Chapter 13: Problem 4
Compute grad \(f,\) then \(D_{\mathrm{a}} f=(\operatorname{grad} f) \cdot \mathrm{u},\) then \(D_{\mathrm{u}} f\) at \(P\). \(f(x, y)=y^{10} \quad \mathbf{u}=(0,-1) \quad P=(1,-1)\)
Short Answer
Expert verified
The directional derivative \( D_{\mathbf{u}} f(P) \) is 10.
Step by step solution
01
Compute the Gradient of f
The gradient of a function, \( f(x, y) \), is a vector of its first partial derivatives. For \( f(x, y) = y^{10} \), we have the following derivatives:- \( \frac{\partial f}{\partial x} = 0 \) because there is no \( x \)-dependence,- \( \frac{\partial f}{\partial y} = 10y^9 \) using the power rule. Therefore, the gradient is \( abla f = (0, 10y^9) \).
02
Evaluate the Gradient at Point P
Substitute \( y = -1 \) into the gradient from Step 1 to evaluate it at point \( P(1, -1) \):- \( abla f(P) = (0, 10(-1)^9) = (0, -10) \).So, the gradient vector at point \( P \) is \( (0, -10) \).
03
Compute the Directional Derivative
The directional derivative \( D_{\mathbf{u}} f \) is calculated as the dot product of the gradient and the direction vector \( \mathbf{u} \). Given \( \mathbf{u} = (0, -1) \), the dot product is:\[ abla f \cdot \mathbf{u} = (0, -10) \cdot (0, -1) = 0 \times 0 + (-10) \times (-1) = 10 \].
04
Conclude the Directional Derivative at Point P
The calculation from Step 3 yields that the directional derivative of \( f \) in the direction of \( \mathbf{u} = (0, -1) \) at the point \( P(1, -1) \) is \( 10 \). Therefore:\( D_{\mathbf{u}} f(P) = 10 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
The gradient is a fundamental concept when dealing with functions of several variables. It provides a vector of the function's partial derivatives.
The gradient points in the direction of the steepest ascent of a function. It is represented as \( abla f \), where \( f \) is the function in consideration.
In the exercise, we have the function \( f(x, y) = y^{10} \). The gradient would then be the combination of partial derivatives with respect to each variable.
The gradient points in the direction of the steepest ascent of a function. It is represented as \( abla f \), where \( f \) is the function in consideration.
In the exercise, we have the function \( f(x, y) = y^{10} \). The gradient would then be the combination of partial derivatives with respect to each variable.
- For \( x \), the partial derivative is \( \frac{\partial f}{\partial x} \). Since \( f(x, y) \) does not include \( x \), this derivative is 0.
- For \( y \), the partial derivative is \( \frac{\partial f}{\partial y} \), calculated as \( 10y^9 \) using the power rule.
Partial Derivatives
Partial derivatives highlight how a function changes as only one variable changes, keeping others constant. This technique is pivotal when analyzing functions with several variables, such as \( f(x, y) \).
To find partial derivatives:
To find partial derivatives:
- Select the variable for differentiation, treating all others as constants.
- Differentiate the function with respect to this chosen variable.
- For \( \frac{\partial f}{\partial x} \), as there is no \( x \) in the function \( f(x, y) = y^{10} \), the derivative is simply 0.
- For \( \frac{\partial f}{\partial y} \), applying the power rule results in \( 10y^9 \).
Dot Product
The dot product is a mathematical operation that takes two vectors and returns a scalar.
It provides insights into the relationship between the directions of these vectors, often used to project one vector along the direction of another.
Given vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]In this exercise, the directional derivative is computed as the dot product of the gradient \( abla f = (0, -10) \) at point \( P \) and the direction vector \( \mathbf{u} = (0, -1) \):
It provides insights into the relationship between the directions of these vectors, often used to project one vector along the direction of another.
Given vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]In this exercise, the directional derivative is computed as the dot product of the gradient \( abla f = (0, -10) \) at point \( P \) and the direction vector \( \mathbf{u} = (0, -1) \):
- The dot product calculation is: \( 0 \times 0 + (-10) \times (-1) = 10 \).
Power Rule
The power rule is a basic yet powerful tool for differentiating functions, especially when dealing with polynomial terms.
For a function \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
This rule simplifies calculus operations by providing a straightforward approach to derivative calculations.
In the context of our exercise, the function \( f(x, y) = y^{10} \) requires using the power rule for its partial derivative with respect to \( y \):
For a function \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
This rule simplifies calculus operations by providing a straightforward approach to derivative calculations.
In the context of our exercise, the function \( f(x, y) = y^{10} \) requires using the power rule for its partial derivative with respect to \( y \):
- Applying the power rule, \( \frac{\partial f}{\partial y} = 10y^{9} \).