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Chapter 11: Problem 30

Use the function $$f(x, y)=3-\frac{x}{3}-\frac{y}{2}$$ Find \(\nabla f(x, y)\)

Short Answer

Expert verified
The gradient of the function \(f(x, y) = 3 - \frac{x}{3} - \frac{y}{2}\) is \(\nabla f(x, y) = \langle -\frac{1}{3}, -\frac{1}{2} \rangle\).

Step by step solution

01

Find the partial derivative of \(f(x, y)\) with respect to \(x\)

The partial derivative of \(f(x, y)\) with respect to \(x\) is calculated by taking the derivative of \(f\) with respect to \(x\), while treating all other variables as constants. This gives \(f_x = -\frac{1}{3}\).
02

Find the partial derivative of \(f(x, y)\) with respect to \(y\)

The partial derivative of \(f(x, y)\) with respect to \(y\) is calculated in a similar way to the \(x\) partial derivative. This time, however, we take the derivative of \(f\) with respect to \(y\), while treating all other variables as constants. This gives \(f_y = -\frac{1}{2}\).
03

Write the gradient

The gradient of \(f(x, y)\) is written as a vector of the two partial derivatives calculated in Steps 1 and 2. So, \(\nabla f(x, y) = \langle f_x, f_y \rangle = \langle -\frac{1}{3}, -\frac{1}{2} \rangle\).

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Key Concepts

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

Gradient
The gradient of a function is like a compass that points in the direction of the greatest rate of increase of the function. For a function of two variables, the gradient is a vector made up of the partial derivatives with respect to each variable.

This vector indicates how the function changes at any point. To find the gradient, we calculate the partial derivatives, as shown in the exercise solution.
  • The gradient tells us how steep the function is at a particular point.
  • It also provides direction; it points towards the maximum increase.
This is especially useful in optimization problems where one needs to find where a function reaches its maximum or minimum value.
Functions of Two Variables
When dealing with functions of two variables, it means that the function relies on two inputs at the same time. In our exercise, these inputs are represented by the variables \(x\) and \(y\).

Specifically, the function \(f(x, y)=3-\frac{x}{3}-\frac{y}{2}\) describes a plane in a 3D space where each point on the plane results from specific values of \(x\) and \(y\).
  • Functions of two variables can represent multiple types of surfaces, such as planes, paraboloids, and more.
  • These functions are commonly used to model real-world phenomena like temperature distribution or elevation in geography.
Understanding these functions is vital in fields like physics and engineering, where two-dimensional graphs can translate into real-world data.
Calculus
Calculus is the branch of mathematics that studies continuous change. It helps us understand how quantities change and interact with each other in dynamic ways. In the context of the given exercise, we primarily deal with derivatives, which measure how a function changes as its input changes.
  • Partial derivatives, like those found in the exercise, are essential tools in calculus for dealing with functions of several variables.
  • They allow us to focus on change with respect to one variable while keeping others constant.
Calculus provides the tools necessary to solve problems about motion, area, volumes, and anything that requires a precise understanding of change. Whether smoothing the edges in graphs or finding optimization points, calculus is a cornerstone for analyzing and understanding both theoretical and practical problems.

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Most popular questions from this chapter

Use the gradient to find a unit normal vector to the graph of the equation at the given point. Sketch your results $$ x e^{y}-y=5,(5,0) $$

Use the gradient to find a unit normal vector to the graph of the equation at the given point. Sketch your results $$ 9 x^{2}+4 y^{2}=40,(2,-1) $$

Acceleration The centripetal acceleration of a particle moving in a circle is \(a=v^{2} / r,\) where \(v\) is the velocity and \(r\) is the radius of the circle. Approximate the maximum percent error in measuring the acceleration due to errors of \(3 \%\) in \(v\) and \(2 \%\) in \(r\)

Find a function \(f\) such that \(\nabla f=e^{x} \cos y \mathbf{i}-e^{x} \sin y \mathbf{j}+z \mathbf{k}\).

Find the gradient of the function and the maximum value of the directional derivative at the given point. $$ \frac{\text { Function }}{w=x y^{2} z^{2}} \frac{\text { Point }}{(2,1,1)} $$
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