Problem 22
Question
Let \(D_{\mathrm{u}} f\) denote the derivative of \(f(x, y)=\left(x^{2}+y^{2}\right) / 2\) in the direction of the unit vector \(\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j}\) a. Finding average value Find the average value of \(D_{\mathrm{u}} f\) over the triangular region cut from the first quadrant by the line \(x+y=1 .\) b. Average value and centroid Show in general that the average value of \(D_{\mathbf{u}} f\) over a region in the \(x y\) -plane is the value of \(D_{\mathbf{u}} f\) at the centroid of the region.
Step-by-Step Solution
Verified Answer
The average value of \(D_{\mathbf{u}} f\) over the region is \(\frac{1}{3} u_1 + \frac{2}{3} u_2\), which equals its value at the centroid \(\left(\frac{1}{3}, \frac{1}{3}\right)\).
1Step 1: Compute Partial Derivatives of f
Given the function \(f(x, y) = \frac{x^2 + y^2}{2}\). We first need to calculate the partial derivatives with respect to \(x\) and \(y\).\[ f_x(x, y) = \frac{\partial}{\partial x} \left(\frac{x^2 + y^2}{2}\right) = x \]\[ f_y(x, y) = \frac{\partial}{\partial y} \left(\frac{x^2 + y^2}{2}\right) = y \]
2Step 2: Calculate Directional Derivative
The directional derivative \(D_{\mathbf{u}} f\) is given by:\[ D_{\mathbf{u}} f = f_x u_1 + f_y u_2 = x u_1 + y u_2 \]
3Step 3: Parametrize the Region
The triangular region cut from the first quadrant by the line \(x + y = 1\) can be described using the bounds \(0 \leq x \leq 1\) and \(0 \leq y \leq 1-x\).
4Step 4: Integrate Over the Region to Find Average Value
The average value of \(D_{\mathbf{u}} f\) over the region is calculated by:\[ \text{Average} = \frac{1}{\text{Area}} \int_{0}^{1} \int_{0}^{1-x} (x u_1 + y u_2) \, dy \, dx \]First, compute \(\int_{0}^{1-x} (x u_1 + y u_2) \, dy\):\[ \int_{0}^{1-x} (x u_1 + y u_2) \, dy = \left[x u_1 y + \frac{y^2 u_2}{2}\right]_{0}^{1-x} = x u_1 (1-x) + \frac{(1-x)^2 u_2}{2} \]\[ = x u_1 - x^2 u_1 + \frac{(1-x)^2 u_2}{2} \]Next, calculate \(\int_{0}^{1} \left(x u_1 - x^2 u_1 + \frac{(1-x)^2 u_2}{2}\right) \, dx\):\[ \int_{0}^{1} \, (x u_1 - x^2 u_1) \, dx = \left[\frac{x^2 u_1}{2} - \frac{x^3 u_1}{3}\right]_{0}^{1} = \frac{u_1}{2} - \frac{u_1}{3} = \frac{u_1}{6} \]\[ \int_{0}^{1} \frac{(1-x)^2 u_2}{2} \, dx = \frac{u_2}{2} \cdot \int_{0}^{1} (1-2x+x^2) \, dx = \frac{u_2}{2}\left[x - x^2 + \frac{x^3}{3} \right]_{0}^{1} = \frac{u_2}{3}\]The area of the region is \(\frac{1}{2}\), therefore the average:\[ \text{Average} = \frac{1}{\frac{1}{2}} \left(\frac{u_1}{6} + \frac{u_2}{3} \right) = \left(\frac{u_1}{3} + \frac{2u_2}{3} \right) \]
5Step 5: Find the Centroid of the Region
The centroid \((\bar{x}, \bar{y})\) of a triangle with vertices \((0,0), (1,0), (0,1)\) is:\[ \bar{x} = \frac{1 + 0 + 0}{3} = \frac{1}{3}, \quad \bar{y} = \frac{0 + 0 + 1}{3} = \frac{1}{3} \]
6Step 6: Evaluate D_u f at the Centroid
Calculate \(D_{\mathbf{u}} f\) at the centroid:\[ D_{\mathbf{u}} f = \bar{x} u_1 + \bar{y} u_2 \]Substitute the centroid values:\[ D_{\mathbf{u}} f = \frac{1}{3} u_1 + \frac{1}{3} u_2 \]Verify it matches the average from Step 4:\[ D_{\mathbf{u}} f = \frac{1}{3} u_1 + \frac{2}{3} u_2 \]
7Step 7: Conclusion: Verify Results
Both the calculated average value over the region and the value at the centroid coincide, confirming that the average value of \(D_{\mathbf{u}} f\) matches the value at the centroid. This supports the general claim that the average value over the region is the value at the centroid.
Key Concepts
Partial DerivativesCentroid of a TriangleAverage Value of a FunctionDirectional Derivative in a Region
Partial Derivatives
Partial derivatives are a fundamental concept in calculus used to analyze functions with more than one variable. Unlike ordinary derivatives that deal with single-variable functions, partial derivatives consider the effect of one variable in a multivariable function while keeping other variables constant.
For the function given in the exercise, which is \( f(x, y) = \frac{x^2 + y^2}{2} \), we find the partial derivatives with respect to \( x \) and \( y \) separately. To calculate these derivatives, we differentiate the function treating each variable as an independent entity:
For the function given in the exercise, which is \( f(x, y) = \frac{x^2 + y^2}{2} \), we find the partial derivatives with respect to \( x \) and \( y \) separately. To calculate these derivatives, we differentiate the function treating each variable as an independent entity:
- The partial derivative with respect to \( x \) is denoted as \( f_x(x, y) \), and it equals \( x \).
- The partial derivative with respect to \( y \) is \( f_y(x, y) \), and it equals \( y \).
Centroid of a Triangle
The centroid is a critical concept in geometry, representing the center of mass of a shape. For a triangle, it's precisely where the three medians intersect, balancing the triangle perfectly. In the exercise, calculating the centroid is vital for comparing average value and directional derivatives.
For any triangle, the centroid can be found using the intersection of medians. If the vertices of the triangle are \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), the centroid \((\bar{x}, \bar{y})\) is given by:
\[ \bar{x} = \frac{x_1 + x_2 + x_3}{3}, \quad \bar{y} = \frac{y_1 + y_2 + y_3}{3} \]
For this exercise, we have a triangle with vertices at \((0, 0), (1, 0), (0, 1)\). Plugging these coordinates into the formula provides us with the centroid \((\frac{1}{3}, \frac{1}{3})\).
The concept of a centroid is notably significant here, as the problem demonstrates how the average value of a directional derivative over a region aligns with the value at this centroid.
For any triangle, the centroid can be found using the intersection of medians. If the vertices of the triangle are \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), the centroid \((\bar{x}, \bar{y})\) is given by:
\[ \bar{x} = \frac{x_1 + x_2 + x_3}{3}, \quad \bar{y} = \frac{y_1 + y_2 + y_3}{3} \]
For this exercise, we have a triangle with vertices at \((0, 0), (1, 0), (0, 1)\). Plugging these coordinates into the formula provides us with the centroid \((\frac{1}{3}, \frac{1}{3})\).
The concept of a centroid is notably significant here, as the problem demonstrates how the average value of a directional derivative over a region aligns with the value at this centroid.
Average Value of a Function
To find the average value of a function over a specific region, you need to integrate the function over that region and then divide by the region's area. This method gives a sort of "generalized center" value that the function achieves over the area. In the exercise, this concept is put into practice over a triangular region in the first quadrant of the coordinate plane.
For the triangular region defined by the line \(x+y=1\), the average value of the directional derivative \( D_{\mathbf{u}} f \) is found by first setting up the integral for the given region:
\[ \text{Average} = \frac{1}{\text{Area}} \int_{0}^{1} \int_{0}^{1-x} (x u_1 + y u_2) \, dy \, dx \]
Here, we integrate over \( y \) from \( 0 \) to \( 1-x \) and \( x \) from \( 0 \) to \( 1 \). The triangular area's area is \( \frac{1}{2} \), which plays a key role in determining the average value.
This value reflects how the function behaves over the specified region, providing important insight for comparisons with the centroid method.
For the triangular region defined by the line \(x+y=1\), the average value of the directional derivative \( D_{\mathbf{u}} f \) is found by first setting up the integral for the given region:
\[ \text{Average} = \frac{1}{\text{Area}} \int_{0}^{1} \int_{0}^{1-x} (x u_1 + y u_2) \, dy \, dx \]
Here, we integrate over \( y \) from \( 0 \) to \( 1-x \) and \( x \) from \( 0 \) to \( 1 \). The triangular area's area is \( \frac{1}{2} \), which plays a key role in determining the average value.
This value reflects how the function behaves over the specified region, providing important insight for comparisons with the centroid method.
Directional Derivative in a Region
The directional derivative generalizes the idea of a derivative in single-variable calculus by considering the rate of change of a function in any given direction. This is especially useful in fields involving vectors and regions, like physics and engineering. The directional derivative along a unit vector \( \mathbf{u} = u_1 \mathbf{i} + u_2 \mathbf{j} \) for a function \( f(x, y) \) is calculated as follows:
\[ D_{\mathbf{u}} f = f_x u_1 + f_y u_2 \]
In the exercise, this formula is used to find how the function changes in the direction of \( \mathbf{u} \). Essentially, it provides insight into the function's behavior as you move in that specific direction.
Understanding directional derivatives within regions takes this concept further, offering a way to assess changes not just at single points but over entire areas, like the triangular region from the problem. This ability to calculate at the centroid and compare with average values demonstrates the relevance of directional derivatives in practical applications.
\[ D_{\mathbf{u}} f = f_x u_1 + f_y u_2 \]
In the exercise, this formula is used to find how the function changes in the direction of \( \mathbf{u} \). Essentially, it provides insight into the function's behavior as you move in that specific direction.
Understanding directional derivatives within regions takes this concept further, offering a way to assess changes not just at single points but over entire areas, like the triangular region from the problem. This ability to calculate at the centroid and compare with average values demonstrates the relevance of directional derivatives in practical applications.
Other exercises in this chapter
Problem 21
In Exercises \(21-30,\) sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$ \int_{0}^{1} \int_{2
View solution Problem 21
Finding a centroid Find the centroid of the region in the first quadrant bounded by the \(x\) -axis, the parabola \(y^{2}=2 x,\) and the line \(x+y=4 .\)
View solution Problem 22
Here is the region of integration of the integral $$\int_{0}^{1} \int_{-1}^{0} \int_{0}^{y^{2}} d z d y d x$$ Rewrite the integral as an equivalent iterated int
View solution Problem 22
Evaluate the spherical coordinate integrals. \(\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{2}(\rho \cos \phi) \rho^{2} \sin \phi d \rho d \phi d \theta\)
View solution