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Chapter 10: Problem 10

Are the following statements true or false? (a) The gradient vector \(\nabla f(a, b)\) is tangent to the contour of \(f\) at \((a, b)\). (b) \(f_{\vec{u}}(a, b)=\|\nabla f(a, b)\| .\) (c) \(f_{\vec{u}}(a, b)\) is parallel to \(\vec{u}\). (d) If \(\vec{u}\) is perpendicular to \(\nabla f(a, b),\) then \(f_{\vec{u}}(a, b)=\langle 0,0\rangle\). (e) If \(\vec{u}\) is a unit vector, then \(f_{\vec{u}}(a, b)\) is a vector. (f) Suppose \(f_{x}(a, b)\) and \(f_{y}(a, b)\) both exist. Then there is always a direction in which the rate of change of \(f\) at \((a, b)\) is zero. (g) If \(f(x, y)\) has \(f_{x}(a, b)=0\) and \(f_{y}(a, b)=0\) at the point \((a, b)\), then \(f\) is constant everywhere. (h) \(\nabla f(a, b)\) is a vector in 3 -dimensional space.

Short Answer

Expert verified
(a) False. The gradient vector \(\nabla f(a, b)\) is perpendicular to the contour lines. (b) False. The expression given is the magnitude of the gradient vector, not the directional derivative. (c) False. The directional derivative is a scalar, not a vector. (d) False. If \(\vec{u}\) is perpendicular to \(\nabla f(a, b)\), then \(f_{\vec{u}} (a, b)\) is a scalar 0, not a vector \(\langle 0,0 \rangle\). (e) False. The directional derivative is always a scalar. (f) True. There is always a direction in which the rate of change of \(f\) at \((a, b)\) is zero. (g) False. Having \(f_{x}(a, b)=0\) and \(f_{y}(a, b)=0\) only implies that \((a, b)\) is a critical point, not that \(f\) is constant everywhere. (h) False. The gradient of a function with two variables, like \(\nabla f(a, b)\), is a vector in 2-dimensional space.

Step by step solution

01

Statement (a)

The gradient vector \(\nabla f(a, b)\) is always perpendicular to the contour lines of a function. Therefore, this statement is false. #b. \(f_{\vec{u}}(a, b)=\|\nabla f(a, b)\|\).#
02

Statement (b)

The directional derivative of a function \(f\) in the direction of \(\vec{u}\) is given by \(f_{\vec{u}}(a, b) = \nabla f(a, b)\cdot\vec{u}\). This statement is false, since the expression given is actually the magnitude of the gradient vector, not the directional derivative. #c. \(f_{\vec{u}}(a, b)\) is parallel to \(\vec{u}\).#
03

Statement (c)

As mentioned in statement (b), the directional derivative of a function \(f\) in the direction of \(\vec{u}\) is given by \(f_{\vec{u}}(a, b) = \nabla f(a, b)\cdot\vec{u}\). The directional derivative is a scalar, not a vector, so it cannot be parallel to any vector. Therefore, this statement is false. #d. If \(\vec{u}\) is perpendicular to \(\nabla f(a, b),\) then \(f_{\vec{u}}(a, b)=\langle 0,0 \rangle\).#
04

Statement (d)

If \(\vec{u}\) is perpendicular to \(\nabla f(a, b)\), then their dot product is zero. Thus, the directional derivative would be \(f_{\vec{u}} (a, b) = \nabla f(a, b) \cdot \vec{u} = 0\). However, the statement claims that the result is a vector \(\langle 0,0\rangle\), which is incorrect. The result should be a scalar, not a vector. This statement is false. #e. If \(\vec{u}\) is a unit vector, then \(f_{\vec{u}}(a, b)\) is a vector.#
05

Statement (e)

As previously mentioned, the directional derivative is always a scalar, regardless of whether or not \(\vec{u}\) is a unit vector. Therefore, this statement is false. #f. Suppose \(f_{x}(a, b)\) and \(f_{y}(a, b)\) both exist. Then there is always a direction in which the rate of change of \(f\) at \((a, b)\) is zero.#
06

Statement (f)

If both partial derivatives exist, then the gradient vector \(\nabla f(a, b) = \langle f_x(a, b), f_y(a, b) \rangle\) also exists. As stated before, the directional derivative of \(f\) in the direction of \(\vec{u}\) is given by \(f_{\vec{u}} (a, b) = \nabla f(a, b) \cdot \vec{u}\). If \(\vec{u}\) is a unit vector perpendicular to \(\nabla f(a, b)\), then their dot product is zero, and the rate of change of \(f\) in that direction is zero. Therefore, this statement is true. #g. If \(f(x,y)\) has \(f_{x}(a, b)=0\) and \(f_{y}(a, b)=0\) at the point \((a, b)\), then \(f\) is constant everywhere.#
07

Statement (g)

If \(f_x(a, b) = 0\) and \(f_y(a, b) = 0\), then the gradient vector is \(\langle 0, 0 \rangle\), which means that at the point \((a, b)\), the function \(f\) has no direction of maximum increase. However, it does not imply that \(f\) is constant everywhere, only that \((a, b)\) is a critical point. Therefore, this statement is false. #h. \(\nabla f(a, b)\) is a vector in 3-dimensional space.#
08

Statement (h)

The gradient of a function with two variables, like \(\nabla f(a, b)\), is always a vector in a 2-dimensional space. It has the form \(\langle f_x(a, b), f_y(a, b) \rangle\), which has only two components. Therefore, this statement is false.

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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, often denoted as \(abla f\), is a fundamental concept in differential calculus. It is crucial for understanding how a function changes at a given point. The gradient vector at a point \((a, b)\) provides the direction of the steepest increase of the function \(f\). Let's imagine hiking on a hill. If you want to find the steepest direction to climb upwards, the gradient vector at your current location points in that direction.
The components of \(abla f(a, b)\) are the partial derivatives \(f_x(a, b)\) and \(f_y(a, b)\), representing the rate of change of \(f\) in the \(x\) and \(y\) directions, respectively.
  • The gradient is perpendicular to the contour lines of \(f\), which are the lines where the function has a constant value.
  • If the gradient vector is \(\textbf{0}\) at a point, this indicates a critical point where the function does not increase or decrease in any direction.
Directional Derivative
The directional derivative provides the rate of change of a function \(f\) in a specific direction defined by a vector \(\vec{u}\). It tells us how fast \(f\) changes as you move along the direction of \(\vec{u}\). The formula for the directional derivative is given by\[ f_{\vec{u}}(a, b) = abla f(a, b) \cdot \vec{u}, \]where \(\cdot\) denotes the dot product. This expression leads to a scalar, not a vector, indicating the magnitude of change in \(f\) along \(\vec{u}\).
  • To maximize the increase in \(f\), move along the direction of \(abla f(a, b)\).
  • If \(\vec{u}\) is perpendicular to the gradient, the directional derivative is zero, signifying no change in \(f\) in that direction.
Contour Lines
Contour lines, or level curves, represent lines on which a function \(f(x, y)\) maintains a constant value. They serve as a valuable tool for visualizing the topology of a function's surface in 3D space.
Visualize contour lines like the rings of a topographic map, illustrating constant elevations in a 3D landscape. As such, they allow you to understand how quickly or slowly the elevation changes.
  • The gradient vector of a function at any point is always perpendicular to the corresponding contour line through that point. This reflects the idea that moving along the contour line means no change in height, while moving perpendicular to it increases your height maximally.
  • A densely packed region of contour lines indicates a steep area, suggesting a rapid change in the function's value over a small distance.
Partial Derivatives
Partial derivatives play a foundational role in understanding how a multivariable function changes with respect to one of these variables, keeping others constant. For a function \(f(x, y)\), the partial derivatives \(f_x\) and \(f_y\) examine its rate of change in the \(x\) and \(y\) directions, respectively.
  • To find \(f_x(a, b)\), differentiate \(f(x, y)\) with respect to \(x\) while treating \(y\) as a constant.
  • Similarly, for \(f_y(a, b)\), differentiate with respect to \(y\) and keep \(x\) constant.
Partial derivatives provide the building blocks for the gradient vector and are essential in evaluating the function's behavior in a local region around a point. Understanding these derivatives helps in assessing critical points, where \(f_x\) and \(f_y\) both are zero, indicating potential local maxima, minima, or saddle points.

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Most popular questions from this chapter

The concentration of salt in a fluid at \((x, y, z)\) is given by \(F(x, y, z)=\) \(2 x^{2}+3 y^{4}+2 x^{2} z^{2} \mathrm{mg} / \mathrm{cm}^{3}\). You are at the point (-1,1,-1) . (a) In which direction should you move if you want the concentration to increase the fastest? direction: _________. (Give your answer as a vector.) (b) You start to move in the direction you found in part (a) at a speed of \(5 \mathrm{~cm} / \mathrm{sec} .\) How fast is the concentration changing? rate of change \(=\) ______________.

For each of the following prompts, provide an example of a function of two variables with the desired properties (with justification), or explain why such a function does not exist. a. A function \(p\) that is defined at \((0,0),\) but \(\lim _{(x, y) \rightarrow(0,0)} p(x, y)\) does not exist. b. A function \(q\) that does not have a limit at \((0,0),\) but that has the same limiting value along any line \(y=m x\) as \(x \rightarrow 0\). c. A function \(r\) that is continuous at \((0,0),\) but \(\lim _{(x, y) \rightarrow(0,0)} r(x, y)\) does not exist. d. A function \(s\) such that \(\lim _{(x, x) \rightarrow(0,0)} s(x, x)=3\) and \(\lim _{(x, 2 x) \rightarrow(0,0)} s(x, 2 x)=6\) for which \(\lim _{(x, y) \rightarrow(0,0)} s(x, y)\) exists. e. A function \(t\) that is not defined at (1,1) but \(\lim _{(x, y) \rightarrow(1,1)} t(x, y)\) does exist.

In this section we argued that if \(f=f(x, y)\) is a function of two variables and if \(f_{x}\) and \(f_{y}\) both exist and are continuous in an open disk containing the point \(\left(x_{0}, y_{0}\right),\) then \(f\) is differentiable at \(\left(x_{0}, y_{0}\right) .\) This condition ensures that \(f\) is differentiable at \(\left(x_{0}, y_{0}\right),\) but it does not define what it means for \(f\) to be differentiable at \(\left(x_{0}, y_{0}\right) .\) In this exercise we explore the definition of differentiability of a function of two variables in more detail. Throughout, let \(g\) be the function defined by \(g(x, y)=\sqrt{|x y|}\) a. Use appropriate technology to plot the graph of \(g\) on the domain \([-1,1] \times[-1,1] .\) Explain why \(g\) is not locally linear at (0,0) b. Show that both \(g_{x}(0,0)\) and \(g_{y}(0,0)\) exist. If \(g\) is locally linear at \((0,0),\) what must be the equation of the tangent plane \(L\) to \(g\) at (0,0)\(?\) c. Recall that if a function \(f=f(x)\) of a single variable is differentiable at \(x=x_{0},\) then $$f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}$$ exists. We saw in single variable calculus that the existence of \(f^{\prime}\left(x_{0}\right)\) means that the graph of \(f\) is locally linear at \(x=x_{0}\). In other words, the graph of \(f\) looks like its linearization \(L(x)=f\left(x_{0}\right)+\) \(f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)\) for \(x\) close to \(x_{0} .\) That is, the values of \(f(x)\) can be closely approximated by \(L(x)\) as long as \(x\) is close to \(x_{0}\). We can measure how good the approximation of \(L(x)\) is to \(f(x)\) with the error function $$E(x)=L(x)-f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)-f(x)$$ As \(x\) approaches \(x_{0}, E(x)\) approaches \(f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(0)-f\left(x_{0}\right)=0\), and so \(L(x)\) provides increasingly better approximations to \(f(x)\) as \(x\) gets closer to \(x_{0} .\) Show that, even though \(g(x, y)=\sqrt{|x y|}\) is not locally linear at \((0,0),\) its error term $$ E(x, y)=L(x, y)-g(x, y) $$ at (0,0) has a limit of 0 as \((x, y)\) approaches \((0,0) .\) (Use the linearization you found in part (b).) This shows that just because an error term goes to 0 as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right),\) we cannot conclude that a function is locally linear at \(\left(x_{0}, y_{0}\right)\). d. As the previous part illustrates, having the error term go to 0 does not ensure that a function of two variables is locally linear. Instead, we need a notation of a relative error. To see how this works, let us return to the single variable case for a moment and consider \(f=f(x)\) as a function of one variable. If we let \(x=x_{0}+h,\) where \(|h|\) is the distance from \(x\) to \(x_{0}\), then the relative error in approximating \(f\left(x_{0}+h\right)\) with \(L\left(x_{0}+h\right)\) is $$\frac{E\left(x_{0}+h\right)}{h}$$ Show that, for a function \(f=f(x)\) of a single variable, the limit of the relative error is 0 as \(h\) approaches 0 . e. Even though the error term for a function of two variables might have a limit of 0 at a point, our example shows that the function may not be locally linear at that point. So we use the concept of relative error to define differentiability of a function of two variables. When we consider differentiability of a function \(f=f(x, y)\) at a point \(\left(x_{0}, y_{0}\right),\) then if \(x=x_{0}+h\) and \(y=y_{0}+k,\) the distance from \((x, y)\) to \(\left(x_{0}, y_{0}\right)\) is \(\sqrt{h^{2}+k^{2}}\)

Find the maximum and minimum values of the function \(f(x, y, z)=\) \(x^{2} y^{2} z^{2}\) subject to the constraint \(x^{2}+y^{2}+z^{2}=64 .\) Maximum value is ____________, occuring at points (positive integer or "infinitely many"). Minimum value is ____________, occuring at points (positive integer or "infinitely many").

Find the absolute maximum and minimum of the function \(f(x, y)=x^{2}+\) \(y^{2}\) subject to the constraint \(x^{4}+y^{4}=6561\). As usual, ignore unneeded answer blanks, and list points in lexicographic order. Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, ) Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, )
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