/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Find the gradient at the point. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the gradient at the point. $$f(x, y, z)=2 x+3 y+4 z, \text { at }(1,1,1)$$

Short Answer

Expert verified
The gradient at the point (1, 1, 1) is (2, 3, 4).

Step by step solution

01

Understand the Gradient Concept

The gradient of a function \( f(x, y, z) \) relates to the direction and rate of the fastest increase of the function in a three-dimensional space. It is a vector composed of the partial derivatives of \( f \) with respect to each variable \( x, y, \) and \( z \). The gradient vector is often represented as \( abla f \) or \( \text{grad } f \).
02

Compute Partial Derivatives

Calculate the partial derivative of \( f(x, y, z) = 2x + 3y + 4z \) with respect to each variable:- \( \frac{\partial f}{\partial x} = 2 \)- \( \frac{\partial f}{\partial y} = 3 \)- \( \frac{\partial f}{\partial z} = 4 \)
03

Form the Gradient Vector

The gradient vector \( abla f \) is formed by combining the partial derivatives:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (2, 3, 4) \]
04

Evaluate the Gradient at Given Point

Since the gradient vector for the function \( f(x, y, z) = 2x + 3y + 4z \) is constant (i.e., doesn't depend on \( x, y, z \)), the gradient at the point \( (1, 1, 1) \) is the same as at any other point, which is \( (2, 3, 4) \).

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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 an essential concept in calculus, especially when dealing with functions of several variables. They show how a multivariable function changes as each individual variable is modified, while keeping the other variables constant. Think of it as the slope of a function in one specific direction.
  • For a function \( f(x, y, z) \), the partial derivative with respect to \( x \) is denoted \( \frac{\partial f}{\partial x} \).
  • Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \), and with respect to \( z \) is \( \frac{\partial f}{\partial z} \).
In the context of the original exercise, we calculated these values as:
  • \( \frac{\partial f}{\partial x} = 2 \)
  • \( \frac{\partial f}{\partial y} = 3 \)
  • \( \frac{\partial f}{\partial z} = 4 \)
This step is crucial because these derivatives are the building blocks of the gradient vector.
Multivariable Calculus
Multivariable calculus is the extension of calculus to functions with more than one variable. Unlike single-variable calculus, multivariable calculus involves considering how functions behave in multidimensional spaces. This makes it applicable to real-world phenomena where several variables interact simultaneously.
For example, if you have a function \( f(x, y, z) = 2x + 3y + 4z \), it maps each point in three-dimensional space to a real number. Handling these functions requires understanding concepts like partial derivatives, gradients, and integrations across surfaces or volumes.
Below are key points in multivariable calculus:
  • **Partial Derivatives:** Determine how the function changes as each independent variable changes individually.
  • **Gradient Vectors:** Represent the direction and rate of maximum increase of the function.
  • **Level Surfaces:** Contours or surfaces where the function has the same value.
These concepts are interlinked and form the foundation for more advanced topics like optimization and differential equations.
Gradient Vector
The gradient vector is a fundamental part of multivariable calculus and points in the direction of the greatest rate of increase of a function. It's a vector made up of the partial derivatives of the function with respect to each of its variables.
In simple terms, if you're climbing a hill, the gradient vector would point directly uphill, giving you the steepest path to climb. In mathematical terms, for a function \( f(x, y, z) \), its gradient is denoted as \( abla f \) and is computed as:
  • \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \)
From our exercise, this results in the gradient vector \( (2, 3, 4) \) for the function \( f(x, y, z) = 2x + 3y + 4z \).
Some important features of the gradient vector include:
  • **Directionality:** Points towards increasing values of the function.
  • **Magnitude:** Indicates how quickly the function's value increases in that direction.
  • **Applications:** Used in optimization to find maxima and minima, and in physics for fields like electromagnetics.

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Most popular questions from this chapter

Find grad \(f\) from the differential. $$d f=2 x d x+10 y d y$$

The gas equation for one mole of oxygen relates its pressure, \(P\) (in atmospheres), its temperature, \(T\) (in \(\mathrm{K}\) ), and its volume, \(V\) (in cubic decimeters, \(\mathrm{dm}^{3}\) ): \(T=16.574 \frac{1}{V}-0.52754 \frac{1}{V^{2}}-0.3879 P+12.187 V P\) (a) Find the temperature \(T\) and differential \(d T\) if the volume is \(25 \mathrm{dm}^{3}\) and the pressure is 1 atmosphere. (b) Use your answer to part (a) to estimate how much the volume would have to change if the pressure increased by 0.1 atmosphere and the temperature remained constant.

Are the statements true or false? Give measons for your answer. If \(f(x, y)\) is a linear function, then \(d f\) is a linear function of \(d x\) and \(d y\)

In a room, the temperature is given by \(T=f(x, t)\) degrees Celsius, where \(x\) is the distance from a heater (in meters) and \(t\) is the elapsed time (in minutes) since the heat has been turned on. A person standing 3 meters from the heater 5 minutes after it has been turned on observes the following: (1) The temperature is increasing by \(1.2^{\circ} \mathrm{C}\) per minute, and (2) As the person walks away from the heater, the temperature decreases by \(2^{\circ} \mathrm{C}\) per meter as time is held constant. Estimate how much cooler or warmer it would be 2.5 meters from the heater after 6 minutes.

Find the gradient of the function. Assume the variables are restricted to a domain on which the function s defined. $$z=(x+y) e^{y}$$

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