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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume \(f\) is differentiable at the points in question. a. The fact that \(f_{x}(2,2)=f_{y}(2,2)=0\) implies that \(f\) has a local maximum, local minimum, or saddle point at (2,2) b. The function \(f\) could have a local maximum at \((a, b)\) where \(f_{y}(a, b) \neq 0\) c. The function \(f\) could have both an absolute maximum and an absolute minimum at two different points that are not critical points. d. The tangent plane is horizontal at a point on a smooth surface corresponding to a critical point.

Short Answer

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#tag_title#Short Answer Question#tag_content#Determine whether the following statement is true or false: The tangent plane is horizontal at critical points of smooth surfaces. Provide an explanation or counterexample for your answer.

Step by step solution

01

Statement a: True or False?

The statement is true. When both partial derivatives are zero, it means there could be a local maximum, local minimum, or saddle point at the given point. To confirm which one it is, we would have to analyze the second partial derivatives using the second partial derivative test. #Step 2: Analyze statement b#
02

Statement b: True or False?

The statement is true. It is possible for a function to have a local maximum at a point where one of its partial derivatives is non-zero. Consider the function \(f(x, y) = -(x^2 + y^3)\). At the point (0, 0), $f_x(0, 0) = 0\( and \)f_y(0, 0) = 0$, yet this is a local maximum because the function is decreasing around the point. #Step 3: Analyze statement c#
03

Statement c: True or False?

The statement is false. An absolute maximum or minimum can only occur at a critical point or on the boundary of the domain. A critical point is defined as a point where the gradient is zero or undefined. Since the statement specifies that both points are not critical points, the scenario described is impossible. #Step 4: Analyze statement d#
04

Statement d: True or False?

The statement is true. The tangent plane is determined by the gradient of the function, which, at a critical point, is either zero or undefined. A horizontal tangent plane occurs when the gradient is zero, hence the tangent plane is horizontal at a critical point of a smooth surface.

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Key Concepts

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

Differentiability
Differentiability is a crucial concept in calculus that indicates how smoothly a function behaves at a certain point. When we say a function is differentiable at a specific point, it means that the function not only is continuous but also has a defined derivative at that point. This derivative gives us the rate at which the function's value is changing with respect to its input variables.

To clarify, if a function is two-variable, the partial derivatives, \(f_x\) and \(f_y\), must exist at that point. These partial derivatives represent how the function changes as each variable is varied while keeping the others constant.

  • Partial Derivatives: These are the derivatives taken with respect to one variable at a time. They show the rate of change along the axes of the variables.
  • Differentiability: Means that all partial derivatives exist and are continuous around the point in question.
Understanding differentiability is foundational, as it sets the stage for analyzing critical points and applying the second partial derivative test.
Second Partial Derivative Test
The second partial derivative test helps us decide the nature of critical points in functions of two variables. After finding the critical points by setting the first partial derivatives to zero, this test evaluates the second partial derivatives at those critical points to determine if they are local maxima, minima, or saddle points.

Let's break down the test into simpler terms:
  • Calculate Second Partial Derivatives: First, find \(f_{xx}, f_{yy},\) and \(f_{xy}\). These are the second derivatives with respect to \(x\), \(y\), and mixed derivatives, respectively.
  • Check Determinant of Hessian (\(D\)): Use \(D = f_{xx}f_{yy} - (f_{xy})^2\) to determine the behavior at the critical point.
  • Analyze Signs:
    • If \(D > 0\) and \(f_{xx} > 0\), it's a local minimum.
    • If \(D > 0\) and \(f_{xx} < 0\), it's a local maximum.
    • If \(D < 0\), then it's a saddle point, indicating that the surface curves differently in different directions.
    • If \(D = 0\), the test is inconclusive.
The second partial derivative test is a powerful tool for getting clarity on the type of behavior a function exhibits at its critical points.
Local Maximum and Minimum
Local maximum and minimum points are locations around which a function takes on its largest or smallest value compared to other nearby points, respectively. These points are part of what we call critical points, the spots where the function might be at an extreme in some local sense.

Here's how you can identify them:
  • Critical Points: Found by setting the first derivative(s) of a function to zero or finding where they don't exist.
  • Local Maximum: A point where the function value is higher than all nearby function values.
  • Local Minimum: A point where the function value is lower than all nearby function values.
Understanding these concepts often involves looking at the graph of the function. Visualizing can be a powerful aid in recognizing these maxima and minima, particularly in complex surfaces, where they give insight into the function's behavior over its domain.
Tangent Plane
In calculus, the tangent plane concept extends the idea of a tangent line to functions of two variables. At any given point on a smooth surface, the tangent plane provides the best linear approximation of the surface near that point.

The tangent plane to a surface defined by a differentiable function \(f(x, y)\) at a point \((a, b)\) can be found using the formula:
\[ z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \]

Where:
  • \(z\) is the function's value on the tangent plane
  • \(f_x(a, b)\) is the partial derivative with respect to \(x\) at \((a, b)\)
  • \(f_y(a, b)\) is the partial derivative with respect to \(y\) at \((a, b)\)
Horizontal Tangent Plane:
When the tangent plane is horizontal, both partial derivatives at the point are zero. This occurs at critical points and often indicates the top of a hill, bottom of a valley, or a saddle point.

The tangent plane plays a vital role in linear approximations and gives us a peek into the behavior of functions over small areas of their domain.

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Most popular questions from this chapter

Travel cost The cost of a trip that is \(L\) miles long, driving a car that gets \(m\) miles per gallon, with gas costs of \(\$ p /\) gal is \(C=L p / m\) dollars. Suppose you plan a trip of \(L=1500 \mathrm{mi}\) in a car that gets \(m=32 \mathrm{mi} / \mathrm{gal},\) with gas costs of \(p=\$ 3.80 / \mathrm{gal}\) a. Explain how the cost function is derived. b. Compute the partial derivatives \(C_{L}, C_{m^{\prime}}\) and \(C_{p^{\prime}}\). Explain the meaning of the signs of the derivatives in the context of this problem. c. Estimate the change in the total cost of the trip if \(L\) changes from \(L=1500\) to \(L=1520, m\) changes from \(m=32\) to \(m=31,\) and \(p\) changes from \(p=\$ 3.80\) to \(p=\$ 3.85\) d. Is the total cost of the trip (with \(L=1500 \mathrm{mi}, m=32 \mathrm{mi} / \mathrm{gal}\). and \(p=\$ 3.80\) ) more sensitive to a \(1 \%\) change in \(L,\) in \(m,\) or in \(p\) (assuming the other two variables are fixed)? Explain.

Use Lagrange multipliers in the following problems. When the constraint curve is unbounded, explain why you have found an absolute maximum or minimum value. Box with minimum surface area Find the dimensions of the rectangular box with a volume of \(16 \mathrm{ft}^{3}\) that has minimum surface area.

Potential functions Potential functions arise frequently in physics and engineering. A potential function has the property that a field of interest (for example, an electric field, a gravitational field, or a velocity field) is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter \(17 .)\) The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(\varphi=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$\mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle$$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=\frac{k Q}{r^{2}} .\) Explain why this relationship is called an inverse square law.

Prove that for the plane described by \(f(x, y)=A x+B y,\) where \(A\) and \(B\) are nonzero constants, the gradient is constant (independent of \((x, y)) .\) Interpret this result.

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.

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