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hitung untuk data klorida 50mg/L
suhu 5 = 7,5
suhu 10 = 6,2
suhu 15 = 5,3
suhu 20 = 4,5
suhu 25 = 3,7
suhu 30 = 2,2

To estimate the concentration of dissolved oxygen for \(T = 22.4^\circ C\) and chloride concentration of 50 mg/L using the given data, we will use the Newton interpolation method.

The data points provided are:
- \(T = 5^\circ C\), \(O_2 = 7.5\)
- \(T = 10^\circ C\), \(O_2 = 6.2\)
- \(T = 15^\circ C\), \(O_2 = 5.3\)
- \(T = 20^\circ C\), \(O_2 = 4.5\)
- \(T = 25^\circ C\), \(O_2 = 3.7\)
- \(T = 30^\circ C\), \(O_2 = 2.2\)

We will use these data points to form the Newton interpolation polynomial and then estimate the oxygen concentration at \(T = 22.4^\circ C\).

Let's perform the calculations step by step.

1. **Form the divided difference table:**

\[
\begin{array}{c|c|c|c|c|c|c}
T & O_2 & \Delta O_2 & \Delta^2 O_2 & \Delta^3 O_2 & \Delta^4 O_2 & \Delta^5 O_2 \\
\hline
5 & 7.5 &  &  &  &  &  \\
10 & 6.2 &  &  &  &  &  \\
15 & 5.3 &  &  &  &  &  \\
20 & 4.5 &  &  &  &  &  \\
25 & 3.7 &  &  &  &  &  \\
30 & 2.2 &  &  &  &  &  \\
\end{array}
\]

2. **Calculate the first-order differences:**

\[
\Delta O_2 = \frac{O_2(T_{i+1}) - O_2(T_i)}{T_{i+1} - T_i}
\]

\[
\begin{array}{c|c|c|c|c|c|c}
T & O_2 & \Delta O_2 & \Delta^2 O_2 & \Delta^3 O_2 & \Delta^4 O_2 & \Delta^5 O_2 \\
\hline
5 & 7.5 &  &  &  &  &  \\
10 & 6.2 & (6.2 - 7.5)/5 = -0.26 &  &  &  &  \\
15 & 5.3 & (5.3 - 6.2)/5 = -0.18 &  &  &  &  \\
20 & 4.5 & (4.5 - 5.3)/5 = -0.16 &  &  &  &  \\
25 & 3.7 & (3.7 - 4.5)/5 = -0.16 &  &  &  &  \\
30 & 2.2 & (2.2 - 3.7)/5 = -0.30 &  &  &  &  \\
\end{array}
\]

3. **Calculate higher-order differences:**

\[
\Delta^2 O_2 = \frac{\Delta O_2(T_{i+1}) - \Delta O_2(T_i)}{T_{i+1} - T_i}
\]

\[
\begin{array}{c|c|c|c|c|c|c}
T & O_2 & \Delta O_2 & \Delta^2 O_2 & \Delta^3 O_2 & \Delta^4 O_2 & \Delta^5 O_2 \\
\hline
5 & 7.5 &  &  &  &  &  \\
10 & 6.2 & -0.26 &  &  &  &  \\
15 & 5.3 & -0.18 & ( -0.18 + 0.26)/5 = 0.016 &  &  &  \\
20 & 4.5 & -0.16 & (-0.16 + 0.18)/5 = 0.004 &  &  &  \\
25 & 3.7 & -0.16 & (-0.16 + 0.16)/5 = 0 &  &  &  \\
30 & 2.2 & -0.30 & (-0.30 + 0.16)/5 = -0.028 &  &  &  \\
\end{array}
\]

Continue calculating higher-order differences in a similar manner. For brevity, I’ll skip detailed intermediate calculations and directly use these differences to estimate \(O_2\) at \(T = 22.4^\circ C\):

Using Newton's interpolation formula, we have:
\[
O_2(T) = O_2(T_0) + (T-T_0) \Delta O_2(T_0) + (T-T_0)(T-T_1) \Delta^2 O_2(T_0) + \cdots
\]

Plugging in \(T_0 = 20^\circ C\), we get:
\[
O_2(22.4) = 4.5 + (22.4 - 20)(-0.16) + (22.4 - 20)(22.4 - 15)(0.004) + \cdots
\]

Let's simplify and calculate the final result:

\[O_2(22.4) = 4.5 + 2.4(-0.16) + 2.4 \cdot 7.4 \cdot 0.004 + \cdots \]

\[O_2(22.4) = 4.5 - 0.384 + 0.07056 + \cdots \]

\[O_2(22.4) \approx 4.5 - 0.384 + 0.07056 \]

\[O_2(22.4) \approx 4.18656\]

Thus, the estimated concentration of oxygen at \(T = 22.4^\circ C\) and chloride concentration of 50 mg/L is approximately \(4.19\) mg/L.

If you have further questions or need additional details, please let me know!

Here are some questions you might want to ask next:
1. How is the Lagrange interpolation method different from Newton's method?
2. Can you show a detailed example of using the Lagrange interpolation method?
3. What are the advantages of using Newton's forward interpolation method?
4. How do you derive the coefficients for higher-order terms in Newton's method?
5. What are some practical applications of interpolation in real-world scenarios?

**Tip:** When performing interpolation, ensure the data points are accurately measured and sufficiently close to the desired value to increase the precision of your estimation.

Learn how to estimate the concentration of dissolved oxygen at 22.4°C with a chloride concentration of 50 mg/L using Newton's interpolation method. This problem includes key data points such as oxygen levels at various temperatures and showcases numerical analysis techniques. Suitable for students in Grades 11-12.

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