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You are at a point \((a, b, c)\) on the surface $$ z=\left(r^{2}-x^{2}-y^{2}\right)^{1 / 2} \quad(z \geq 0) $$ Assuming both \(a\) and \(b\) are positive, in what direction must you move (a) so that the rate of change of \(z\) will be zero? (b) so that the rate of increase of \(z\) will be greatest? (c) so that the rate of decrease of \(z\) will be greatest? Draw a sketch to show the geometric significance of your answers.

Short Answer

Expert verified
(a) Move perpendicular to \( \nabla z \). (b) Move in the direction of \( \nabla z \). (c) Move opposite to \( \nabla z \).

Step by step solution

01

Find the Gradient Vector

The gradient of a function gives the direction of the greatest rate of increase. For the given surface, we have the equation \( z = \sqrt{r^2 - x^2 - y^2} \). The gradient can be found using partial derivatives.\[abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial z}{\partial z} \right)\]First, let's find each partial derivative:1. \( \frac{\partial z}{\partial x} = \frac{-x}{\sqrt{r^2 - x^2 - y^2}} \)2. \( \frac{\partial z}{\partial y} = \frac{-y}{\sqrt{r^2 - x^2 - y^2}} \)3. \( \frac{\partial z}{\partial z} = 1 \)Hence, \( abla z = \left( \frac{-x}{z}, \frac{-y}{z}, 1 \right) \).
02

Direction for Zero Change of z

For the rate of change of \( z \) to be zero, the direction of movement should be perpendicular to the gradient vector. This is because the gradient vector gives the greatest increase, and moving perpendicular to it causes neither an increase nor a decrease.The direction for zero change can be any direction in the plane that is perpendicular to \( \left( \frac{-x}{z}, \frac{-y}{z}, 1 \right) \).
03

Direction for Greatest Increase of z

The rate of increase of \( z \) is greatest in the direction of the gradient vector. Therefore, to achieve the maximum increase, move in the direction \( \left( \frac{-x}{z}, \frac{-y}{z}, 1 \right) \).
04

Direction for Greatest Decrease of z

For the greatest decrease in \( z \), move in the opposite direction of the gradient vector. Thus, the direction for the greatest decrease is \( \left( \frac{x}{z}, \frac{y}{z}, -1 \right) \).
05

Geometric Sketch

To visualize the situation, imagine the surface as part of a sphere centered at the origin. At any point \((a,b,c)\), the gradient \( abla z \) points directly outward along the surface. The directions perpendicular to the gradient lie in the tangent plane at the point, representing directions where the change in \( z \) is zero.

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Key Concepts

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

Gradient Vector
The gradient vector is a crucial concept in vector calculus, especially regarding functions of several variables. It serves as a vector field indicating the direction of steepest ascent on a surface or in a space.
In mathematical terms, the gradient of a function \( f \), denoted as \( abla f \), is a vector composed of all its first-order partial derivatives. For a function \( f(x, y, z) \), the gradient vector is represented by:
\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \]
This vector indicates how changing each variable affects the function \( f \) locally.
  • The direction of the gradient vector conveys the path in which the function ascends most rapidly.
  • The magnitude of the gradient specifies the rate of this increase.
Understanding the gradient vector helps determine directions of increase and decrease, which is fundamental when analyzing functions like those in physics and engineering, where surfaces represent potential energy, heat, or pressure.
Rate of Change
The rate of change describes how a quantity varies with another variable, crucial for understanding dynamic systems. In the context of a multi-variable function like \( z = \sqrt{r^2 - x^2 - y^2} \), the rate of change tells us how \( z \) fluctuates as \( x \), \( y \), or \( r \) vary.
This is essential when analyzing physical phenomena or systems' responses to modifications in one or more parameters.
  • A positive rate of change indicates an increase in \( z \) as variables modify.
  • A negative rate points to a reduction in \( z \).
  • A zero rate of change suggests that \( z \) remains constant despite alterations in parameters.
By examining the rate of change, we can decipher how sensitive a function is to specific variables, guiding decision-making and predicting outcomes.
Partial Derivatives
Partial derivatives extend the concept of derivatives to functions of multiple variables. They measure how a function changes as one variable moves slightly, holding others constant.
For instance, the partial derivative \( \frac{\partial z}{\partial x} \) with respect to \( x \) gives the rate of change of \( z \) as only \( x \) varies:
\[ \frac{\partial z}{\partial x} = \frac{-x}{\sqrt{r^2 - x^2 - y^2}} \] This derivative indicates how the surface slopes relative to the \( x \)-axis.
  • Computing partial derivatives for each variable allows building the gradient vector.
  • They are fundamental in optimizing functions and understanding surface behavior.
Thus, partial derivatives are vital tools for analyzing functions' sensitivity to each independent variable, especially in fields like economics, engineering, and physics.
Directional Derivative
The directional derivative indicates how a function changes as you move in a specific direction from a particular point. It expands the idea of derivatives to any direction, providing more versatility in analysis.
Mathematically, the directional derivative of \( f \) in the direction of a unit vector \( \mathbf{u} \) is \( D_\mathbf{u} f \), calculated as:
\[ D_\mathbf{u} f = abla f \cdot \mathbf{u} \] Here, \( abla f \) is the gradient, and \( \mathbf{u} \) is the direction vector.
  • If the directional derivative is zero, the function does not change in that direction, as seen when moving perpendicular to the gradient vector.
  • A positive directional derivative indicates moving in a direction of increasing function value.
  • A negative value reveals a decrease along that path.
Understanding directional derivatives is key in fields where small directional changes profoundly impact, such as fluid dynamics, electromagnetics, and optimization problems.

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Most popular questions from this chapter

(a) Calculate \(\mathbf{F}=\nabla f\) for each of the following scalar functions: (i) \(f=x y z\). (ii) \(f=x^{2}+y^{2}+z^{2}\). (iii) \(f=x y+y z+x z\). (iv) \(f=3 x^{2}-4 z^{2}\). (v) \(f=e^{-x} \sin y\). (b) Verify that $$ \oint_{C} \mathbf{F} \cdot \hat{\mathbf{t}} d s=0 $$ for one or more of the functions \(\mathbf{F}\) determined in part (a) choosing for the curve \(C\) : (i) the square in the \(x y\)-plane with vertices at \((0,0),(1,0)\), \((1,1)\), and \((0,1)\). (ii) the triangle in the \(y z\)-plane with vertices at \((0,0),(1,0)\), and \((0,1)\). (iii) the circle of unit radius centered at the origin and lying in the \(x z\)-plane. (c) Verify by direct calculation that \(\nabla \times \mathbf{F}=0\) for one or more of the functions \(\mathbf{F}\) determined in part (a).

(a) Consider a surface \(z=f(x, y)\). Let \(\mathbf{u}\) be a vector of arbitrary length tangent to the surface at a point \(P(x, y, z)\) in the direction of the unit vector \(\hat{\mathbf{p}}=\mathbf{i} p_{x}+\mathbf{j} p_{y}\) as indicated in the figure. Use the directional derivative to show that $$ \mathbf{u}=\hat{\mathbf{p}}+\mathbf{k}(\hat{\mathbf{p}} \cdot \nabla f) $$ where \(\nabla f\) is evaluated at \((x, y) .[\) Note: Since the length of \(\mathbf{u}\) is arbitrary, your result may differ from the preceding by some positive multiplicative constant.] (b) Let \(v\) be a second vector of arbitrary length tangent to the surface at \(P\) but in the direction of the unit vector \(\hat{\mathbf{q}}=\mathbf{i} q_{x}+\mathbf{j} q_{y}\) \((\hat{\mathbf{p}} \neq \hat{\mathbf{q}}) .\) Then from (a) we have $$ \mathbf{v}=\hat{\mathbf{q}}+\mathbf{k}(\hat{\mathbf{q}} \cdot \nabla f) $$ Show that $$ \mathbf{u} \times \mathbf{v}=[\mathbf{k} \cdot(\hat{\mathbf{p}} \times \hat{\mathbf{q}})](\mathbf{k}-\nabla f) $$ and use this to rederive Equation (II-4) for the unit vector \(\hat{\mathbf{n}}\) normal to the surface \(z=f(x, y)\) at \((x, y, z)\). This shows that the result derived in the text for \(\hat{\mathbf{n}}\) is unique (apart from sign) even though it was obtained with the special choices \(\hat{\mathbf{p}}=\mathbf{i}\) and \(\hat{\mathbf{q}}=\mathbf{j}\).

Here is a "proof" that there is no such thing as magnetism. One of Maxwell's equations tells us that $$ \nabla \cdot \mathbf{B}=0 $$ where \(\mathbf{B}\) is any magnetic field. Then using the divergence theorem, we find $$ \iint_{S} \mathbf{B} \cdot \hat{\mathbf{n}} d S=\iiint_{V} \nabla \cdot \mathbf{B} d V=0 $$ Because B has zero divergence, we know (see Problem III-24) there exists a vector function, call it \(\mathbf{A}\), such that $$ \mathbf{B}=\nabla \times \mathbf{A} $$ Combining these last two equations, we get $$ \iint_{s} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Next we apply Stokes' theorem and the preceding result to find $$ \oint_{C} \mathbf{A} \cdot \hat{\mathbf{t}} d s=\iint_{S} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Thus we have shown that the circulation of \(\mathbf{A}\) is path independent. It follows that we can write \(\mathbf{A}=\nabla \psi\), where \(\psi\) is some scalar function. Since the curl of the gradient of a function is zero, we arrive at the remarkable fact that $$ \mathbf{B}=\nabla \times \nabla \psi=0 $$ that is, all magnetic fields are zero! Where did we go wrong? [Taken from G. Arfken, Amer. J. Phys., 27, 526 (1959).]

The unit vector normal to the surface \(z=f(x, y)\) is given by $$ \hat{\mathbf{n}}=\left(-\mathbf{i} \frac{\partial f}{\partial x}-\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k}\right) \mid \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}} $$ [see Equation (II-4)]. We have also established that \(\nabla F\) is a vector normal to the surface \(F(x, y, z)=\) const. (page 140 ) so that \(\nabla F /|\nabla F|\) is a unit vector normal to the surface \(F(x, y, z)=\) const. Show that these two expressions for the unit normal vector are identical if \(F(x, y, z)=\) const. and \(z=f(x, y)\) describe the same surface.

(a) Find the charge density \(\rho(x, y, z)\) that produces the electric field $$ \mathbf{E}=g(\mathbf{i} x+\mathbf{j} \mathbf{y}+\mathbf{k} z) $$ where \(g\) is a constant. (b) Find an electrostatic potential \(\Phi\) such that \(-\nabla \Phi\) is the field \(\mathbf{E}\) given in (a). (c) Verify that \(\nabla^{2} \Phi=-\rho / \epsilon_{0}\)

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