Question

I just need the last part. T_3min

HW7_3 Ball bearings are hardened through a process known as quenching-submersion of the heated ball bearing in oil or water in order to cool it rapidly. The data below represent the temperature of the ball at various points in the cooling process | time (sec) 0 30 | 60 | 150 | 240 Temp (C) 800 457 269 | 79 46 Plot the data, along with a) interpolating polynomial, (use polyfit and polyval) b) piecewise cubic interpolation (not-a-knot) and c) piecewise cubic interpolation (shape-preserving) – all in one figure window. Include a legend for the three methods used: polynomial interpolation, not-a-knot, shape-preserving. Then choose what you believe to be the best of these to predict the temperature after 3 minutes of cooling. Store the result in the variable T 3min. Paste the script file into Cody, which will check the correctness of T_3min by a hidden test.

Answer 1

Code:

clc
clear

t = [0 30 60 150 240];
T = [800 457 269 79 46];
tq = 0:240;

pa = polyfit(t, T, 3);
Ta = polyval(pa, tq);
Tb = interp1(t, T, tq, 'pchip');
Tc = interp1(t, T, tq, 'spline');

plot(t, T, 'ro');
hold on
grid on
plot(tq, Ta, '-b');
plot(tq, Tb, '--m');
plot(tq, Tc, '-.k');
legend('data', '3 degree polynomial', 'piecewise cubic', 'not-a-knot');
xlabel('time(sec)');
ylabel('Temp(C)');
%% clearly from the graph not-a-knot predict best
disp("Temp after 3 min:")
disp(Tc(find(tq == 180)))

Output:

Temp after 3 min:
72.8395

>>

o data - 3 degree polynomial ---piecewise cubic -------not-a-knot Temp(C) R 50 100 150 time (sec) 200 250