/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 For the given vector \(\vec{v}\)... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 33

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\langle 5,5\rangle $$

Short Answer

Expert verified
The magnitude is approximately \(7.07\) and the angle \(\theta = 45^{\circ}\).

Step by step solution

01

Find the Magnitude

To find the magnitude of the vector \( \vec{v} = \langle 5, 5 \rangle \), we use the formula for the magnitude of a vector \( \langle x, y \rangle \), which is \[ \|\vec{v}\| = \sqrt{x^2 + y^2} \]Substitute the given values:\[ \|\vec{v}\| = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} \]\[ \|\vec{v}\| = 5\sqrt{2} \approx 7.07 \] (rounded to two decimal places).
02

Determine the Angle using Trigonometric Functions

To find the angle \( \theta \) we will use the trigonometric functions. For a vector \( \vec{v} = \langle x, y \rangle \), the angle can be found using \[ \tan(\theta) = \frac{y}{x} \]Substitute the values:\[ \tan(\theta) = \frac{5}{5} = 1 \]Hence, \( \theta \) is such that \( \tan(\theta) = 1 \).
03

Find the Angle in Degrees

Since \( \tan(\theta) = 1 \) and considering the range \( 0 \leq \theta < 360^{\circ} \), the angle \( \theta \) for which the tangent is 1 is \( \theta = 45^{\circ} \). This is because it lies in the first quadrant where both sine and cosine are positive.
04

Confirm that \( \vec{v} \) Can Be Expressed in Trigonometric Form

Since we found \( \theta = 45^{\circ} \), substitute into the trigonometric representation:\[ \vec{v} = \|\vec{v}\| \langle \cos(45^{\circ}), \sin(45^{\circ}) \rangle \]\[ \langle \cos(45^{\circ}), \sin(45^{\circ}) \rangle = \langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \]Substitute magnitude:\[ \vec{v} = 7.07 \cdot \langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \approx \langle 5, 5 \rangle \]This confirms the trigonometric representation.

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Key Concepts

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

Trigonometric functions
Trigonometric functions are essential in mathematics, especially when dealing with angles and vectors. They help us understand the relationships between the angles and lengths in right-angled triangles, among other things. There are three primary trigonometric functions: sine (9)
  • Sine (9f)9:- 9gives the opposite side's length based on the hypotenuse9s length in a right-angled triangle.
  • Cosine (9gg)9 - which relates the adjacent side's length to the hypotenuse.
  • Tangent (9t)9 - which is the ratio of the opposite side to the adjacent side.
In the context of vectors like 9gg9g9gg9t "9gv = 98aangle{98, 9}", these functions help describe the vector in terms of the magnitude and the direction, simplifying complex operations like rotations and transformations. Trigonometric functions are foundational for many real-life applications, such as computer graphics, engineering, and physics. It's worth noting that these functions often tie into the unit circle, a fundamental concept where angles are represented on a circular plane, aiding in understanding periodic phenomena.
Angle determination
Determining angles is vital in vector analysis, as it helps define the direction in which a vector points. When working with a vector 9gg9g9gg9t"9gv = 98aangle{98, 9}", the angle 99must be calculated precisely to ensure the correct vector representation in polar coordinates.
To find the angle low, we typically use the tangent function, expressed as 9tan(9f). The equation 9t"tan(9f) = 99times9t"
  • where y is the vertical component of the vector.
  • x is the horizontal component.
In our scenario with vector 9gg9g9gg9t" "9gv = 98aangle{5, 5}", the computation is straightforward: 9td 1/1 equals 1 implies that 9tg45ictions with angles such as 9dcommonly repeat every 180 degrees. Recognizing this helps in accommodating different situations, like identifying correct vectors in specific quadrants given a circular plane (i.e., the unit circle). In practical applications, gauge angle precision is critical when designing anything from mechanical parts to electrical circuits.
Magnitude formula
The magnitude of a vector represents its size or length, and it's an essential element in understanding vectors' physical impact. The formula to calculate the magnitude of a vector 9gg9g9gg9t"9gv = 98aangle{x, y}", is:\[\|\vec{v}\| = \sqrt{x^2 + y^2}\]It grabs both horizontal and vertical components to yield the total vector length.
For vector 9gg9g9gg9t"9gv = 98aangle{5, 5}", substituting values into the formula yields 9g
  • \(\sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50}\)
This simplifies to \(5\sqrt{2} \approx 7.07\), showcasing how the vector's actual length in multi-dimensional space is determined.
Knowing a vector's magnitude is valuable in showcasing its effectiveness or reach in various directions. This principle finds applications in physics and engineering, ensuring optimization in designs such as propulsion, load balancing, and signal processing. Understanding the magnitude helps predict behavior when vectors act in combination or opposition, allowing for better informed decisions in scientific advancements.

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Most popular questions from this chapter

In Exercises \(1-20\), plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=2 t \\ y=t^{2} \end{array} \text { for }-1 \leq t \leq 2\right. $$

In Exercises \(1-20\), plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=3 \cos (t) \\ y=3 \sin (t) \end{array} \text { for } 0 \leq t \leq \pi\right. $$

In Exercises \(41-50\), use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. The region which lies inside of the circle \(r=3 \cos (\theta)\) but outside of the circle \(r=\sin (\theta)\)

In Exercises \(21-24\), plot the set of parametric equations with the help of a graphing utility. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=t^{3}-3 t \\ y=t^{2}-4 \end{array} \text { for }-2 \leq t \leq 2\right. $$

In Exercises \(1-20\), use the pair of vectors \(\vec{v}\) and \(\vec{w}\) to find the following quantities. $$ \vec{v}=\left\langle\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle \text { and } \vec{w}=\left\langle\frac{1}{2},-\frac{\sqrt{3}}{2}\right\rangle $$
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