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[M] In a wind tunnel experiment, the force on a projectile due to air resistance was measured at different velocities: \(\begin{array}{llllll}{\text { Velocity }(100 \mathrm{ft} / \mathrm{sec})} & {0} & {2} & {4} & {6} & {8} & {10} \\ {\text { Force }(100 \mathrm{lb})} & {0} & {2.90} & {14.8} & {39.6} & {74.3} & {119}\end{array}\) Find an interpolating polynomial for these data and estimate the force on the projectile when the projectile is traveling at 750 \(\mathrm{ft} / \mathrm{sec} .\) Use \(p(t)=a_{0}+a_{1} t+a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}\) \(+a_{5} t^{5} .\) What happens if you try to use a polynomial of degree less than 5\(?\) (Try a cubic polynomial, for instance.) \(^{5}\) \(^{5}\) Exercises marked with the symbol I \(\mathbf{M} ]\) are designed to be worked with the aid of a "Matrix program" (a computer program, such as MATLAB, Maple, Mathematica, Math Cad, or Derive, or a programmable calculator with matrix capabilities, such as those manufactured by Texas Instruments or Hewlett-Packard).

Short Answer

Expert verified
Use a degree 5 polynomial for exact interpolation of given data; lower degrees increase error.

Step by step solution

01

Organize the Data Points

The given data points are (0, 0), (2, 2.90), (4, 14.8), (6, 39.6), (8, 74.3), and (10, 119). These points will be used to find the coefficients of the interpolating polynomial. Each point is of the form (velocity, force).
02

Set Up the Polynomial Equation

We need to find a polynomial of the form \(p(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 + a_4 t^4 + a_5 t^5\). Since we have 6 data points, we need a polynomial of degree 5 to interpolate these points exactly.
03

Construct the System of Equations

For each data point \((t_i, y_i)\), substitute into the polynomial equation to form a system of linear equations. With 6 points, this will yield 6 equations with 6 unknowns (\(a_0, a_1, a_2, a_3, a_4, a_5\)). For example, at \(t=0\), the equation is \(y_0 = a_0\). Repeat for other points.
04

Solve the System Using a Matrix Method

Use a matrix program or a calculator with matrix capabilities to solve the system of linear equations for the coefficients \(a_0, a_1, ..., a_5\). The solution will provide the coefficients of the polynomial that fits the data.
05

Interpolate to Estimate the Force at 750 ft/sec

Convert 750 ft/sec to the scaled units (remember units are given in 100 ft/sec), which is 7.5. Substitute \(t = 7.5\) into the polynomial \(p(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 + a_4 t^4 + a_5 t^5\) using the coefficients found to estimate the force at this velocity.
06

Evaluate Using a Polynomial of Lower Degree

Try using a polynomial of lower degree, such as 3, and repeat the process by solving a new system of equations with fewer terms. This will illustrate the impact on accuracy when fewer terms are used.
07

Conclusion on Degree of Polynomial

Using a degree 5 polynomial fits all data points exactly due to having 6 data points. Using a lower degree polynomial will generally increase the interpolation error as it cannot accurately pass through all points.

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Key Concepts

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

Matrix Methods
Matrix methods are powerful tools used to solve systems of linear equations, which is exactly what we need for polynomial interpolation. In this context, the matrix method helps us organize and solve the equations to find the coefficients of the polynomial. Here’s how it works in a nutshell:
  • Firstly, we convert our polynomial equation into a matrix form. This involves rewriting the equation for each data point and lining them up into rows and columns.
  • Each row represents an equation derived from substituting the data into the polynomial formula, while the columns correspond to the unknowns (the polynomial coefficients).
  • The resulting matrix equation has the form \(Ax = b\), where \(A\) is the matrix of coefficients, \(x\) is the vector of unknown coefficients \((a_0, a_1, ..., a_5)\), and \(b\) is the vector of force values.
  • We then use a matrix-solving technique such as Gaussian elimination, LU decomposition, or a digital tool like MATLAB to find \(x\), giving us the coefficients needed for our polynomial.
Linear Equations
Linear equations are the building blocks of our polynomial interpolation process. Each data point provides us with a linear equation by plugging the velocity into the polynomial equation. For example, with a polynomial of degree 5, the general form is:
  • For a data point \(t_i, y_i\), the equation becomes \(y_i = a_0 + a_1 t_i + a_2 t_i^2 + a_3 t_i^3 + a_4 t_i^4 + a_5 t_i^5\).
  • For each of the 6 data points, you derive a linear equation, giving a system of 6 equations.
To solve this, we use matrix methods to tackle the system, finding the coefficients that allow the polynomial to pass through all data points exactly. These linear equations are merely one part of translating the physical phenomenon measured into an accurate mathematical model.
Degree of Polynomial
The degree of the polynomial is a crucial aspect when fitting data points. In our case, having 6 data points implies using a 5th-degree polynomial to fit the curve accurately through every single point. Here's why the degree is significant:
  • A polynomial's degree corresponds to the maximum number of change in direction it can have.
  • For 6 points, a 5th-degree polynomial is required to make a curve that can "bend" to go through each point.
  • If we use a lower degree polynomial, it cannot fully adapt its shape to align with all the measured values, leading to errors.
By ensuring the polynomial degree matches the number of data points minus one, we maximize accuracy.
Data Fitting
Data fitting involves finding a mathematical function that closely matches a set of data points. Polynomial interpolation is a specific type of data fitting using polynomials to achieve this result. Here's how it applies in our problem:
  • The aim is to establish a single polynomial function that passes exactly through the given data points.
  • This allows us not only to replicate the existing data but also to estimate or predict values at points not directly measured.
  • By finding the right polynomial, we essentially generalize the underlying relationship between velocity and force discovered during the experiment.
Data fitting through polynomial interpolation is immensely helpful in various scientific fields as it offers a means to visualize the trend and make predictions beyond the set of initial data.

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Most popular questions from this chapter

Let \(T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) be a linear transformation. Show that if \(T\) maps two linearly independent vectors onto a linearly dependent set, then the equation \(T(\mathbf{x})=0\) has a nontrivial solution. [Hint: Suppose \(\mathbf{u}\) and \(\mathbf{v}\) in \(\mathbb{R}^{n}\) are linearly independent and yet \(T(\mathbf{u})\) and \(T(\mathbf{v})\) are linearly dependent. Then \(c_{1} T(\mathbf{u})+c_{2} T(\mathbf{v})=\mathbf{0}\) for some weights \(c_{1}\) and \(c_{2},\) not both zero. Use this equation.]

Let \(\mathbf{q}_{1}, \mathbf{q}_{2}, \mathbf{q}_{3},\) and \(\mathbf{v}\) represent vectors in \(\mathbb{R}^{5},\) and let \(x_{1}, x_{2}\) and \(x_{3}\) denote scalars. Write the following vector equation as a matrix equation. Identify any symbols you choose to use. \(x_{1} \mathbf{q}_{1}+x_{2} \mathbf{q}_{2}+x_{3} \mathbf{q}_{3}=\mathbf{v}\)

Could a set of three vectors in \(\mathbb{R}^{4}\) span all of \(\mathbb{R}^{4} ?\) Explain. What about \(n\) vectors in \(\mathbb{R}^{m}\) when \(n\) is less than \(m ?\)

Suppose \(A\) is an \(m \times n\) matrix with the property that for all \(\mathbf{b}\) in \(\mathbb{R}^{m}\) the equation \(A \mathbf{x}=\mathbf{b}\) has at most one solution. Use the definition of linear independence to explain why the columns of \(A\) must be linearly independent.

1\( produces ore that contains 20 metric tons of copper and 550 kilograms of silver, while one day's opera… # A mining company has two mines. One day's operation at mine \)\\# 1\( produces ore that contains 20 metric tons of copper and 550 kilograms of silver, while one day's operation at mine \)\\# 2\( produces ore that contains 30 metric tons of copper and 500 kilograms of silver. Let \)\mathbf{v}_{1}=\left[\begin{array}{c}{20} \\ {550}\end{array}\right]\( and \)\mathbf{v}_{2}=\left[\begin{array}{c}{30} \\ {500}\end{array}\right] .\( Then \)\mathbf{v}_{1}\( and \)\mathbf{v}_{2}\( represent the "output per day" of mine \)\\# 1\( and mine \)\\# 2\( , respectively. a. What physical interpretation can be given to the vector 5 \)\mathbf{v}_{1} ?\( b. Suppose the company operates mine \)\\# 1\( for \)x_{1}\( days and mine \)\\# 2\( for \)x_{2}\( days. Write a vector equation whose solution gives the number of days each mine should operate in order to produce 150 tons of copper and 2825 kilograms of silver. Do not solve the equation. c. \)[\mathbf{M}]\( Solve the equation in \)(b)$

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