A projectile has been shot into the atmosphere. Find the polynomial that interpolates the speed and acceleration in function of time.

Question

This is the table (tempo = time, velocidade = speed, aceleracao = acceleration)

I used Newton's interpolation to get the polynomial (5th degree).

I got $$102 + 138(x-15) + (-124/15)(x-15)^2 + (289/225)(x-15)^2(x-30) + (-0.0689)(x-15)^2(x-30)^2 + (0.0048)(x-15)^2(x-30)^2(x-45)$$

I tested this in my calculator and the values for p(15) and p(30) are correct but p(45) is completely wrong.

Then I got the derivative to test the derivatives $$14341.5 - 2226.13 x + 122.896 x^2 - 2.8676 x^3 + 0.024 x^4$$

p'(15) and p'(30) are correct, albeit with a slight error, but p'(45) is very off (I got 132 point something).

Is this normal? Is it possible to get a better estimate for this?

My calculations:

Answer 1

If you do the table correctly, with $p(x_2)=420$, you get $$\small\begin{array}{c|c|c|} p[x_0]=102&p[x_0,x_0]=138&p[x_0,x_0,x_1]=-8.2667&p[x_0,x_0,x_1,x_1]=1.28444\\ &p[x_0,x_1]=14&p[x_0,x_1,x_1]=11&p[x_0,x_1,x_1,x_2]=-0.74844\\ p[x_1]=312&p[x_1,x_1]=179&p[x_1,x_1,x_2]=-11.4533&p[x_1,x_1,x_2,x_2]=1.36267\\ &p[x_1,x_2]=7.2&p[x_1,x_2,x_2]=8.9867\\ p[x_2]=420&p[x_2,x_2]=142 \end{array}$$ cont. $$\small\begin{array}{c|c|c|} &p[x_0,x_0,x_1,x_1,x_2]= -0.067763&p[x_0,x_0,x_1,x_1,x_2,x_2]=0.004604\\ &p[x_0,x_1,x_1,x_2,x_2]=0.070370& \end{array}$$

Answer 2

Given the 6 degrees of freedom and the six constraints, the interpolation can be exact, if it is possible at all (which is always a possibility in practice, but unlikely for a homework problem). Any small inaccuracies you see are the result of either errors in your work or the fact that you're calculating with decimal approximations. I'm not sure where your -0.0689 and and 0.0048 terms came from. You should double-check your computations using precise fractions -- note that since all the constraints are integral, all your powers of 15 in denominators should cancel out.

If it turns out there is still an issue after checking your computations, I'll suggest an alternate cross-check method. Letting the 6 coefficients of a fifth degree polynomial be unknowns, the 6 constraints you have from a system of linear equations in those coefficients. Converting this into a matrix equation, you should be able to solve it through Gaussian elimination. You can make life easier for yourself by also having the polynomial be a function of not $t$ but $z = (t-15)/15$, so that all the constraints are at $z = 0$, $1$, and $2$.

Answer 3

For me, the problem comes for the conversion to decimals (notice that $45^5=184528125$ and any error on the last coefficient will have a dramatic impact.

Since I suppose that you used matrix calculations, you must have arrived at $$v=-68178+\frac{69497 t}{5}-\frac{26902 t^2}{25}+\frac{88873 t^3}{2250}-\frac{11633 t^4}{16875}+\frac{259 t^5}{56250}$$ which reproduces exactly the values of speed and acceleration.

Deriving, we have $$\frac{69497}{5}-\frac{53804 t}{25}+\frac{88873 t^2}{750}-\frac{46532 t^3}{16875}+\frac{259 t^4}{11250}$$ which ,concerted to decimals, gives $$13899.4-2152.16 t+118.497 t^2-2.75745 t^3+0.0230222 t^4$$ which are quite different to the ones you posted.