v = 2i + 3j[/tex]

A: Gradient of f

[tex]\nabla f = (\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}) = (y,x) = y\hat{i} + x\hat{j}[/tex]

B: Gradient of f at point P:

Just put the coordinates of p in above formula:

[tex]\nabla f = -4\hat{i} -4\hat{j}[/tex]

C: The directional derivative of f and P in direction of v:

The directional derivative is found by dot product of [tex]\nabla f(P) \: \rm and \: \rm v[/tex]:

[tex]\nabla f(P)v = [-4,4][2,3]^T = -20\\[/tex]

D: The maximum rate of change of f at P is calculated by evaluating the magnitude of gradient vector at P:

[tex]| \nabla f(P)| = \sqrt{(-4)^2 + (-4)^2} = 4\sqrt{2}[/tex]

E: The (unit) direction vector in which the maximum rate of change occurs at P is:

[tex]v = (\dfrac{-4}{4\sqrt{2}}, \dfrac{-4}{4\sqrt{2}}) = -\dfrac{1}{\sqrt{2}}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{j}[/tex]

That vector v is the needed unit vector in this case.

we divided by [tex]4\sqrt{2}[/tex] to make that vector as of unit length.

Learn more about vectors here:

https://brainly.com/question/12969462

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