---
title: "Derive Hermite cubic curve and blending functions"
subject: "Computer Aided Design"
chapter: 3
key: "semester_4/computer_aided_design:hermite-cubic-derivation"
source: https://www.bctengineer.com/q/semester_4/computer_aided_design/hermite-cubic-derivation
site: BCT Engineering
---

# Derive Hermite cubic curve and blending functions

Computer Aided Design · Chapter 3: Mathematical Representation of Curves

- Asked in: year not given, 8 marks
- Video: [B.Tech|| ME|| CAD/CAM||Hermite Curve(Part 1)](https://www.youtube.com/watch?v=yhkcJdYevSA) — Platform4me
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/computer_aided_design/hermite-cubic-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-3

## Question

Derive the equation of a Hermite cubic curve in terms of its end points and end tangent vectors. Obtain the blending functions and write the equation in matrix form. State the properties of the curve.

## Short answer

A Hermite cubic is defined by endpoints P0, P1 and tangents P0', P1'. The curve is P(u)=F1 P0 + F2 P1 + F3 P0' + F4 P1' with blending functions F1=2u^3-3u^2+1, F2=-2u^3+3u^2, F3=u^3-2u^2+u, F4=u^3-u^2. It interpolates endpoints, tangents at ends are controlled, and segments join with C1 continuity if tangents match.

## Answer

A Hermite cubic segment is defined by two end points $\mathbf P_0,\mathbf P_1$ and two end tangent vectors $\mathbf P_0',\mathbf P_1'$.

#### Derivation
A cubic parametric curve is

$$
\mathbf P(u)=\mathbf a_3u^3+\mathbf a_2u^2+\mathbf a_1u+\mathbf a_0,\qquad 0\le u\le1
$$

with tangent

$$
\mathbf P'(u)=3\mathbf a_3u^2+2\mathbf a_2u+\mathbf a_1
$$

Apply the four boundary conditions:

$$
\begin{aligned}
\mathbf P(0)&=\mathbf a_0=\mathbf P_0 \\
\mathbf P(1)&=\mathbf a_3+\mathbf a_2+\mathbf a_1+\mathbf a_0=\mathbf P_1 \\
\mathbf P'(0)&=\mathbf a_1=\mathbf P_0' \\
\mathbf P'(1)&=3\mathbf a_3+2\mathbf a_2+\mathbf a_1=\mathbf P_1'
\end{aligned}
$$

Solving:

$$
\begin{aligned}
\mathbf a_0&=\mathbf P_0,\qquad \mathbf a_1=\mathbf P_0' \\
\mathbf a_2&=3(\mathbf P_1-\mathbf P_0)-2\mathbf P_0'-\mathbf P_1' \\
\mathbf a_3&=2(\mathbf P_0-\mathbf P_1)+\mathbf P_0'+\mathbf P_1'
\end{aligned}
$$

Substituting and collecting terms in $\mathbf P_0,\mathbf P_1,\mathbf P_0',\mathbf P_1'$:

$$
\mathbf P(u)=F_1\mathbf P_0+F_2\mathbf P_1+F_3\mathbf P_0'+F_4\mathbf P_1'
$$

#### Blending (Hermite) functions
$$
\begin{aligned}
F_1&=2u^3-3u^2+1 \\
F_2&=-2u^3+3u^2 \\
F_3&=u^3-2u^2+u \\
F_4&=u^3-u^2
\end{aligned}
$$

#### Matrix form
$$
\mathbf P(u)=\begin{bmatrix}u^3&u^2&u&1\end{bmatrix}
\begin{bmatrix}2&-2&1&1\\-3&3&-2&-1\\0&0&1&0\\1&0&0&0\end{bmatrix}
\begin{bmatrix}\mathbf P_0\\\mathbf P_1\\\mathbf P_0'\\\mathbf P_1'\end{bmatrix}
$$

#### Properties
- Passes through both end points (interpolates).
- Tangent direction and magnitude at the ends are controlled; a larger tangent magnitude makes the curve stay longer in that direction.
- $F_1+F_2=1$ for every $u$.
- Two segments are joined with $C^1$ continuity by taking the same tangent at the common point.
- Local control: changing one segment does not change the others.
- Drawback: tangent vectors are not intuitive to the designer, which led to the Bezier form.

## Check yourself

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

1. What four conditions determine a Hermite cubic?
   A. Four points on the curve
   B. Two end points and two curvatures
   C. Two end points and two end tangent vectors
   D. Three points and one tangent
   Answer: C. Hermite uses P(0), P(1), P'(0), P'(1). (Key point: Boundary conditions)

2. From the conditions, what are a0 and a1?
   A. a0 = P0', a1 = P0
   B. a0 = P1, a1 = P1'
   C. a0 = 0, a1 = P0
   D. a0 = P0, a1 = P0'
   Answer: D. P(0) = a0 = P0 and P'(0) = a1 = P0'. (Key point: Coefficients)

3. Which is F1, the blending function for P0?
   A. 2u^3 - 3u^2 + 1
   B. -2u^3 + 3u^2
   C. u^3 - 2u^2 + u
   D. u^3 - u^2
   Answer: A. F1 = 2u^3 - 3u^2 + 1; F2 = -2u^3 + 3u^2 applies to P1. (Key point: Blending functions)

4. How can two Hermite segments be joined with C1 continuity?
   A. Use equal end points only
   B. Use the same tangent at the common point
   C. Use zero tangent
   D. Not possible
   Answer: B. Same position and same tangent at the joint gives C1 continuity. (Key point: Continuity)

5. What is the drawback of Hermite curves?
   A. They do not pass through end points
   B. They lack local control
   C. Tangent vectors are not intuitive for the designer
   D. They need many points
   Answer: C. Tangent vectors are hard to specify intuitively, which led to the Bezier form. (Key point: Drawback)

## Related in Computer Aided Design

- [Hermite curve points and tangents](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Ahermite-curve-numerical) — Computer Aided Design, chapter 3
- [Cubic Bezier point and tangent calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-curve-numerical) — Computer Aided Design, chapter 3
- [Convert Hermite data to Bezier cubic](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Ahermite-to-bezier-conversion) — Computer Aided Design, chapter 3
- [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
- [Bezier curve: equation and properties](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-curve-properties) — Computer Aided Design, chapter 3
- [Bezier surface equation and properties](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-surface-properties) — Computer Aided Design, chapter 4

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- [Derive cubic Bezier equation and properties](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-cubic-equation-properties) — Computer Graphics, chapter 5
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