Problem 7
Question
In Problems 7-12, find \(\nabla f\). $$ f(x, y, z)=x^{2}-3 x y+2 z $$
Step-by-Step Solution
Verified Answer
\(\nabla f = (2x - 3y, -3x, 2)\)
1Step 1: Understand the Gradient Notation
The gradient of a function, denoted by \(abla f\), is a vector that contains the partial derivatives of the function with respect to each of its variables. For a function \(f(x, y, z)\), the gradient is given by \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\).
2Step 2: Compute Partial Derivative with Respect to x
To find \(\frac{\partial f}{\partial x}\), differentiate \(f(x, y, z) = x^2 - 3xy + 2z\) with respect to \(x\). This gives \(\frac{\partial f}{\partial x} = 2x - 3y\).
3Step 3: Compute Partial Derivative with Respect to y
To find \(\frac{\partial f}{\partial y}\), differentiate \(f(x, y, z) = x^2 - 3xy + 2z\) with respect to \(y\). This gives \(\frac{\partial f}{\partial y} = -3x\).
4Step 4: Compute Partial Derivative with Respect to z
To find \(\frac{\partial f}{\partial z}\), differentiate \(f(x, y, z) = x^2 - 3xy + 2z\) with respect to \(z\). This gives \(\frac{\partial f}{\partial z} = 2\).
5Step 5: Write the Gradient Vector
Combine the results of the partial derivatives to write the gradient vector: \(abla f = (2x - 3y, -3x, 2)\).
Key Concepts
Partial DifferentiationMultivariable CalculusVector Calculus
Partial Differentiation
Partial differentiation is a fundamental concept in multivariable calculus, focusing on how a function changes with respect to just one variable while keeping the others constant. This is particularly useful when dealing with functions of several variables, like our case here with \(f(x, y, z) = x^2 - 3xy + 2z\).
To perform partial differentiation, you differentiate the function with respect to one variable, treating all other variables as constants. For example, to find the partial derivative of our function with respect to \(x\), denoted by \(\frac{\partial f}{\partial x}\), you look at changes in \(f\) while keeping \(y\) and \(z\) constant. The process is similar for \(y\) and \(z\).
To perform partial differentiation, you differentiate the function with respect to one variable, treating all other variables as constants. For example, to find the partial derivative of our function with respect to \(x\), denoted by \(\frac{\partial f}{\partial x}\), you look at changes in \(f\) while keeping \(y\) and \(z\) constant. The process is similar for \(y\) and \(z\).
- \(\frac{\partial f}{\partial x}\) = differentials only in terms of \(x\)
- \(\frac{\partial f}{\partial y}\) = differentials only in terms of \(y\)
- \(\frac{\partial f}{\partial z}\) = differentials only in terms of \(z\)
Multivariable Calculus
Multivariable calculus extends calculus to functions of more than one variable. It includes topics like partial differentiation and multiple integrals, all building on single-variable calculus principles. This branch of mathematics is essential for analyzing situations where several varying inputs affect outputs.
In our current function, \(f(x, y, z)\), we are dealing with a three-variable situation. This means we must consider how a change in each variable uniquely affects \(f\).
In our current function, \(f(x, y, z)\), we are dealing with a three-variable situation. This means we must consider how a change in each variable uniquely affects \(f\).
- Concept of level curves: These are traces representing constant values of a function or path in a plane. In higher dimensions, they become level surfaces.
- Chain Rule in several variables: Similar to single-variable calculus but extended to handle partial derivatives.
- Optimizing multivariable functions: Finding maximum and minimum values by setting gradients to zero (critical points analysis).
Vector Calculus
Vector calculus is another advanced part of calculus that deals with vector fields—functions that assign a vector to each point in a space. This is part of how we use the gradient to understand our function.
In the context of \(f(x, y, z)\), the gradient \(abla f\) is a vector that points in the direction of the greatest rate of increase of \(f\). This vector is composed of the partial derivatives \((\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})\).
In the context of \(f(x, y, z)\), the gradient \(abla f\) is a vector that points in the direction of the greatest rate of increase of \(f\). This vector is composed of the partial derivatives \((\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})\).
- Gradient Vector \(abla f\): A vector representation of all partial derivatives, signaling the slope of a function in space.
- Directional Derivatives: Use the gradient to find function rate of change in any specified direction.
- Applications in physics and engineering: Gradient can represent things like potential fields, where objects tend to move from one potential point to another based on the gradient.
Other exercises in this chapter
Problem 6
Evaluate each line integral. \(\int_{C}\left(x^{2}+y^{2}+z^{2}\right) d s ; \quad C\) is the curve \(x=4 \cos t\), \(y=4 \sin t, z=3 t, 0 \leq t \leq 2 \pi .\)
View solution Problem 7
In Problems \(7-12\), use Stokes's Theorem to calculate \(\oint_{C} \mathbf{F} \cdot \mathbf{T} d s\). \(\mathbf{F}=2 z \mathbf{i}+x \mathbf{j}+3 y \mathbf{k} ;
View solution Problem 7
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^
View solution Problem 7
\(\int_{C} y d x+x^{2} d y\); \(C\) is the curve \(x=2 t, \quad y=t^{2}-1\), \(0 \leq t \leq 2\).
View solution