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Magazine Chapter 13: Problem 1
Find \(D_{u} f\) for \(f=x^{2}+x y+y^{2}\) in the direction of \(u=\langle 2,1\rangle\) at the point \((1,1) .\)
Short Answer
Expert verified
The directional derivative is \( \frac{9\sqrt{5}}{5} \).
Step by step solution
01
Find the Gradient of f
The function is given as \( f(x, y) = x^2 + xy + y^2 \). Calculate the gradient \( abla f \) which consists of the partial derivatives of \( f \) with respect to \( x \) and \( y \). - \( \frac{\partial f}{\partial x} = 2x + y \)- \( \frac{\partial f}{\partial y} = x + 2y \)So, the gradient is \( abla f = \langle 2x + y, x + 2y \rangle \).
02
Evaluate Gradient at Specific Point
The point given is \((1,1)\).Plug \( x=1 \) and \( y=1 \) into the gradient \( abla f \).- \( abla f(1,1) = \langle 2(1) + 1, 1 + 2(1) \rangle = \langle 3, 3 \rangle \).
03
Normalize the Direction Vector
The direction vector is given as \( u = \langle 2, 1 \rangle \).First, find the magnitude of \( u \):\[\| u \| = \sqrt{2^2 + 1^2} = \sqrt{5}\]Normalize \( u \) to obtain the unit vector \( \hat{u} \):\[\hat{u} = \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle\]
04
Compute Directional Derivative
The directional derivative \( D_{u} f \) of \( f \) at the point \((1,1)\) in the direction of \( u \) is given by the dot product of \( abla f(1,1) \) and \( \hat{u} \).Calculate the dot product:\[D_{u} f = abla f(1,1) \cdot \hat{u} = \langle 3, 3 \rangle \cdot \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle\]\[D_{u} f = 3 \cdot \frac{2}{\sqrt{5}} + 3 \cdot \frac{1}{\sqrt{5}} = \frac{6}{\sqrt{5}} + \frac{3}{\sqrt{5}} = \frac{9}{\sqrt{5}}\]
05
Simplify Directional Derivative
To express \( D_{u} f \) in a simplified form, multiply both the numerator and the denominator by \( \sqrt{5} \) to rationalize the denominator:\[D_{u} f = \frac{9}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{9\sqrt{5}}{5}\]
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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 captures the multivariable rate of change and is a crucial concept in vector calculus. It's like a compass pointing in the direction of the steepest ascent of a surface represented by the function. From the function given, which is \( f(x, y) = x^2 + xy + y^2 \), we calculate the gradient \( abla f \).
This involves finding partial derivatives with respect to each variable. The component for \( x \) is \( \frac{\partial f}{\partial x} = 2x + y \) and for \( y \) is \( \frac{\partial f}{\partial y} = x + 2y \).
Therefore, the gradient is a vector represented by \( \langle 2x + y, x + 2y \rangle \). At a particular point, such as \((1, 1)\), substitution gives us \( abla f(1, 1) = \langle 3, 3 \rangle \).
This vector tells us how much the function increases when moving from point \((1, 1)\) in any direction.
This involves finding partial derivatives with respect to each variable. The component for \( x \) is \( \frac{\partial f}{\partial x} = 2x + y \) and for \( y \) is \( \frac{\partial f}{\partial y} = x + 2y \).
Therefore, the gradient is a vector represented by \( \langle 2x + y, x + 2y \rangle \). At a particular point, such as \((1, 1)\), substitution gives us \( abla f(1, 1) = \langle 3, 3 \rangle \).
This vector tells us how much the function increases when moving from point \((1, 1)\) in any direction.
Partial Derivatives
Partial derivatives break down the changes in a multivariable function by focusing on the change with respect to one variable at a time while keeping the others constant.
Consider the function \( f(x, y) = x^2 + xy + y^2 \). The partial derivative with respect to \( x \) involves treating \( y \) as a constant. This gives \( \frac{\partial f}{\partial x} = 2x + y \). Likewise, the partial derivative with respect to \( y \) treats \( x \) as constant, resulting in \( \frac{\partial f}{\partial y} = x + 2y \).
These derivatives help form the gradient vector. At \((1, 1)\), these values become 3 and 3, showing how the function value changes in each coordinate direction at that specific point. Partial derivatives are foundational for understanding how components of a vector field interact and change relative to one another.
Consider the function \( f(x, y) = x^2 + xy + y^2 \). The partial derivative with respect to \( x \) involves treating \( y \) as a constant. This gives \( \frac{\partial f}{\partial x} = 2x + y \). Likewise, the partial derivative with respect to \( y \) treats \( x \) as constant, resulting in \( \frac{\partial f}{\partial y} = x + 2y \).
These derivatives help form the gradient vector. At \((1, 1)\), these values become 3 and 3, showing how the function value changes in each coordinate direction at that specific point. Partial derivatives are foundational for understanding how components of a vector field interact and change relative to one another.
Vector Normalization
Normalizing a vector involves converting it into a unit vector, maintaining its direction but giving it a magnitude of one.
For a vector \( u = \langle 2, 1 \rangle \), we first find its magnitude \( \| u \| \) which is the square root of the sum of its squared components: \( \sqrt{2^2 + 1^2} = \sqrt{5} \).
The normalized vector or unit vector \( \hat{u} \) is obtained by dividing each component of \( u \) by its magnitude, resulting in \( \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle \).
This unit vector helps keep the direction information of \( u \) while making calculations, like the dot product with a gradient, straightforward and manageable.
For a vector \( u = \langle 2, 1 \rangle \), we first find its magnitude \( \| u \| \) which is the square root of the sum of its squared components: \( \sqrt{2^2 + 1^2} = \sqrt{5} \).
The normalized vector or unit vector \( \hat{u} \) is obtained by dividing each component of \( u \) by its magnitude, resulting in \( \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle \).
This unit vector helps keep the direction information of \( u \) while making calculations, like the dot product with a gradient, straightforward and manageable.
Dot Product
The dot product is a mathematical operation that multiplies two vectors, resulting in a scalar that is a measure of their directional similarity.
To compute the dot product of \( abla f(1,1) = \langle 3, 3 \rangle \) and the normalized vector \( \hat{u} = \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle \), each corresponding component is multiplied and then summed up: \( 3 \times \frac{2}{\sqrt{5}} \) and \( 3 \times \frac{1}{\sqrt{5}} \).
This results in \( D_{u} f = \frac{6}{\sqrt{5}} + \frac{3}{\sqrt{5}} = \frac{9}{\sqrt{5}} \). To further simplify, the directional derivative becomes \( \frac{9\sqrt{5}}{5} \) when the radicals in the denominator are rationalized.
This simplified expression measures the rate of change of the function in the specified direction \( u \), offering insight into the behavior of the surface described by \( f \).
To compute the dot product of \( abla f(1,1) = \langle 3, 3 \rangle \) and the normalized vector \( \hat{u} = \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle \), each corresponding component is multiplied and then summed up: \( 3 \times \frac{2}{\sqrt{5}} \) and \( 3 \times \frac{1}{\sqrt{5}} \).
This results in \( D_{u} f = \frac{6}{\sqrt{5}} + \frac{3}{\sqrt{5}} = \frac{9}{\sqrt{5}} \). To further simplify, the directional derivative becomes \( \frac{9\sqrt{5}}{5} \) when the radicals in the denominator are rationalized.
This simplified expression measures the rate of change of the function in the specified direction \( u \), offering insight into the behavior of the surface described by \( f \).
