Question

A field is given as \(\mathbf{G}=\left[25 /\left(x^{2}+y^{2}\right)\right]\left(x \mathbf{a}_{x}+y \mathbf{a}_{y}\right) .\) Find \((a)\) a unit vector in the direction of \(\mathbf{G}\) at \(P(3,4,-2) ;(b)\) the angle between \(\mathbf{G}\) and \(\mathbf{a}_{x}\) at \(P\); (c) the value of the following double integral on the plane \(y=7\). $$ \int_{0}^{4} \int_{0}^{2} \mathbf{G} \cdot \mathbf{a}_{y} d z d x $$

Short answer

Question: Find the following for the vector field \(\mathbf{G} = \frac{25}{(x^2 + y^2)}(x\mathbf{a}_x + y\mathbf{a}_y)\) at point \(P(3, 4, -2)\). (a) A unit vector in the direction of \(\mathbf{G}\) at point \(P\). (b) The angle between \(\mathbf{G}\) and \(\mathbf{a}_x\) at point \(P\). (c) The value of the given double integral on the plane \(y=7\). Answer: (a) The unit vector in the direction of \(\mathbf{G}\) at point \(P\) is \(\frac{1}{\sqrt{3^2 + 4^2}}\left(3\mathbf{a}_x + 4\mathbf{a}_y\right)\). (b) The angle between \(\mathbf{G}\) and \(\mathbf{a}_x\) at point \(P\) is \(\arccos\left(\frac{3}{\sqrt{3^2 + 4^2}}\right)\). (c) The value of the double integral on the plane \(y=7\) is \(175\arctan{\frac{4}{7}}\).

Step-by-step solution

Find the vector field \(\mathbf{G}\) at point \(P\)

At point \(P(3,4,-2)\), substitute the values of \(x\) and \(y\) into the vector field \(\mathbf{G}\). $$ \mathbf{G}(3,4) = \left[\frac{25}{(3^2 + 4^2)}\right]\left(3\mathbf{a}_x + 4\mathbf{a}_y\right) $$

Find the magnitude of \(\mathbf{G}\) at point \(P\)

Find the magnitude of \(\mathbf{G}(3,4)\): $$ \|\mathbf{G}(3,4)\| = \left\|\left[\frac{25}{(3^2 + 4^2)}\right]\left(3\mathbf{a}_x + 4\mathbf{a}_y\right)\right\| = \frac{25}{5}\sqrt{3^2 + 4^2} = 5\sqrt{3^2 + 4^2} $$

Find a unit vector in the direction of \(\mathbf{G}\) at point \(P\)

Divide the vector \(\mathbf{G}(3,4)\) by its magnitude to find the unit vector: $$ \hat{\mathbf{G}}(3,4) = \frac{\mathbf{G}(3,4)}{\|\mathbf{G}(3,4)\|} = \frac{1}{\sqrt{3^2 + 4^2}}\left(3\mathbf{a}_x + 4\mathbf{a}_y\right) $$

Find the angle between \(\mathbf{G}\) and \(\mathbf{a}_x\) at point \(P\)

Using the dot product formula, we have: $$ \cos \theta = \frac{\mathbf{G}(3,4) \cdot \mathbf{a}_x}{\|\mathbf{G}(3,4)\|\|\mathbf{a}_x\|} $$ Since \(\mathbf{G}(3,4) \cdot \mathbf{a}_x = \frac{25}{(3^2 + 4^2)}(3\mathbf{a}_x + 4\mathbf{a}_y) \cdot \mathbf{a}_x = \frac{75}{25}\) and \(\|\mathbf{a}_x\| = 1\), we get $$ \cos \theta = \frac{3}{\sqrt{3^2 + 4^2}} $$ Thus, the angle \(\theta\) between \(\mathbf{G}\) and \(\mathbf{a}_x\) at point \(P\) is given by \(\theta = \arccos\left(\frac{3}{\sqrt{3^2 + 4^2}}\right)\).

Evaluate the double integral

Calculate the value of the given double integral on the plane \(y=7\): $$ \int_{0}^{4} \int_{0}^{2} \mathbf{G}(x,7) \cdot \mathbf{a}_y\ dz\ dx = \int_{0}^{4} \left[\frac{25}{(x^2 + 7^2)}\right]\left(7\mathbf{a}_y\right) \cdot \mathbf{a}_y dx $$ Since \(\mathbf{a}_y \cdot \mathbf{a}_y = 1\), therefore, $$ \int_{0}^{4} \left[\frac{25}{(x^2 + 7^2)}\right]\left(7\mathbf{a}_y\right) \cdot \mathbf{a}_y dx = \int_{0}^{4} \frac{175}{x^2 + 49} dx $$ Now, integrate the function with respect to \(x\): $$ \int_{0}^{4} \frac{175}{x^2 + 49} dx = \left[175\arctan{\left(\frac{x}{7}\right) }\right]_{0}^{4} = 175(\arctan{\frac{4}{7}} - \arctan{0}) = 175\arctan{\frac{4}{7}} $$ Hence, the value of the double integral on the plane \(y=7\) is \(175\arctan{\frac{4}{7}}\).

Key concepts

Unit Vector Calculation

When dealing with vectors, it is often useful to have a vector that just indicates direction without any reference to the magnitude. That's where a unit vector comes into play. A unit vector has a magnitude of one and points in the direction of the vector in question. This standardized length helps in various vector operations.

To calculate a unit vector, you simply divide the vector by its own magnitude. In the given textbook problem, the vector field \textbf{G} at point P is divided by its magnitude to find the corresponding unit vector. The calculation boils down to using the formula:
\[\begin{equation}\hat{\mathbf{G}}(x,y) = \frac{\mathbf{G}(x,y)}{\|\mathbf{G}(x,y)\|}\end{equation}\]By ensuring that the vector is of unit length, we can then focus on directional properties, such as the direction of a force or velocity, without the added complication of differing magnitudes.

Angle Between Vectors

The concept of the angle between two vectors extends our geometric understanding of angles onto a multitude of mathematically defined spaces. The angle provides essential insights into the relative position and direction of vectors with respect to one another. To find the angle between two vectors, we use the dot product, which is an algebraic operation that combines two equal-length sequences of numbers into a single number. The dot product of two vectors also relates to the cosine of the angle between them, which can be used to find the exact value of this angle. The formula to calculate the angle \( \theta \) is given by:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\|\|\mathbf{B}\|} \]
where \(\mathbf{A}\) and \(\mathbf{B}\) are vectors. In the problem, the angle between the vector field \(\mathbf{G}\) and the unit vector \(\mathbf{a}_x\) at point P is found using this formula, leading to understanding how vector \(\mathbf{G}\) is oriented relative to the coordinate axis represented by \(\mathbf{a}_x\).

Double Integral Evaluation

Double integrals are an essential tool in mathematics, especially when evaluating the area under a surface or the volume of a solid. To compute a double integral, one typically iterates the process of integration twice, once for each variable. In the context of vector fields, the evaluation of a double integral helps to determine quantities such as flux or work done by a force over a certain area.

The task may sound daunting, but let's simplify it. The double integral from our exercise essentially adds up all the values of the dot product of the vector field \(\mathbf{G}\) and the unit vector \(\mathbf{a}_y\) over the specified range of \(x\) and \(z\), on the plane where \(y = 7\). Symbolically, this is written as:
\[\int_{a}^{b} \int_{c}^{d} \mathbf{G}(x, y) \cdot \mathbf{a}_y dz dx\]
The inner integral sums over one variable while holding the other constant, and then the outer integral sums those results over the remaining variable. By going step by step—as the solution dictates—we first solve the inner integral and then the outer integral to find our final answer. Properly evaluating a double integral provides a numerical result, which could represent a physical quantity such as flux in the plane of interest.