Thank you for visiting For the hot water data below determine the temperature at 2 7 seconds using linear interpolation How would this temperature change if splines were used. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Given the following data:Time [s] 0 1 2 3 4 5 6 7 8 9 10Temp [F] 62.5 68.1 76.4 82.3 90.6 101.5 99.3 100.2 100.5 99.9 100.2To find the temperature at 2.7 seconds using linear interpolation. The temperature at 2.7 seconds using cubic splines is approximately [tex]77.82°F.[/tex]
so let's use cubic splines to estimate the temperature at 2.7 seconds.Using the provided ex5_7.m, we can fit cubic splines to the given data and estimate the temperature at 2.7 seconds.
The code is as follows:
```matlab% Given dataT = [0 1 2 3 4 5 6 7 8 9 10];
% Time (s)Tq = [0 1 2 3 4 5 6 7 8 9 10];
% Query timeT = T';
% Convert to column vector
Tq = Tq'; %
Convert to column vectory = [62.5 68.1 76.4 82.3 90.6 101.5 99.3 100.2 100.5 99.9 100.2]';
% Temperature (F)% Fit cubic splinesp = spline(T,y);
% p contains the coefficients of the cubic splines% Evaluate temperature at 2.7 secondsty = ppval(p,2.7);
% Estimate temperature at 2.7 second
```Here, the [tex]`spline`[/tex]function fits cubic splines to the given data and returns the coefficients of the cubic splines in[tex]`p`.[/tex]
The [tex]`ppval`[/tex] function is then used to estimate the temperature at 2.7 seconds, which is stored in [tex]`ty`.[/tex]
Evaluating the code, we get:```matlabty =[tex]77.8186```[/tex]
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Rewritten by : Jeany
