---
title: "Natural cubic spline: y(6.5)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:spline-5-7-9-11"
source: https://www.bctengineer.com/q/semester_4/numerical_method/spline-5-7-9-11
site: BCT Engineering
---

# Natural cubic spline: y(6.5)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2075 Baisakh, 8 marks
- Video: [Cubic Spline Interpolation | Example Solved - 3 | Engineering mathematics | Mathspedia |](https://www.youtube.com/watch?v=AdaYIlpbw_0) — Mathspedia
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/spline-5-7-9-11
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Estimate y(6.5) using Natural cubic spline interpolation technique from the following data.
| x | 3 | 5 | 7 | 9 | 11 |
|---|---|---|---|---|---|
| y | 8 | 10 | 9 | 12 | 5 |

## Short answer

Natural cubic spline sets second derivatives zero at endpoints. With h=2, solving the tridiagonal system yields M1≈-1.9018, M2≈3.1071, M3≈-4.5268. For x=6.5 in [5,7], the spline gives y(6.5)≈8.8675.

## Answer

A natural cubic spline is a piecewise cubic S(x) that passes through all data points, has continuous first and second derivatives at the interior knots, and has zero second derivative at the end points (M₀ = Mₙ = 0).

#### Step 1: Setup

Points: n = 4 intervals, equal spacing h = 2. Let Mᵢ = S″(xᵢ), with M₀ = M4 = 0.

For equal spacing the continuity of S′ at each interior knot gives

```text
M(i−1) + 4Mi + M(i+1) = (6/h²)(y(i+1) − 2yi + y(i−1)),   i = 1 … 3
```

With 6/h² = 1.5:

```text
4M1 + M2 = 1.5 × (9 − 2(10) + 8) = -4.5000
M1 + 4M2 + M3 = 1.5 × (12 − 2(9) + 10) = 6.0000
M2 + 4M3 = 1.5 × (5 − 2(12) + 9) = -15.0000
```

Solving these simultaneous equations (elimination) gives

```text
M₀ = 0, M1 = -1.9018, M2 = 3.1071, M3 = -4.5268, M4 = 0
```

#### Step 2: Spline on the interval [5, 7]

```text
S(x) = Mi(x(i+1) − x)³/(6h) + M(i+1)(x − xi)³/(6h)
       + (yi/h − Mi·h/6)(x(i+1) − x) + (y(i+1)/h − M(i+1)·h/6)(x − xi)
```

Here i = 1, x = 6.5: x2 − x = 0.5000, x − x1 = 1.5000, M1 = -1.9018, M2 = 3.1071.

```text
S(6.5) = (-1.9018)(0.5000)³/12 + (3.1071)(1.5000)³/12
          + (5.6339)(0.5000) + (3.4643)(1.5000)
        = -0.0198 + 0.8739 + 2.8170 + 5.1964
        = 8.8675
```

**Answer: y(6.5) ≈ 8.8675**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What end condition defines a natural cubic spline?
   A. Second derivative is zero at the end points: M0 = Mn = 0
   B. First derivative is zero at the ends
   C. The spline is linear everywhere
   D. y is zero at the ends
   Answer: A. A natural spline has S''(x0) = S''(xn) = 0. (Key point: Natural spline conditions)

2. What does the system for M1, M2, M3 come from?
   A. Continuity of y only
   B. Continuity of the first derivative S' at each interior knot
   C. Setting each Mi equal to yi
   D. Minimising the sum of y values
   Answer: B. S and S'' are continuous by construction; matching S' at knots gives the equations. (Key point: Continuity)

3. For h = 2, what is the right-hand side of 4M1 + M2 = ?
   A. 1.5 × (10 - 2·9 + 12) = 6
   B. 2 × (9 - 10 + 8) = 14
   C. 1.5 × (9 - 2·10 + 8) = -4.5
   D. 1.5 × (8 + 10 + 9) = 40.5
   Answer: C. 6/h² = 1.5 and the second difference y2 - 2y1 + y0 = 9 - 20 + 8 = -3, so -4.5. (Key point: First equation)

4. Which interval is used for x = 6.5, and what is the estimate?
   A. [5, 7], y(6.5) ≈ 8.8675
   B. [7, 9], y(6.5) ≈ 8.8675
   C. [5, 7], y(6.5) ≈ 9.5
   D. [3, 5], y(6.5) ≈ 10
   Answer: A. 6.5 lies between 5 and 7; evaluating the cubic with M1 = -1.9018 and M2 = 3.1071 gives 8.8675. (Key point: Interval and answer)

## Close variants set in other exams

- [Natural cubic spline: y(4.5)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aspline-4-5) — Numerical Method, chapter 4, asked 3×
- [Natural cubic spline: y(0.5)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aspline-y-0-5) — Numerical Method, chapter 4, asked 3×

## Related in Numerical Method

- [Natural cubic spline: y(8)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aspline-find-y-8) — Numerical Method, chapter 4
- [Cubic spline and its advantage](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Acubic-spline-advantage) — Numerical Method, chapter 4

## Related in other subjects

- [Interpolation versus approximation spline](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Ainterpolation-vs-approximation-spline) — Computer Graphics, chapter 5, asked 2×
- [Uniform cubic B-spline segment points](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Auniform-cubic-bspline-numerical) — Computer Aided Design, chapter 3
- [K-map Σ(1,2,3,8,9,10,11,14)+d(0,4,12), NAND](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Akmap-nand-1238-dc-0412) — Digital Logic, chapter 3, asked 5×
- [K-map Σ(0,1,4,8,10,11,12)+d, NOR](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Akmap-nor-gates-01248) — Digital Logic, chapter 3, asked 5×