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Chapter 1: Problem 4

A circle, centered at the origin with a radius of 2 units, lies in the \(x y\) plane. Determine the unit vector in rectangular components that lies in the \(x y\) plane, is tangent to the circle at \((-\sqrt{3}, 1,0)\), and is in the general direction of increasing values of \(y\).

Short Answer

Expert verified
Question: Find a unit vector in the \(xy\) plane that is tangent to the given circle with center (0, 0, 0) and radius 2 at the point \((-\sqrt{3}, 1, 0)\) such that the tangent vector has an increasing \(y\) direction. Answer: A possible unit vector is \(\left(\frac{1}{\sqrt{1^2 + k^2}}, \frac{k}{\sqrt{1^2 + k^2}}\right)\), where \(k>0\).

Step by step solution

01

Find the direction vector of the tangent at \((-\sqrt{3}, 1, 0)\).

To find the direction vector of the tangent, we first need to determine a point on the circle that is very close to the given point \((-\sqrt{3}, 1, 0)\). Let it be \((-\sqrt{3}+ dx, 1+ dy, 0)\). As the point lies on the circle with a radius of 2 units, it must satisfy the equation \((x^2 + y^2 = 4)\). So, \(((- \sqrt{3} + dx)^2 + (1 + dy)^2 = 4 \) We can now find the direction vector, \(\overrightarrow{d}\), as: \(\overrightarrow{d} = ((-\sqrt{3} + dx ) - (-\sqrt{3}), (1 + dy) -1) = (dx, dy)\)
02

Use the equation of the circle for a point close to \((-\sqrt{3}, 1, 0)\).

As derived in Step 1: \(((- \sqrt{3} + dx)^2 + (1 + dy)^2 = 4\) Taking the derivative with respect to \(x\), we get: \(2(-\sqrt{3} + dx)\cdot1 + 2(1 + dy) \cdot \frac{dy}{dx} = 0\) This gives us the value of slope \(\frac{dy}{dx}\).
03

Calculate the slope and determine the direction vector.

Simplifying the equation from Step 2, we get: \(\frac{dy}{dx} = \frac{\sqrt{3} - dx}{1 + dy}\) Using the fact that the tangent vector has an increasing \(y\) direction, the slope \(\frac{dy}{dx}>0\). So the direction vector is \((1, \frac{dy}{dx}) = (1, k)\), where \(k = \frac{\sqrt{3} - dx}{1 + dy}\) and \(k > 0\).
04

Find the magnitude of the direction vector.

The magnitude of the direction vector, \(||(1,k)||\), is given by: \(||\overrightarrow{d}|| = \sqrt{1^2 + k^2}\)
05

Calculate the unit vector.

To find the unit vector, we must divide the direction vector by its magnitude: \(\hat{d} = \frac{\overrightarrow{d}}{||\overrightarrow{d}||} = \left(\frac{1}{\sqrt{1^2 + k^2}}, \frac{k}{\sqrt{1^2 + k^2}}\right)\) Since we are looking for a unit vector in the direction of increasing values of \(y\), we must choose a \(k>0\). \(\hat{d} = \left(\frac{1}{\sqrt{1^2 + k^2}}, \frac{k}{\sqrt{1^2 + k^2}}\right)\) for \(k>0\) There can be multiple values of \(k>0\) that give unit vectors tangent to the circle at the given point and in the increasing direction of \(y\).

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Most popular questions from this chapter

Given point \(P\left(r=0.8, \theta=30^{\circ}, \phi=45^{\circ}\right)\) and \(\mathbf{E}=1 / r^{2}\left[\cos \phi \mathbf{a}_{r}+\right.\) \(\left.(\sin \phi / \sin \theta) \mathbf{a}_{\phi}\right]\), find \((a) \mathbf{E}\) at \(P ;(b)|\mathbf{E}|\) at \(P ;(c)\) a unit vector in the direction of \(\mathbf{E}\) at \(P\)

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 \(y=7\). $$ \int_{0}^{4} \int_{0}^{2} \mathbf{G} \cdot \mathbf{a}_{y} d z d x $$

Express in cylindrical components: \((a)\) the vector from \(C(3,2,-7)\) to \(D(-1,-4,2) ;(b)\) a unit vector at \(D\) directed toward \(C ;(c)\) a unit vector at \(D\) directed toward the origin.

Find \((a)\) the vector component of \(\mathbf{F}=10 \mathbf{a}_{x}-6 \mathbf{a}_{y}+5 \mathbf{a}_{z}\) that is parallel to \(\mathbf{G}=0.1 \mathbf{a}_{x}+0.2 \mathbf{a}_{v}+0.3 \mathbf{a}_{z} ;(b)\) the vector component of \(\mathbf{F}\) that is perpendicular to \(\mathbf{G} ;(c)\) the vector component of \(\mathbf{G}\) that is perpendicular to \(\mathbf{F}\).

Find the acute angle between the two vectors \(\mathbf{A}=2 \mathbf{a}_{x}+\mathbf{a}_{y}+3 \mathbf{a}_{z}\) and \(\mathbf{B}=\mathbf{a}_{x}-3 \mathbf{a}_{y}+2 \mathbf{a}_{z}\) by using the definition of \((a)\) the dot product; \((b)\) the cross product.
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