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$$ \text { In Problems 1-10, find the gradient } \nabla f \text {. } $$ $$ f(x, y, z)=x^{2} y+y^{2} z+z^{2} x $$

Short Answer

Expert verified
The gradient is \( \nabla f = \langle 2xy + z^2, x^2 + 2yz, y^2 + 2zx \rangle.\)

Step by step solution

01

Understanding the Gradient

The gradient of a function, denoted as \( abla f \), is a vector that contains the partial derivatives of the function with respect to each of its variables. For the function \( f(x, y, z) = x^2y + y^2z + z^2x \), we need to find \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), and \( \frac{\partial f}{\partial z} \).
02

Finding \( \frac{\partial f}{\partial x} \)

To find the partial derivative of \( f \) with respect to \( x \), treat \( y \) and \( z \) as constants. Differentiate \( x^2y + y^2z + z^2x \) with respect to \( x \), resulting in: \[ \frac{\partial f}{\partial x} = 2xy + z^2. \]
03

Finding \( \frac{\partial f}{\partial y} \)

To find the partial derivative of \( f \) with respect to \( y \), treat \( x \) and \( z \) as constants. Differentiate \( x^2y + y^2z + z^2x \) with respect to \( y \), resulting in: \[ \frac{\partial f}{\partial y} = x^2 + 2yz. \]
04

Finding \( \frac{\partial f}{\partial z} \)

To find the partial derivative of \( f \) with respect to \( z \), treat \( x \) and \( y \) as constants. Differentiate \( x^2y + y^2z + z^2x \) with respect to \( z \), resulting in: \[ \frac{\partial f}{\partial z} = y^2 + 2zx. \]
05

Writing the Gradient Vector

The gradient vector \( abla f \) is composed of the partial derivatives found in the previous steps. Thus, the gradient of \( f \) is: \[ abla f = \langle 2xy + z^2, x^2 + 2yz, y^2 + 2zx \rangle. \]

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Key Concepts

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

Partial Derivatives
In calculus, the concept of partial derivatives is crucial when dealing with functions of multiple variables. A partial derivative of a function with respect to one of the variables involves taking the derivative while keeping all other variables constant. This is different from ordinary derivatives where the function depends on only one variable.

For instance, consider the function \( f(x, y, z) = x^2y + y^2z + z^2x \). To find the partial derivative with respect to \( x \), you treat \( y \) and \( z \) as constants and only differentiate with respect to \( x \).

  • Partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = 2xy + z^2 \).
  • Partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = x^2 + 2yz \).
  • Partial derivative with respect to \( z \): \( \frac{\partial f}{\partial z} = y^2 + 2zx \).

These derivatives help in understanding how the function changes as we vary one variable at a time.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. It covers topics such as partial derivatives, gradients, and multiple integrals. This type of calculus is significant in fields that require analysis of more than two dimensions.

A function in multivariable calculus, like \( f(x, y, z) = x^2y + y^2z + z^2x \), has three variables: \( x \), \( y \), and \( z \). Each variable can independently influence the outcome of the function, making the study of how each one affects the function particularly interesting.

Using techniques learned in multivariable calculus:
  • Derivatives of functions with more than one variable can be calculated.
  • Slope and curvature in higher dimensions can be understood.
  • Optimization problems involving several variables can be approached.

Multivariable calculus thus opens the door to analyzing complex systems and phenomena that occur in real life.
Vector Calculus
Vector calculus involves using vectors to describe and solve problems in physical and mathematical sciences. It includes operations like dot products, cross products, and gradient operations. Vector fields are also a central concept.

The gradient, a foundational concept in vector calculus, is a vector that contains all partial derivatives of a multivariable function. For the function \( f(x, y, z) = x^2y + y^2z + z^2x \), the gradient \( abla f \) reveals the direction of the steepest increase of the function.

The gradient found is:
  • \( abla f = \langle 2xy + z^2, x^2 + 2yz, y^2 + 2zx \rangle \)

This vector points in the direction where the function \( f \) increases most rapidly. Vector calculus is essential for modeling and analyzing physical systems in engineering and physics, where these concepts give insights into fields like electromagnetism and fluid dynamics.

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Most popular questions from this chapter

If the temperature of a plate at the point \((x, y)\) is \(T(x, y)=10+x^{2}-y^{2}\), find the path a heat-seeking particle (which always moves in the direction of greatest increase in temperature) would follow if it starts at \((-2,1)\). Hint: The particle moves in the direction of the gradient $$ \nabla T=2 x \mathbf{i}-2 y \mathbf{j} $$ We may write the path in parametric form as $$ \mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j} $$ and we want \(x(0)=-2\) and \(y(0)=1\). To move in the required direction means that \(\mathbf{r}^{\prime}(t)\) should be parallel to \(\nabla T\). This will be satisfied if $$ \frac{x^{\prime}(t)}{2 x(t)}=-\frac{y^{\prime}(t)}{2 y(t)} $$ together with the conditions \(x(0)=-2\) and \(y(0)=1\). Now solve this differential equation and evaluate the arbitrary constant of integration.

In Problems 7-12, find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\). $$ w=e^{x^{2}+y^{2}} ; x=s \sin t, y=t \sin s $$

Suppose that the temperature \(T\) at the point \((x, y, z)\) depends only on the distance from the origin. Show that the direction of greatest increase in \(T\) is either directly toward the origin or directly away from the origin.

In what direction u does \(f(x, y)=\sin (3 x-y)\) decrease most rapidly at \(\mathbf{p}=(\pi / 6, \pi / 4)\) ?

In determining the specific gravity of an object, its weight in air is found to be \(A=36\) pounds and its weight in water is \(W=20\) pounds, with a possible error in each measurement of \(0.02\) pound. Find, approximately, the maximum possible error in calculating its specific gravity \(S\), where \(S=A /(A-W)\).

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