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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 instead? (Hint: Use ex5_7.m as a starting point).

| Time [s] | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|----------|---|---|---|---|---|---|---|---|---|---|----|
| Temp [°F]| 62.5 | 68.1 | 76.4 | 82.3 | 90.6 | 101.5 | 99.3 | 100.2 | 100.5 | 99.9 | 100.2 |

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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