/*! 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 87 An aircraft is flying at a const... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An aircraft is flying at a constant altitude with velocity magnitude \(r_{1}\) (relative to the air) and angle \(\theta_{1}\) (in a twodimensional coordinate system). The magnitude and direction of the wind are \(r_{2}\) and \(\theta_{2},\) respectively. Suppose that the wind angle is uniformly distributed between 10 and 20 degrees and all other parameters are constant. Determine the probability density function of the magnitude of the resultant vector \(r=\left[r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\left(\cos \theta_{1}-\cos \theta_{2}\right)\right]^{0.5}\)

Short Answer

Expert verified
The PDF of the resultant velocity magnitude \( r \) is influenced uniformly due to the balanced effect of \( \theta_2 \) over 10 degrees.

Step by step solution

01

Identify Variables and Relationship

Given parameters include the aircraft velocity magnitude \( r_1 \), its direction \( \theta_1 \), wind velocity magnitude \( r_2 \), and its direction \( \theta_2 \). We need to find the probability density function (PDF) of the resultant velocity magnitude \( r \). The angle \( \theta_2 \) has given bounds of 10 to 20 degrees and is assumed to be uniformly distributed. The resultant vector's magnitude is calculated using:\[r = \left[r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}(\cos(\theta_1) - \cos(\theta_2))\right]^{0.5}\]
02

Define Problem Constraints

Since \( \theta_2 \) is uniformly distributed between 10 and 20 degrees, we first need to focus on this parameter. The uniform distribution implies that the probability density function (PDF) for \( \theta_2 \) is constant: \( f(\theta_2) = \frac{1}{10} \) for \( \theta_2 \) within the 10 to 20-degree range. With this information, \( \theta_2 \) must be converted from degrees to radians for calculations.
03

Substitute and Simplify

Substitute \( \theta_2 \) from the uniform distribution into the magnitude equation and analyze:\[r = \left[r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}\left(\cos(\theta_1) - \cos(\theta_2)\right)\right]^{0.5}\]Focus on the component \( \cos(\theta_2) \) as it is the variable dependent on \( \theta_2 \). The function \( \cos(\theta_2) \) will vary as \( \theta_2 \) changes, affecting the resulting \( r \).
04

Analyze Effect on Resultant Magnitude r

The resultant magnitude \( r \) is affected by the cosine difference in the formula. As \( \theta_2 \) spans 10 to 20 degrees, \( \cos(\theta_2) \) spans a certain range. Given the cos function's slowly changing nature over the small angle difference, \( r \) will have minimal variation.
05

Determine Probability Density Function for r

Given the uniform spread of \( \theta_2 \), one must integrate the effect of \( \theta_2 \) on \( r \). However, since the change in cosine value over a 10-degree range is minimal, and no specific distribution is heavily defined for \( r \) besides the randomness of \( \theta_2 \), a basic resultant is computed. This suggests a continuous variation of \( r \) with possibly minuscule differences in frequency across small \( r \) intervals.
06

Conclusion and Formulation

Because \( \theta_2 \) influences the system minimally over its 10-degree range, and under the uniform distribution condition, the density of \( r \) defined by \( f(r) \) remains affected by the form, not a distinct value distribution. This results in a general linear profile attribute, indicating equidistribution in its bounded segment.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Uniform Distribution
In statistics, a uniform distribution is where every outcome in a given interval is equally likely. When applied to the angle \( \theta_2 \) in the context of the aircraft navigation problem, it suggests that all angles between 10 and 20 degrees have the same probability of occurrence. This equates to a constant probability density function (PDF) within that interval. For the uniform distribution, the probability density function for any angle \( \theta_2 \) within the range is \( f(\theta_2) = \frac{1}{b-a} \) where \( a \) and \( b \) define the interval. Here, \( a = 10 \) degrees and \( b = 20 \) degrees, so the PDF is \( f(\theta_2) = \frac{1}{10} \). This implies a consistent probability for each degree, ensuring that \( \theta_2 \) affects the system evenly across its range.
Resultant Vector Magnitude
The resultant vector magnitude, \( r \), is a critical component in determining the final velocity of the aircraft as it navigates through the wind. It's derived by combining the aircraft's velocity vector with the wind's velocity vector. Using the formula: \[ r = \sqrt{r_1^2 + r_2^2 + 2r_1r_2 \cos(\theta_1 - \theta_2)} \] This calculation finds the combined effect of the two velocities taking into account their directionality. Yet, since \( \theta_2 \) is uniformly distributed and the change is mild over the given interval, the impact on \( r \) remains subtle. This keeps the overall resultant almost constant, notwithstanding the minor fluctuations in angle due to wind.
Trigonometric Functions
Trigonometric functions play a pivotal role in calculating vector magnitudes, most notably with the cosine function affecting the resultant vector's magnitude. In this exercise, the relationship \( r = \sqrt{r_1^2 + r_2^2 + 2r_1r_2 \cos(\theta_1 - \theta_2)} \) is used where \( \cos(\theta_1 - \theta_2) \) adjusts the magnitude depending on the angle between the velocity vector of the aircraft and the wind. As \( \theta_2 \) variances over its uniform distribution, \( \cos(\theta_2) \) slightly shifts while having only a minimal impact due to trigonometric properties over small angle changes. Understanding these properties is crucial for anticipating the behavior of the resultant vector in real-world applications like aircraft navigation.
Aircraft Navigation
Aircraft navigation involves finding the most efficient path considering external factors like wind. By navigating with a clear understanding of both vectors and their combined magnitude, pilots can strategize routes for optimized flight paths. Through the exercise problem, we see how the aircraft's velocity integrates with wind velocity to produce the resultant vector that guides navigation. The uniform distribution of \( \theta_2 \) highlights the variability of wind direction without excessive complexity, representative of actual flight conditions. This model aids in illustrating the dynamics airplanes encounter, ensuring a balance between natural forces and navigational tactics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that \(X\) and \(Y\) are independent continuous random variables. Show that \(\sigma_{X Y}=0 .\)

An article in Knee Surgery Sports Traumatology, Arthroscopy ["Effect of Provider Volume on Resource Utilization for Surgical Procedures" (2005, Vol. 13, pp. \(273-279\) ) showed a mean time of 129 minutes and a standard deviation of 14 minutes for ACL reconstruction surgery for high-volume hospitals (with more than 300 such surgeries per year). If a high-volume hospital needs to schedule 10 surgeries, what are the mean and variance of the total time to complete these surgeries? Assume that the times of the surgeries are independent and normally distributed.

Suppose that \(X\) and \(Y\) have a bivariate normal distribution with \(\sigma_{X}=0.04, \sigma_{Y}=0.08, \mu_{X}=3.00, \mu_{Y}=7.70,\) and \(\rho=0 .\) Determine the following: (a) \(P(2.95

\(X\) and \(Y\) are independent, normal random variables with \(E(X)=2, V(X)=5, E(Y)=6,\) and \(V(Y)=8 .\) Determine the following: (a) \(E(3 X+2 Y)\) (b) \(V(3 X+2 Y)\) (c) \(P(3 X+2 Y<18)\) (d) \(P(3 X+2 Y<28)\)

In the manufacture of electroluminescent lamps, several different layers of ink are deposited onto a plastic substrate. The thickness of these layers is critical if specifications regarding the final color and intensity of light are to be met. Let \(X\) and \(Y\) denote the thickness of two different layers of ink. It is known that \(X\) is normally distributed with a mean of 0.1 millimeter and a standard deviation of 0.00031 millimeter, and \(Y\) is normally distributed with a mean of 0.23 millimeter and a standard deviation of 0.00017 millimeter. The value of \(\rho\) for these variables is equal to \(0 .\) Specifications call for a lamp to have a thickness of the ink corresponding to \(X\) in the range of 0.099535 to 0.100465 millimeter and \(Y\) in the range of 0.22966 to 0.23034 millimeter. What is the probability that a randomly selected lamp will conform to specifications?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.