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Chapter 14: Problem 7

find \(\nabla f\) at the given point. \(\begin{array}{c}f(x, y, z)=x^{2}+y^{2}-2 z^{2}+z \ln x, \quad(1,1,1)\end{array}\)

Short Answer

Expert verified
The gradient \(\nabla f\) at the point \((1, 1, 1)\) is \([3, 2, -4]\).

Step by step solution

01

Identify the Gradient

To find \(abla f\), we need to compute the gradient of the function \(f(x, y, z) = x^2 + y^2 - 2z^2 + z \ln x\). The gradient, \(abla f\), is a vector of the partial derivatives of \(f\) with respect to \(x\), \(y\), and \(z\).
02

Compute Partial Derivative with respect to x

Compute the partial derivative of \(f\) with respect to \(x\):\[ \frac{\partial f}{\partial x} = \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2z^2) + \frac{d}{dx}(z \ln x) = 2x + 0 + 0 + \frac{z}{x} = 2x + \frac{z}{x} \]
03

Compute Partial Derivative with respect to y

Compute the partial derivative of \(f\) with respect to \(y\):\[ \frac{\partial f}{\partial y} = \frac{d}{dy}(x^2) + \frac{d}{dy}(y^2) - \frac{d}{dy}(2z^2) + \frac{d}{dy}(z \ln x) = 0 + 2y + 0 + 0 = 2y \]
04

Compute Partial Derivative with respect to z

Compute the partial derivative of \(f\) with respect to \(z\):\[ \frac{\partial f}{\partial z} = \frac{d}{dz}(x^2) + \frac{d}{dz}(y^2) - \frac{d}{dz}(2z^2) + \frac{d}{dz}(z \ln x) = 0 + 0 - 4z + \ln x = -4z + \ln x \]
05

Evaluate the Gradient at the Given Point

Evaluate the gradient \(abla f\) at the point \((1, 1, 1)\). Substitute \(x = 1\), \(y = 1\), and \(z = 1\) into the partial derivatives:\[ abla f = \left[ 2(1) + \frac{1}{1}, 2(1), -4(1) + \ln(1) \right] = [3, 2, -4] \]
06

Conclusion

The gradient of the function \(f(x, y, z) = x^2 + y^2 - 2z^2 + z \ln x\) at the point \((1, 1, 1)\) is \([3, 2, -4]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are a fundamental concept in calculus, particularly when dealing with functions of multiple variables. Imagine a function that depends on more than one variable, say, \( f(x, y, z) \). Partial derivatives allow us to investigate how the function changes as each variable changes while keeping the others constant. This is a little different from regular derivatives in single-variable calculus, where we only focus on one variable.
Here are some key points about partial derivatives:
  • To find the partial derivative of a function with respect to one of its variables, treat all other variables as constants and differentiate normally.
  • Notation for the partial derivative of \( f(x, y, z) \) with respect to \( x \) is \( \frac{\partial f}{\partial x} \).
  • Partial derivatives can be used to construct a gradient vector, which helps us understand the direction and rate of maximum increase of a function.
Understanding partial derivatives is crucial to fully grasping the behavior of functions in higher dimensions.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions with more than one variable. While single-variable calculus deals with functions such as \( f(x) \), multivariable calculus deals with functions like \( f(x, y, z) \), involving several inputs.
This branch of calculus introduces new topics, some of which are:
  • Partial Derivatives, which help understand changes in functions concerning multiple variables.
  • Double and Triple Integrals, which extend the idea of finding areas and volumes under curves in higher dimensions.
  • Vector Calculus, which includes concepts like curl and divergence, critical for fields like physics and engineering.
The advent of multivariable calculus allows us to model and solve real-world problems involving multiple changing quantities. Waterflow over a geographical area or temperature distribution in a room are examples where multivariable calculus plays a crucial role.
Gradient Vector
The gradient vector is a powerful tool in multivariable calculus. It is a vector that points in the direction of the greatest rate of increase of a scalar function, like \( f(x, y, z) \). More than just pointing out direction, it also tells us how steep the increase is. This steepness is represented by the gradient's magnitude.
Key features of a gradient vector include:
  • It is composed of the partial derivatives of a function with respect to each variable. For example, for function \( f(x, y, z) \), the gradient is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
  • The gradient vector can help locate the maximum and minimum points of a function by setting it to zero, a process crucial in optimization problems.
  • It's applicable in multiple fields such as physics for understanding potential fields, or machine learning for optimization algorithms.
Understanding the gradient is central to any serious exploration of multivariable calculus, as it encapsulates both direction and the rate of change.

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Most popular questions from this chapter

Find the value of \(\partial z / \partial x\) at the point \((1,1,1)\) if the equation $$x y+z^{3} x-2 y z=0$$ defines \(z\) as a function of the two independent variables \(x\) and \(y\) and the partial derivative exists.

Use a CAS to plot the implicitly defined level surfaces in Exercises \(73-76 .\) $$ 4 \ln \left(x^{2}+y^{2}+z^{2}\right)=1 $$

Two dependent variables Express \(v_{x}\) in terms of \(u\) and \(y\) if the equations \(x=v \ln u\) and \(y=u \ln v\) define \(u\) and \(v\) as functions of the independent variables \(x\) and \(y,\) and if \(v_{x}\) exists. (Hint: Differentiate both equations with respect to \(x\) and solve for \(v_{x}\) by eliminating \(u_{x} .\) .

Find the linearizations \(L(x, y, z)\) of the functions at the given points. $$ f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}} \quad \text {at} $$ $$ \text { a. }(1,0,0) \quad \text { b. }(1,1,0) \quad \text { c. }(1,2,2) $$

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Determine the distance from the line \(y=x+1\) to the parabola \(y^{2}=x .\) (Hint: Let \((x, y)\) be a point on the line and \((w, z)\) a point on the parabola. You want to minimize \((x-w)^{2}+(y-z)^{2}.)\)
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