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If a function \(f(x, y)\) has continuous second partial derivatives throughout an open region \(R,\) must the first-order partial derivatives of \(f\) be continuous on \(R ?\) Give reasons for your answer.

Short Answer

Expert verified
Yes, the first-order partial derivatives must be continuous on \( R \).

Step by step solution

01

Understanding the Given Conditions

We are given that the function \( f(x, y) \) has continuous second partial derivatives throughout an open region \( R \). This means that the second derivatives \( \frac{\partial^2 f}{\partial x^2}, \frac{\partial^2 f}{\partial y^2}, \frac{\partial^2 f}{\partial x \partial y}, \) and \( \frac{\partial^2 f}{\partial y \partial x} \) are continuous on \( R \).
02

Examining the Implications for First-order Derivatives

The continuity of second partial derivatives implies each second derivative is both continuous and exists. This guarantees that first-order partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) exist and have continuous partial derivatives.
03

Applying Clairaut's Theorem for Mixed Derivatives

Clairaut's Theorem states that if the mixed second partial derivatives \( \frac{\partial^2 f}{\partial x \partial y} \) and \( \frac{\partial^2 f}{\partial y \partial x} \) are continuous at a point, they are equal. This continuous interchangeability enforces that the first-order derivatives are at least continuously differentiable, thus continuous.
04

Conclusion

If all second partial derivatives of \( f \) are continuous on \( R \), it implies the first-order partial derivatives are not only well-defined but also continuous. Therefore, yes, the first-order partial derivatives must be continuous throughout \( R \).

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Key Concepts

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

Clairaut's Theorem
Clairaut's Theorem is a fundamental result in differential calculus concerning the equality of mixed partial derivatives. It essentially states that if the second mixed partial derivatives of a function are continuous over an open region, then these mixed derivatives are equal. In mathematical terms, this means that for a function \( f(x, y) \), the derivatives \( \frac{\partial^2 f}{\partial x \partial y} \) and \( \frac{\partial^2 f}{\partial y \partial x} \) are the same if these derivatives are continuous at a point or throughout some region.
This theorem is vital because it simplifies the study of functions with multiple variables and provides a crucial criterion for the interchangeability of differentiation order. Without the hypothesis of continuity, the equality might not hold, so it's central in ensuring that functions behave well under differentiation. If you're tackling problems involving partial derivatives, having Clairaut's Theorem in your toolbox is invaluable for verifying the existence and equality of these mixed derivatives.
First-order partial derivatives
The first-order partial derivatives of a function measure how the function changes with respect to changes in one of its variables, holding the other variables constant. For a function \( f(x, y) \), the first-order partial derivatives are \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
These derivatives play a crucial role in understanding the slope or gradient of a function in multi-dimensional space. Just like regular derivatives, they help in finding tangent lines to the surface defined by the function, indicating how the function rises or falls as you move in the direction of one of the axes.
With the condition of continuous second partial derivatives, as given in the original exercise, it guarantees that these first-order partial derivatives are not only existing but are smooth or continuous over the region considered. This continuity ensures that there are no sharp corners or jumps; thus, the function behaves in a regular and predictable manner across the specified domain.
Continuity of functions
Continuity of functions ensures that small changes in input will result in small changes in output, without abrupt jumps or holes in the function's graph. For partial derivatives, continuity means that the behavior of these derivatives is smooth over the function's domain.
This concept is important in multi-variable calculus because it influences whether derivatives, particularly higher-order and mixed ones, exist and can be differentiated further. A function having continuous partial derivatives means it is aesthetically well-behaved, making it a favorite in many practical applications because it is analytic and easier to work with.
  • Continuous functions are predictable and reliable in modeling real-world problems.
  • They allow for the interchange of differentiation orders due to Clairaut’s Theorem.
  • Smooth functions avoid issues like cusps and asymptotic behaviors, making them suitable for computer modeling and simulations.
For the function in the exercise, knowing that its second partial derivatives are continuous assures us that the first-order derivatives are continuous. This characteristic is fundamental for accurately analyzing and applying such functions in various scientific fields.

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Most popular questions from this chapter

Change along a helix Find the derivative of \(f(x, y, z)=\) \(x^{2}+y^{2}+z^{2}\) in the direction of the unit tangent vector of the helix $$\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+t \mathbf{k}$$ at the points where \(t=-\pi / 4,0,\) and \(\pi / 4 .\) The function \(f\) gives the square of the distance from a point \(P(x, y, z)\) on the helix to the origin. The derivatives calculated here give the rates at which the square of the distance is changing with respect to \(t\) as \(P\) moves through the points where \(t=-\pi / 4,0,\) and \(\pi / 4\) .

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Function: \(f(x, y)=x y\) Curves: i. The line \(x=2 t, \quad y=t+1\) ii. The line segment \(x=2 t, \quad y=t+1, \quad-1 \leq t \leq 0\) iii. The line segment \(x=2 t, \quad y=t+1, \quad 0 \leq t \leq 1\)

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0 .\) b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2} .\) d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x^{2}-x y+y^{2}-z^{2}-1=0\) and \(x^{2}+y^{2}-1=0\)

Extrema on a curve of intersection Find the extreme values of \(f(x, y, z)=x^{2} y z+1\) on the intersection of the plane \(z=1\) with the sphere \(x^{2}+y^{2}+z^{2}=10\) .

Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes \(y+2 z=12\) and \(x+y=6 .\)

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