Problem 16
Question
Let \(f(x, y)=2 x+x^{2} y+y \sin y\) and let \(\mathbf{u}=a \mathbf{i}+b \mathbf{j}\) be a unit vector. a. Express \(D_{\mathbf{u}} f(1,0)\) in terms of \(a\) and \(b\). b. Using the result of part (a), find the values of \(a\) and \(b\) for which \(D_{\mathbf{u}} f(1,0)\) is maximum.
Step-by-Step Solution
Verified Answer
a. \(D_{\mathbf{u}} f(1,0) = 2a + b\); b. \(a = \frac{2}{\sqrt{5}}, b = \frac{1}{\sqrt{5}}\).
1Step 1: Calculate the Partial Derivatives
First, find the partial derivatives of the function \( f(x, y) = 2x + x^2 y + y \sin y \). Start with \( f_x(x, y) \):\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(2x + x^2 y + y \sin y) = 2 + 2xy\]Now find \( f_y(x, y) \):\[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(2x + x^2 y + y \sin y) = x^2 + \sin y + y \cos y\]
2Step 2: Evaluate the Partial Derivatives at (1,0)
Substitute \( x = 1 \) and \( y = 0 \) into the partial derivatives:For \( f_x(1, 0) \):\[f_x(1, 0) = 2 + 2(1)(0) = 2\]For \( f_y(1, 0) \):\[f_y(1, 0) = (1)^2 + \sin(0) + 0 \cos(0) = 1\]
3Step 3: Write the Directional Derivative Formula
The directional derivative of \( f \) at \( (1,0) \) in the direction of a unit vector \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} \) is given by:\[D_{\mathbf{u}} f(1,0) = f_x(1,0) \cdot a + f_y(1,0) \cdot b\]
4Step 4: Substitute the Evaluated Derivatives
Substitute \( f_x(1, 0) = 2 \) and \( f_y(1, 0) = 1 \) into the directional derivative formula:\[D_{\mathbf{u}} f(1,0) = 2a + b\]
5Step 5: Maximize the Directional Derivative
The function \( D_{\mathbf{u}} f(1,0) = 2a + b \) is maximized when \( a \mathbf{i} + b \mathbf{j} \) is in the direction of the gradient vector \( abla f(1,0) \). The gradient is given by:\[abla f(1,0) = \langle 2, 1 \rangle \]For maximum value, \( \mathbf{u} \) should point in the direction of \( abla f(1,0) \):\( a = \frac{2}{\sqrt{5}} \), \( b = \frac{1}{\sqrt{5}} \).
6Step 6: Verify the Unit Vector Condition
Make sure the vector \( \mathbf{u} \) is indeed a unit vector:\[\sqrt{a^2 + b^2} = \sqrt{\left(\frac{2}{\sqrt{5}}\right)^2 + \left(\frac{1}{\sqrt{5}}\right)^2} = \sqrt{\frac{4}{5} + \frac{1}{5}} = \sqrt{1} = 1\]This confirms that \( a = \frac{2}{\sqrt{5}} \) and \( b = \frac{1}{\sqrt{5}} \) maintain the condition that \( \mathbf{u} \) is a unit vector.
Key Concepts
Gradient VectorPartial DerivativesUnit Vector
Gradient Vector
The gradient vector, denoted as \( abla f(x, y) \), is a crucial concept in multivariable calculus. It represents the direction in which a function increases most quickly.
In simple terms, if you think of a hill, the gradient vector points in the direction of the steepest slope upward.
For a function like \( f(x, y) = 2x + x^2 y + y \sin y \), the gradient vector at a point gives us a full picture of how the function's value changes around that point.
The gradient vector consists of partial derivatives of the given function.
These are the rates at which the function changes as you move along the x-axis and y-axis respectively.
In our specific function, the gradient vector is:
This shows that the steepest ascent at the point (1,0) is in the direction \( \langle 2, 1 \rangle \).
Understanding the gradient vector helps us find the directional derivative, which tells us how the function changes in any given direction represented by a unit vector.
In simple terms, if you think of a hill, the gradient vector points in the direction of the steepest slope upward.
For a function like \( f(x, y) = 2x + x^2 y + y \sin y \), the gradient vector at a point gives us a full picture of how the function's value changes around that point.
The gradient vector consists of partial derivatives of the given function.
These are the rates at which the function changes as you move along the x-axis and y-axis respectively.
In our specific function, the gradient vector is:
- \( f_x(x, y) = 2 + 2xy \)
- \( f_y(x, y) = x^2 + \sin y + y \cos y \)
This shows that the steepest ascent at the point (1,0) is in the direction \( \langle 2, 1 \rangle \).
Understanding the gradient vector helps us find the directional derivative, which tells us how the function changes in any given direction represented by a unit vector.
Partial Derivatives
Partial derivatives are the building blocks of the gradient vector. They measure how a function changes with respect to one variable while keeping the other variables constant.
In the context of our function \( f(x, y) = 2x + x^2 y + y \sin y \), we compute the partial derivatives with respect to x and y.
\( \frac{\partial f}{\partial x}(1,0) = 2 \) and \( \frac{\partial f}{\partial y}(1,0) = 1 \).
This informs us that, at the point (1,0), changing x has a greater immediate effect on \( f \) compared to y.
The partial derivatives thus provide insight into the local behaviour of \( f \) and guide our understanding of how directions affect \( f \), leading to the concept of the directional derivative.
In the context of our function \( f(x, y) = 2x + x^2 y + y \sin y \), we compute the partial derivatives with respect to x and y.
- \( \frac{\partial f}{\partial x} \) captures how \( f \) changes as we tweak x while y stays fixed. For our function, \( \frac{\partial f}{\partial x} = 2 + 2xy \).
- \( \frac{\partial f}{\partial y} \) assesses the change in \( f \) as y varies, keeping x constant. Here, it's \( \frac{\partial f}{\partial y} = x^2 + \sin y + y \cos y \).
\( \frac{\partial f}{\partial x}(1,0) = 2 \) and \( \frac{\partial f}{\partial y}(1,0) = 1 \).
This informs us that, at the point (1,0), changing x has a greater immediate effect on \( f \) compared to y.
The partial derivatives thus provide insight into the local behaviour of \( f \) and guide our understanding of how directions affect \( f \), leading to the concept of the directional derivative.
Unit Vector
A unit vector is any vector with a magnitude of 1. It is used to define the direction in which we want to evaluate the rate of change, or the directional derivative, of a function.
In this exercise, the unit vector \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} \) is used.
To ensure \( \mathbf{u} \) is a unit vector, the condition \( \sqrt{a^2 + b^2} = 1 \) must hold true.
This signifies that the length of the vector does not change the direction calculation.
Instead, it standardizes it, focusing solely on the direction.
For maximizing the directional derivative of \( f \) at a point, \( \mathbf{u} \) should align with the gradient vector at that point.
From our gradient vector \( abla f(1,0) = \langle 2, 1 \rangle \), we derive:
Understanding unit vectors is essential as they help translate problems of change into geometric terms—making math approachable.
In this exercise, the unit vector \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} \) is used.
To ensure \( \mathbf{u} \) is a unit vector, the condition \( \sqrt{a^2 + b^2} = 1 \) must hold true.
This signifies that the length of the vector does not change the direction calculation.
Instead, it standardizes it, focusing solely on the direction.
For maximizing the directional derivative of \( f \) at a point, \( \mathbf{u} \) should align with the gradient vector at that point.
From our gradient vector \( abla f(1,0) = \langle 2, 1 \rangle \), we derive:
- \( a = \frac{2}{\sqrt{5}} \)
- \( b = \frac{1}{\sqrt{5}} \)
Understanding unit vectors is essential as they help translate problems of change into geometric terms—making math approachable.
Other exercises in this chapter
Problem 16
Determine \(d f\). $$ f(x, y)=\cos (x+y)+\cos (x-y) $$
View solution Problem 16
Find the direction in which \(f\) increases most rapidly at the given point, and find the maximal directional derivative at that point. $$ f(x, y)=e^{2 x}(\cos
View solution Problem 16
Find the first partial derivatives of the function. $$ f(u, v, w)=\frac{1}{\sqrt{u^{2}+v^{2}+w^{2}}} $$
View solution Problem 16
Determine whether \(\lim _{(x, y) \rightarrow(0,0)} \frac{x y}{x^{2}+y^{2}}\) exists.
View solution