How to plot trajectory curve given x and y points as well as minima and maxima of the polynomial?

I've spent hours trying to figure out how to do this in matlab, which is frusturating considering it was so easy to find the solution by hand:

The three position (horizontal, vertical) points (0,80), (120,0), (140,5) were given in the word problem; this data was used to create three equations using the polynomial template shown in equation one. Since the point (0,80) is the maximum of the slope and (120,0) is the minimum, the derivative of the height function at these tangent points is zero; this adds two more equations to the system. With a total of five equations, the resulting solution to the system is a fourth degree polynomial. The coefficients of this polynomial were calculated using matricies.

edit:

Here's what I've got so far

How do I set the derivative to zero for the min and max points in my plot?

Answer 1

[KSSV]

syms a b c d e ;
eqns = [e == 80 ; 
        207360000*a+1728000*b+1440*c+120*d+e == 0 ;
        384160000*a+2744000*b+19600*c+140*d+e == 5 ; 
        d == 0 ;
        6912000*a+43200*b+240*c+d==0] ; 

S = solve(eqns,[a,b,c,d,e]) ;         
S.a 
S.b
S.c
S.d
S.e

[Newslin Heiller]

I'm trying to do something more like this (Horner's Method for Evaluating Polynomials):

x0=[0 120 140];

y0=[80 0 5];

xs=linspace(0,1,150)

n=length(x0)-1;

ylim([0 100]);

pp_pchip=pchip(x0,y0);

coeffs_deriv = zeros(n,3);

for k = 1:n

c = pp_pchip.coefs(k,:);

coeffs_deriv(k,:) = polyder(c);

end

pp_deriv = mkpp(x0,coeffs_deriv);

clf;

ys_deriv = ppval(pp_deriv,xs);

ydata_deriv = ppval(pp_deriv,x0);

plot(xs,ys_deriv,'g','linewidth',2);

hold on;

plot(x0,ydata_deriv,'k.','markersize',30);

title('First derivative (pchip)','fontsize',18);

Answer 2

[Image Analyst]

If you have x and y (hx), you can get the coefficients with polyfit like this

coefficients = polyfit(x, hx, 4);

Now you can get separate coefficients if you want to:

a = coefficients(1);
b = coefficients(2);
c = coefficients(3);
d = coefficients(4);
e = coefficients(5);

These are the same a, b, c, and d in both the h(x) and h'(x) equations, obviously.

[Newslin Heiller]

I'm trying to do something more like this (Horner's Method for Evaluating Polynomials):

x0=[0 120 140];
y0=[80 0 5];
xs=linspace(0,1,150)
n=length(x0)-1;
ylim([0 100]);
pp_pchip=pchip(x0,y0);
coeffs_deriv = zeros(n,3);
for k = 1:n
    c = pp_pchip.coefs(k,:);
    coeffs_deriv(k,:) = polyder(c);
end
pp_deriv = mkpp(x0,coeffs_deriv);
clf;
ys_deriv = ppval(pp_deriv,xs);
ydata_deriv = ppval(pp_deriv,x0);
plot(xs,ys_deriv,'g','linewidth',2);
hold on;
plot(x0,ydata_deriv,'k.','markersize',30);
title('First derivative (pchip)','fontsize',18);

[Image Analyst]

OK, so, did you actually try it? Is it working for you?

[Newslin Heiller]

No, I'm still need to figure out how to define the min and max points.

[Image Analyst]

Did you use the max() and min() functions? Like for example this:

maxValue = max(ys_deriv);

If not, then why not?