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Determine whether the following statements are true and give an explanation or counterexample. a. If \(f(x, y)=x^{2}+y^{2}-10,\) then \(\nabla f(x, y)=2 x+2 y\) b. Because the gradient gives the direction of maximum increase of a function, the gradient is always positive. c. The gradient of \(f(x, y, z)=1+x y z\) has four components. d. If \(f(x, y, z)=4,\) then \(\nabla f=\mathbf{0}\)

Short Answer

Expert verified
Question: Determine which statements are true or false and provide a counterexample if needed. a. The gradient of the function \(f(x, y) = x^2 + y^2 - 10\) is equal to \(2x + 2y\). b. The gradient is always positive. c. The gradient of the function \(f(x, y, z) = 1 + xyz\) has four components. d. For the function \(f(x, y, z) = 4\), we have \(\nabla f = \mathbf{0}\). Answer: a. False. The gradient of the function \(f(x, y) = x^2 + y^2 - 10\) is \(\nabla f(x, y) = (2x, 2y)\). b. False. The gradient is not always positive. Counterexample: \(f(x) = -x^2\) has a negative gradient at \(x>0\). c. False. The gradient of the function \(f(x, y, z) = 1 + xyz\) has three components: \(\nabla f(x, y, z) = (yz, xz, xy)\). d. True. For the constant function \(f(x, y, z) = 4\), we have \(\nabla f = \mathbf{0}\).

Step by step solution

01

Statement a - Gradient calculation

To find the gradient (\(\nabla f\)), we compute the partial derivatives of the function with respect to \(x\) and \(y\). For \(f(x, y)=x^{2}+y^{2}-10\), we have the following partial derivatives: \(\frac{\partial f}{\partial x} = 2x\) \(\frac{\partial f}{\partial y} = 2y\) So, the gradient \(\nabla f(x, y) = (2x, 2y)\). Therefore, the statement a is false since \(\nabla f(x,y) \neq 2x+2y\).
02

Statement b - Gradient always positive

The gradient is not tied with being "always positive" as it is a vector. A vector can have positive or negative components or even be equal to zero. The statement b is false. A counterexample could be the function \(f(x) = -x^2\), which has a negative gradient at \(x>0\).
03

Statement c - Number of gradient components

We need to find the gradient \(\nabla f(x, y, z)\) of the function \(f(x, y, z)=1+x y z\). The partial derivatives are: \(\frac{\partial f}{\partial x} = yz\) \(\frac{\partial f}{\partial y} = xz\) \(\frac{\partial f}{\partial z} = xy\) So, \(\nabla f(x, y, z) = (yz, xz, xy)\). It has three components, not four. Thus, statement c is false.
04

Statement d - Gradient for constant function

For the function \(f(x, y, z)=4\), which is a constant function (not dependent on \(x\), \(y\), or \(z\)), we have the following partial derivatives: \(\frac{\partial f}{\partial x} = 0\) \(\frac{\partial f}{\partial y} = 0\) \(\frac{\partial f}{\partial z} = 0\) Therefore, \(\nabla f(x, y, z) = \mathbf{0}\), and statement d is true.

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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 the realm of calculus, especially when dealing with functions that have more than one variable. When we talk about the partial derivative of a function with respect to one of its variables, we essentially focus on how the function changes as this single variable changes, while keeping other variables constant.

For instance, consider the function \(f(x, y) = x^2 + y^2 - 10\).
  • The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 2x\), which shows how \(f\) changes as \(x\) changes, holding \(y\) constant.
  • The partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = 2y\), indicating how \(f\) changes with \(y\) keeping \(x\) unchanged.
Remember, calculating partial derivatives is much like differentiating regular functions, but you only vary one variable at a time. This concept is very crucial in vector calculus and for understanding gradients.
Vector Calculus
Vector calculus is a branch of mathematics that encompasses both vector algebra and calculus, dealing primarily with vector fields that are complex in nature. The gradient is one of the primary operations in vector calculus, giving crucial insights into the behavior of functions that map from \(\mathbb{R}^n\) to \(\mathbb{R}\).

Understanding Gradients

The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. If you think of a function as a hill, the gradient points uphill. The magnitude of this vector indicates how steep the hill is. For a function \(f(x, y, z)\), the gradient \(abla f\) is composed of its partial derivatives:
  • If \(f(x, y, z) = 1 + xyz\), then the gradient is \(abla f = (yz, xz, xy)\).
  • It tells us how the function \(f\) behaves in multi-dimensional space, providing insights similar to the slope in single-variable calculus but in higher dimensions.
Furthermore, the gradient is not inherently positive or negative; it's a direction with components that can take any sign based on the function's nature.
Constant Function
A constant function is quite intuitive—it's a function that does not depend on its input variables. Such a function outputs the same value no matter what values the input variables take. For instance, the function \(f(x, y, z) = 4\) remains confined to the value \(4\) regardless of \(x\), \(y\), or \(z\).

Understanding Gradients of Constant Functions

In the world of calculus, the gradient of a constant function is quite straightforward. Since there is no change based on any variable, all partial derivatives of a constant function are zero. Thus, its gradient is \(\mathbf{0}\), or the zero vector.
  • This makes sense because constant functions exhibit no change or direction of increase—after all, the function’s value is the same everywhere!
  • A zero gradient also serves as confirmation that such functions stand unmoved by the variables, reflecting perfect flatness in any dimensional space they inhabit.
This concept is important as it brings clarity when analyzing more complex functions with flat regions.

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Most popular questions from this chapter

Use the Second Derivative Test to prove that if \((a, b)\) is a critical point of \(f\) at which \(f_{x}(a, b)=f_{y}(a, b)=0\) and \(f_{x x}(a, b)<0

Given positive numbers \(x_{1}, \ldots, x_{n},\) prove that the geometric mean \(\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n}\) is no greater than the arithmetic mean \(\left(x_{1}+\cdots+x_{n}\right) / n\) in the following cases. a. Find the maximum value of \(x y z,\) subject to \(x+y+z=k\) where \(k\) is a real number and \(x>0, y>0\), and \(z>0 .\) Use the result to prove that $$(x y z)^{1 / 3} \leq \frac{x+y+z}{3}.$$ b. Generalize part (a) and show that $$\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n} \leq \frac{x_{1}+\cdots+x_{n}}{n}.$$

(1946 Putnam Exam) Let \(P\) be a plane tangent to the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) at a point in the first octant. Let \(T\) be the tetrahedron in the first octant bounded by \(P\) and the coordinate planes \(x=0, y=0\), and \(z=0 .\) Find the minimum volume of \(T\). (The volume of a tetrahedron is one-third the area of the base times the height.)

Let \(w=f(x, y, z)=2 x+3 y+4 z\) which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\) \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\)

Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0\) and \(h(x, y, z)=0\).

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