Deciphering this problem statement: Air resistance on projectiles

[cookiemnstr510510]

Homework Statement

See attached image

Relevant Equations

basic augmented matricies

I am struggling with our equation and where to plug in my velocities and forces? After looking at my book I don't understand why p(t) would be where I plug in force, and why variable "t" is where we plug our velocity in?

When the problem says "what happens if you try to use a polynomial of degree less than 5" do they mean have an equation like:
p(t)=a0+a1t+a2t^2+a3t^3 ?

Any help would be appreciated.
Thanks

 

[haruspex]

Homework Statement:: See attached image
Homework Equations:: basic augmented matricies

I don't understand why p(t) would be where I plug in force, and why variable "t" is where we plug our velocity in?

p is a function, not a variable. ##p(t)=a0+a1t+a2t^2+a3t^3## says that if you supply the input value t to the function p it will return the value ##a0+a1t+a2t^2+a3t^3##. t just stands for any arbitrary input you might wish to make.
You want to express a force, F, as a polynomial function of wind speed, w. That means you are looking for a polynomial p such that ##F=p(w)##. So in p(t) the input p is to take the values you have for w, and the polynomial is to return something close to the observed values of F: ##F_i\approx p(w_i)##

 

[archaic]

Homework Statement:: See attached image
Homework Equations:: basic augmented matricies

When the problem says "what happens if you try to use a polynomial of degree less than 5" do they mean have an equation like:
p(t)=a0+a1t+a2t^2+a3t^3 ?

Yes.
By now, I think you have figured that you must find those constants in that polynomial. How do you think you can do that? Look at what you have been given.

 

[jbriggs444]

There is a crank-and-grind approach to constructing an interpolating polynomial through a given set of data points. The Wiki page gives it, introduced by the words: "Alternatively, we may write down the polynomial immediately in terms of Lagrange polynomials:"

The idea behind that construction is that one can easily write down a polynomial which is zero at all but one of the given data points. For instance, a polynomial which is zero at every given data point other than t=4 is:$$(t-0)(t-2)(t-6)(t-8)(t-10)$$ You want to scale that polynomial up or down so that its result, when evaluated at t=4 is equal to the given data value for t=4. i.e. it needs to evaluate to 14.8. You determine the required scaling factor by taking the given data value (14.8) and dividing it by the result of evaluating the polynomial at t=4.

i.e. you evaluate:$$\frac{(t-0)(t-2)(t-6)(t-8)(t-10)}{(4-0)(4-2)(4-6)(4-8)(4-10)}14.8$$ as a polynomial. That is, it should come out as$$\text{something }t^5 + \text{something }t^4 + \text{something }t^3 + \text{something }t^2 + \text{something }t + \text{something}$$ You then repeat for each of the other data points and add all six polynomials together. Because of the way they were constructed, each individual formula is a fifth degree polynomial, is zero at all but one point and takes on the correct value at that data point. The sum of all of them must therefore be a fifth (or fewer) degree polynomial that takes the correct value at all the data points.

It is a fair bit of mind-numbing work.

Edit: Switched to use "t" as the dummy variable in the polynomial as the question suggests.

 

[DEvens]

The manual process suggested by jbriggs444 will certainly work.

For something less painful, you might consider using a spreadsheet such as MS Excel, and looking up the "generalized least square" method.

 

[gmax137]

A word of caution regarding the use of Excel: you can ask it for a 5th degree polynomial trendline and "display equation on Chart" but the default will display only 2 decimal places for the coefficients. For a 5th degree polynomial that is just not enough to use it to get good values. If you go the trendline route, you need to format the trendline equation to show 5 or 6 figures.

 

[DEvens]

A word of caution regarding the use of Excel: you can ask it for a 5th degree polynomial trendline and "display equation on Chart" but the default will display only 2 decimal places for the coefficients. For a 5th degree polynomial that is just not enough to use it to get good values. If you go the trendline route, you need to format the trendline equation to show 5 or 6 figures.

I was not talking about using the trend line function. Though, of course, that's possible. The digits displayed can be formatted to show the desired number of digits.

I was talking about using the spreadsheet to do the generalized least square fit to a polynomial of your choice. You make a column with all 1's, a column with x, a column with x-squared, a column with x-cubed, etc. Then you build your polynomial out of A times the first column, B times the second, etc. Then you minimize the sum of the squares of the differences with your data by adjusting the parameters. You get the same notion as a least-square fit to a straight line, but to a polynomial. There's a keen little matrix equation that you need to solve to get the minimum. And such matrix equations, up to a few 100 data points and up to a dozen columns, is easily within Excel's ability.

Excel also has a built-in polynomial fitting function. This will work directly on the spreadsheet, no graph required.

 

[gmax137]

I was not talking about using the trend line function.

I know you weren't .

But I have seen the trendline feature used a lot for this, and I have had to show numerous co-workers that it needs to be done with caution. Many people are fooled by the "R-squared = 0.9999" and just copy, or cut and paste, the displayed coefficients. The pitfall is, many of the polynomials have a negative coefficient in the leading or second term, so they depend on subtracting large numbers that don't differ by much. To get an accurate subtraction you need to carry a lot of digits.

I suppose this caution applies to the manually-constructed polynomial as well? Don't round the coefficients as you go.