Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 13: Problem 9
Find the directional derivative of \(f\) at \(P\) in the direction of \(\mathbf{a} .\) $$ f(x, y)=4 x^{3} y^{2} ; P(2,1) ; \mathbf{a}=4 \mathbf{i}-3 \mathbf{j} $$
Short Answer
Expert verified
The directional derivative is 0.
Step by step solution
01
Find the Gradient of the Function
The gradient of the function \( f(x, y) = 4x^3y^2 \) is found by computing the partial derivatives.- Partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 12x^2y^2 \).- Partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 8x^3y \).Thus, the gradient is \( abla f = (12x^2y^2, 8x^3y) \).
02
Evaluate the Gradient at P(2,1)
Substitute the coordinates of point \( P(2,1) \) into the gradient.- \( \frac{\partial f}{\partial x} \) at \( P \) is \( 12 \times 2^2 \times 1^2 = 48 \).- \( \frac{\partial f}{\partial y} \) at \( P \) is \( 8 \times 2^3 \times 1 = 64 \).The evaluated gradient at \( P \) is \( abla f(2,1) = (48, 64) \).
03
Normalize the Direction Vector \( \mathbf{a} \)
The vector \( \mathbf{a} = 4\mathbf{i} - 3\mathbf{j} \) needs to be normalized. The magnitude of \( \mathbf{a} \) is \( \sqrt{4^2 + (-3)^2} = 5 \). Thus, the normalized vector is \( \mathbf{u} = \frac{1}{5}(4, -3) \).
04
Compute the Directional Derivative
The directional derivative of \( f \) in the direction of \( \mathbf{u} \) at \( P \) is given by the dot product \( abla f(P) \cdot \mathbf{u} \).Substitute \( abla f(2,1) = (48, 64) \) and \( \mathbf{u} = (\frac{4}{5}, -\frac{3}{5}) \):\[ abla f(P) \cdot \mathbf{u} = 48 \times \frac{4}{5} + 64 \times \left(-\frac{3}{5}\right) = \frac{192}{5} - \frac{192}{5} = 0 \]}],
05
Short Answer
The directional derivative of \( f \) at \( P \) in the direction of \( \mathbf{a} \) is 0.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
The gradient of a function is an essential concept when dealing with multivariable calculus. It is a vector that consists of the partial derivatives of the function with respect to each of its variables.
For a function like \( f(x, y) = 4x^3y^2 \), we need to find how the function changes as each variable changes.
For a function like \( f(x, y) = 4x^3y^2 \), we need to find how the function changes as each variable changes.
- Partial derivative with respect to \( x \): This calculates the rate of change of \( f \) as \( x \) changes while keeping \( y \) constant. For our function, this is \( \frac{\partial f}{\partial x} = 12x^2y^2 \).
- Partial derivative with respect to \( y \): Similarly, this finds the rate of change as \( y \) varies with \( x \) constant. We get \( \frac{\partial f}{\partial y} = 8x^3y \).
Partial Derivatives
Partial derivatives are the building blocks when analyzing functions of multiple variables. They involve differentiating the function with respect to one variable at a time while treating others as constants.
This method of differentiation helps in understanding how individual variables impact a function independently.
This method of differentiation helps in understanding how individual variables impact a function independently.
- Example in \( f(x, y) = 4x^3y^2 \): When finding \( \frac{\partial f}{\partial x} \), we differentiate concerning \( x \) while treating \( y \) as a constant, resulting in \( 12x^2y^2 \).
- Similarly, \( \frac{\partial f}{\partial y} \) is obtained by keeping \( x \) constant and differentiating with respect to \( y \), giving \( 8x^3y \).
Vector Normalization
Vector normalization is a process used to scale a non-zero vector so that it has a magnitude of one, or in other words, it becomes a unit vector.
This is particularly useful when we want to find directional derivatives, as direction vectors need to be normalized.
To normalize a vector like \( \mathbf{a} = 4\mathbf{i} - 3\mathbf{j} \):
This is particularly useful when we want to find directional derivatives, as direction vectors need to be normalized.
To normalize a vector like \( \mathbf{a} = 4\mathbf{i} - 3\mathbf{j} \):
- First, calculate the magnitude \( ||\mathbf{a}|| = \sqrt{4^2 + (-3)^2} = 5 \).
- Then, divide each component of \( \mathbf{a} \) by its magnitude. This results in the unit vector \( \mathbf{u} = \left( \frac{4}{5}, -\frac{3}{5} \right) \).
Dot Product
The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number.
In the context of directional derivatives, the dot product helps measure how much a vector points in the direction of another.
In the context of directional derivatives, the dot product helps measure how much a vector points in the direction of another.
- When two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are involved, the dot product is calculated as \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + \ldots + a_nb_n \).
- In this exercise, to find the directional derivative of \( f \), compute the dot product of the gradient at \( P \) and the vector \( \mathbf{u} \), \( abla f(2,1) \cdot \mathbf{u} \).
