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Find the directional derivative of \(f\) at \(P\) in the direction of \(\mathbf{v}\); that is, find \(D_{\mathbf{u}} f(P) . \quad\) where \(\quad \mathbf{u}=\frac{\mathbf{v}}{|\mathbf{v}|}\). $$ f(x, y)=x^{2}+2 x y+3 y^{2}: \quad P(2,1), \mathbf{v}=\langle 1,1\rangle $$

Short Answer

Expert verified
The directional derivative \( D_{\mathbf{u}}f(P) \) is \( 8\sqrt{2} \).

Step by step solution

01

Find the Gradient of f

The gradient of a function \(f(x, y)\), denoted as \( abla f \), is the vector of partial derivatives \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y) = x^2 + 2xy + 3y^2 \), we find the partial derivatives:\[ \frac{\partial f}{\partial x} = 2x + 2y \]\[ \frac{\partial f}{\partial y} = 2x + 6y \]Thus, the gradient at any point \((x, y)\) is \( abla f = (2x + 2y, 2x + 6y) \).
02

Evaluate the Gradient at Point P

To find \( abla f \) at point \( P(2, 1) \), substitute \( x = 2 \) and \( y = 1 \) into the gradient:\[ abla f(2, 1) = (2(2) + 2(1), 2(2) + 6(1)) = (6, 10) \].
03

Find the Unit Vector \( \mathbf{u} \)

The direction vector \( \mathbf{v} = \langle 1, 1 \rangle \). Its magnitude is\[ |\mathbf{v}| = \sqrt{1^2 + 1^2} = \sqrt{2} \].The unit vector in the direction of \( \mathbf{v} \) is \( \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \).
04

Compute the Dot Product

The directional derivative \( D_{\mathbf{u}}f(P) \) is given by the dot product of the gradient and the unit vector \( \mathbf{u} \). Compute:\[ (6, 10) \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = 6\left(\frac{1}{\sqrt{2}}\right) + 10\left(\frac{1}{\sqrt{2}}\right) \].Simplify:\[ \frac{6}{\sqrt{2}} + \frac{10}{\sqrt{2}} = \frac{16}{\sqrt{2}} = 8\sqrt{2} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient
The gradient of a function is like a roadmap showing the direction and rate of fastest increase. In mathematical terms, the gradient is a vector formed by all the partial derivatives of the function, indicating how the function changes as each variable changes.

To find the gradient of a function like \( f(x, y) = x^2 + 2xy + 3y^2 \), compute the partial derivatives with respect to each variable. These partial derivatives, or small changes, tell us how \( f \) changes as \( x \) or \( y \) changes, respectively.
  • The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x + 2y \).
  • The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2x + 6y \).

Therefore, the gradient is \( abla f = (2x + 2y, 2x + 6y) \). When you evaluate this at a particular point, like \((2, 1)\), it tells you how the function \( f \) changes at that exact spot; in this case, \( abla f(2, 1) = (6, 10) \). This means at point \((2, 1)\), the function increases 6 units for every unit \( x \) moves in one direction and 10 units for each unit \( y \) moves in that same direction.
Partial Derivatives
Partial derivatives are a way of looking at how a function changes as you tweak one variable at a time while keeping others constant. They give us the components needed to form the gradient vector. For example, if you have a function of two variables, like \( f(x, y) \), you can derive how \( f \) changes with just \( x \) changing and then separately with \( y \) changing.
  • \( \frac{\partial f}{\partial x} \) tells us how \( f \) changes as \( x \) changes, keeping \( y \) constant.
  • \( \frac{\partial f}{\partial y} \) tells us how \( f \) changes as \( y \) changes, keeping \( x \) constant.

These derivatives help us understand the trajectory of a function in a multi-dimensional space. Calculating them for a function like \( f(x, y) = x^2 + 2xy + 3y^2 \) shows that:
  • \( \frac{\partial f}{\partial x} = 2x + 2y \)
  • \( \frac{\partial f}{\partial y} = 2x + 6y \)

Each partial derivative gives us detailed insight into how the function behaves in relation to each variable, serving as a building block for more complex calculations involving gradients and directional derivatives.
Unit Vector
A unit vector is essentially a "pure direction" since its magnitude is always 1. When calculating a directional derivative, we need a unit vector to specify which direction we are moving along the function's surface.

If you have a vector \( \mathbf{v} = \langle 1, 1 \rangle \), you first find its magnitude using the formula \( |\mathbf{v}| = \sqrt{a^2 + b^2} \). For \( \mathbf{v} \) in our example, the magnitude is \( \sqrt{1^2 + 1^2} = \sqrt{2} \). To find the corresponding unit vector \( \mathbf{u} \), we scale \( \mathbf{v} \) to have a magnitude of 1 by dividing each component by \( |\mathbf{v}| \):
  • \( \mathbf{u} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \).

This unit vector is crucial for computing the directional derivative since it provides the exact direction for evaluating how fast a function \( f \) increases or decreases. In other words, \( \mathbf{u} \) helps us focus on a specific line of movement, ensuring that the derivative only pertains to that path, without influence from vector scale or other directions.

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