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Chapter 9: Problem 12

$$\begin{aligned} &\mathbf{u}=\frac{\sqrt{2}}{2} \mathbf{i}+\frac{\sqrt{2}}{2} \mathbf{j} ; \quad \nabla f=\left(4+y^{2}\right) \mathbf{i}+(2 x y-5) \mathbf{j} ; \quad \nabla f(3,-1)=5 \mathbf{i}-11 \mathbf{j}\\\ &D_{\mathbf{u}} f(3,-1)=\frac{5 \sqrt{2}}{2}-\frac{11 \sqrt{2}}{2}=-3 \sqrt{2} \end{aligned}$$

Short Answer

Expert verified
The directional derivative is \(-3\sqrt{2}\).

Step by step solution

01

Identify Given Information

From the exercise, we have the vector \( \mathbf{u} = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \), and the gradient of \( f \) at any point \( (x,y) \) is given by \( abla f = (4 + y^2) \mathbf{i} + (2xy - 5) \mathbf{j} \). The specific point of interest is \( (3, -1) \), where \( abla f(3,-1) = 5 \mathbf{i} - 11 \mathbf{j} \). We need to find \( D_{\mathbf{u}} f(3, -1) \).
02

Evaluate the Gradient at (3, -1)

Confirm that \( abla f(3,-1) \) is correctly given. Substituting \( (3, -1) \) into \( abla f \), we have: \[ (4 + (-1)^2) \mathbf{i} + (2 \cdot 3 \cdot (-1) - 5) \mathbf{j} = 5 \mathbf{i} - 11 \mathbf{j} \] Therefore, \( abla f(3,-1) = 5 \mathbf{i} - 11 \mathbf{j} \) is correct.
03

Calculate Directional Derivative Formula

The directional derivative of a function \( f \) in the direction of a unit vector \( \mathbf{u} \) is calculated using \( D_{\mathbf{u}} f = abla f \cdot \mathbf{u} \). In this case, substitute \( abla f(3,-1) = 5 \mathbf{i} - 11 \mathbf{j} \) and \( \mathbf{u} = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \).
04

Compute the Dot Product

Using the expression for the directional derivative, compute the dot product: \[ D_{\mathbf{u}} f(3,-1) = (5 \mathbf{i} - 11 \mathbf{j}) \cdot \left( \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \right) \] Evaluating this, \[ = 5 \times \frac{\sqrt{2}}{2} - 11 \times \frac{\sqrt{2}}{2} \] \[ = \frac{5\sqrt{2}}{2} - \frac{11\sqrt{2}}{2} \] \[ = -3\sqrt{2} \]
05

Conclude the Result

The directional derivative \( D_{\mathbf{u}} f(3,-1) \) is the result of the dot product, giving us \( -3\sqrt{2} \). This is the rate of change of the function \( f \) at point \( (3, -1) \) in the direction specified by \( \mathbf{u} \).

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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 fundamental concept in calculus, especially when dealing with functions of several variables. It represents the direction and magnitude of the greatest rate of increase of a function. For a function \( f(x, y) \), the gradient \( abla f \) is a vector given by:
  • \( abla f = \left( \frac{\partial f}{\partial x} \right) \mathbf{i} + \left( \frac{\partial f}{\partial y} \right) \mathbf{j} \)
This expression means that the gradient vector comprises the partial derivatives of the function with respect to each variable. In our exercise, the gradient is defined as \( abla f = (4 + y^2) \mathbf{i} + (2xy - 5) \mathbf{j} \).

For a specific point, like \( (3,-1) \), the gradient vector becomes constant because we're substituting specific values into the partial derivatives. This step simplifies it to \( 5 \mathbf{i} - 11 \mathbf{j} \), which is the gradient evaluated at the point \( (3, -1) \). This vector not only points in the direction where the function increases fastest but also tells us the rate of that increase.
Dot Product
The dot product is an essential operation in vector algebra, allowing us to measure the angle and relationship between two vectors. It is computed as the sum of the products of the corresponding entries of the vectors. For two vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} \), the dot product is:
  • \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \)
This operation results in a scalar rather than a vector.

In the context of the exercise, the dot product is used to find the directional derivative, which measures how the function changes as we move in a given direction. Here, we calculate \( (5 \mathbf{i} - 11 \mathbf{j}) \cdot \left( \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \right) \). This calculates to \( \frac{5\sqrt{2}}{2} - \frac{11\sqrt{2}}{2} = -3\sqrt{2} \), determining the rate of change in the direction of the unit vector \( \mathbf{u} \).
Unit Vector
Unit vectors serve an important role in defining directions without affecting magnitudes. A unit vector has a magnitude of 1, making it ideal for scaling.
  • To convert any vector \( \mathbf{v} \) into a unit vector \( \mathbf{u} \), divide by its magnitude: \( \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} \)
In the exercise, the unit vector \( \mathbf{u} \) is given as \( \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \), already a unit vector because its magnitude
  • \( ||\mathbf{u}|| = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = 1 \)
Unit vectors are crucial in determining directions, and they provide the necessary scaling factor when calculating directional derivatives. In this case, the directional derivative was calculated in the direction of \( \mathbf{u} \), verifying it remains a unit vector throughout the exercise.

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Most popular questions from this chapter

The density is \(\rho=k x\). \(m=\int_{0}^{\pi} \int_{0}^{\sin x} k x d y d x=\left.\int_{0}^{\pi} k x y\right|_{0} ^{\sin x} d x=\int_{0}^{\pi} k x \sin x d x\) Integration by parts \(=\left.k(\sin x-x \cos x)\right|_{0} ^{\pi}=k \pi\) \(M_{y}=\int_{0}^{\pi} \int_{0}^{\sin x} k x^{2} d y d x=\left.\int_{0}^{\pi} k x^{2} y\right|_{0} ^{\sin x} d x=\int_{0}^{\pi} k x^{2} \sin x d x\) Integration by parts \(=\left.k\left(-x^{2} \cos x+2 \cos x+2 x \sin x\right)\right|_{0} ^{\pi}=k\left[\left(\pi^{2}-2\right)-2\right]=k\left(\pi^{2}-4\right)\) \(M_{x}=\int_{0}^{\pi} \int_{0}^{\sin x} k x y d y d x=\left.\int_{0}^{\pi} \frac{1}{2} k x y^{2}\right|_{0} ^{\sin x} d x=\int_{0}^{\pi} \frac{1}{2} k x \sin ^{2} x d x=\int_{0}^{\pi} \frac{1}{4} k x(1-\cos 2 x) d x\) \(=\frac{1}{4} k\left[\int_{0}^{\pi} x d x-\int_{0}^{\pi} x \cos 2 x d x\right]\) Integration by parts \(=\frac{1}{4} k\left[\left.\frac{1}{2} x^{2}\right|_{0} ^{\pi}-\left.\frac{1}{4}(\cos 2 x+2 x \sin 2 x)\right|_{0} ^{\pi}\right]=\frac{1}{4} k\left(\frac{1}{2} \pi^{2}\right)=\frac{1}{8} k \pi^{2}\) \(\bar{x}=M_{y} / m=\frac{k\left(\pi^{2}-4\right)}{k \pi}=\pi-4 / \pi ; \bar{y}=M_{x} / m=\frac{k \pi^{2} / 8}{k \pi}=\pi / 8 .\) The center of mass is \((\pi-4 / \pi, \pi / 8)\).

\(I_{x}=\int_{0}^{1} \int_{0}^{y-y^{2}} 2 x y^{2} d x d y=\left.\int_{0}^{1} x^{2} y^{2}\right|_{0} ^{y-y^{2}} d y=\int_{0}^{1}\left(y-y^{2}\right)^{2} y^{2} d y\) \(=\int_{0}^{1}\left(y^{4}-2 y^{5}+y^{6}\right) d y=\left.\left(\frac{1}{5} y^{5}-\frac{1}{3} y^{6}+\frac{1}{7} y^{7}\right)\right|_{0} ^{1}=\frac{1}{105}\).

We are given \(\sigma=k z .\) Now \(z_{x}-\frac{x}{\sqrt{16-x^{2}-y^{2}}}, \quad z_{y}=-\frac{y}{\sqrt{16-x^{2}-y^{2}}}\) \(d S=\sqrt{1+\frac{x^{2}}{16-x^{2}-y^{2}}+\frac{y^{2}}{16-x^{2}-y^{2}}} d A=\frac{4}{\sqrt{16-x^{2}-y^{2}}} d A\) Using polar coordinates, $$\begin{aligned} Q &=\iint_{S} k z d S=k \iint_{R} \sqrt{16-x^{2}-y^{2}} \frac{4}{\sqrt{16-x^{2}-y^{2}}} d A=4 k \int_{0}^{2 \pi} \int_{0}^{3} r d r d \theta \\ &=\left.4 k \int_{0}^{2 \pi} \frac{1}{2} r^{2}\right|_{0} ^{3} d \theta=4 k \int_{0}^{2 \pi} \frac{9}{2} d \theta=36 \pi k \end{aligned}$$

We are given \(\rho(x, y, z)=k\). $$\begin{aligned} I_{x} &=\int_{0}^{1} \int_{z^{2}}^{2-z} \int_{0}^{3} k\left(y^{2}+z^{2}\right) d y d x d z=\left.k \int_{0}^{1} \int_{z^{2}}^{2-z}\left(\frac{1}{3} y^{3}+y z^{2}\right)\right|_{0} ^{3} d x d z=k \int_{0}^{1} \int_{z^{2}}^{2-z}\left(9+3 z^{2}\right) d x d z \\ &=\left.k \int_{0}^{1}\left(9 x+3 x z^{2}\right)\right|_{z^{2}} ^{2-z} d z=k \int_{0}^{1}\left(18-9 z-3 z^{2}-3 z^{3}-3 z^{4}\right) d z \\ &=\left.k\left(18 z-\frac{9}{2} z^{2}-z^{3}-\frac{3}{4} z^{4}-\frac{3}{5} z^{5}\right)\right|_{0} ^{1}=\frac{223}{20} k \\ m &=\int_{0}^{1} \int_{z^{2}}^{2-z} \int_{0}^{3} k d y d x d z=k \int_{0}^{1} \int_{z^{2}}^{2-z} 3 d x d z=3 k \int_{0}^{1}\left(2-z-z^{2}\right) d z=\left.3 k\left(2 z-\frac{1}{2} z^{2}-\frac{1}{3} z^{3}\right)\right|_{0} ^{1}=\frac{7}{2} k \\ R_{g} &=\sqrt{\frac{I_{x}}{m}}=\sqrt{\frac{223 k / 20}{7 k / 2}}=\sqrt{\frac{223}{70}} \end{aligned}$$

\(\operatorname{div} \mathbf{F}=2(z-1) .\) Using cylindrical coordinates, $$\begin{aligned}\iint_{S} \mathbf{F} \cdot \mathbf{n} d S &=\iiint_{D} 2(z-1) V=\int_{0}^{2 \pi} \int_{0}^{4} \int_{1}^{5} 2(z-1) d z r d r d \theta=\left.\int_{0}^{2 \pi} \int_{0}^{4}(z-1)^{2}\right|_{1} ^{5} r d r d \theta \\\&=\int_{0}^{2 \pi} \int_{0}^{4} 16 r d r d \theta=\left.\int_{0}^{2 \pi} 8 r^{2}\right|_{0} ^{4} d \theta=128 \int_{0}^{2 \pi} d \theta=256 \pi\end{aligned}$$
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