---
title: "Hermite curve points and tangents"
subject: "Computer Aided Design"
chapter: 3
key: "semester_4/computer_aided_design:hermite-curve-numerical"
source: https://www.bctengineer.com/q/semester_4/computer_aided_design/hermite-curve-numerical
site: BCT Engineering
---

# Hermite curve points and tangents

Computer Aided Design · Chapter 3: Mathematical Representation of Curves

- Asked in: year not given, 8 marks
- Video: [S 2.39 Numerical on Hermite Cubic Spline to find Parametric Equations and Points on Spline (1/2)](https://www.youtube.com/watch?v=dQJGSaS93t4) — Dr. Khizar Ahmed Pathan
- 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-curve-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-3

## Question

A Hermite cubic curve has end points P0 = (0, 0) and P1 = (4, 2), and end tangent vectors P0' = (4, 4) and P1' = (4, -4). Find the coordinates of the curve and its tangent vector at u = 0.25, 0.5 and 0.75. Verify the end conditions.

## Short answer

Using Hermite blending functions: P(0.25)=(1.00,1.0625), P(0.5)=(2.00,2.00), P(0.75)=(3.00,2.4375). Tangent vectors: (4,4.25) at u=0.25, (4,3) at u=0.5, (4,0.25) at u=0.75.

## Answer

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

with $F_1=2u^3-3u^2+1$, $F_2=-2u^3+3u^2$, $F_3=u^3-2u^2+u$, $F_4=u^3-u^2$. The tangent uses the derivatives $F_1'=6u^2-6u$, $F_2'=-6u^2+6u$, $F_3'=3u^2-4u+1$, $F_4'=3u^2-2u$.

Since $\mathbf P_0=(0,0)$:

$$
\mathbf P(u)=F_2(4,2)+F_3(4,4)+F_4(4,-4)
$$

#### At u = 0.5
$F_2=0.5$, $F_3=0.125-0.5+0.5=0.125$, $F_4=0.125-0.25=-0.125$.

$$
\begin{aligned}
x&=0.5(4)+0.125(4)-0.125(4)=2.0 \\
y&=0.5(2)+0.125(4)-0.125(-4)=1+0.5+0.5=2.0
\end{aligned}
$$

#### At u = 0.25
$F_2=-0.03125+0.1875=0.15625$; $F_3=0.015625-0.125+0.25=0.140625$; $F_4=0.015625-0.0625=-0.046875$.

$x=0.15625(4)+0.140625(4)-0.046875(4)=0.625+0.5625-0.1875=1.000$

$y=0.15625(2)+0.140625(4)+0.046875(4)=0.3125+0.5625+0.1875=1.0625$

#### At u = 0.75
$F_2=-0.84375+1.6875=0.84375$; $F_3=0.421875-1.125+0.75=0.046875$; $F_4=0.421875-0.5625=-0.140625$.

$x=0.84375(4)+0.046875(4)-0.140625(4)=3.000$

$y=0.84375(2)+0.046875(4)+0.140625(4)=1.6875+0.1875+0.5625=2.4375$

#### Tangent vectors
Using $\mathbf P'(u)=F_2'(4,2)+F_3'(4,4)+F_4'(4,-4)$:

| $u$ | $x$ | $y$ | $P'_x$ | $P'_y$ |
|---|---|---|---|---|
| 0 | 0 | 0 | 4 | 4 |
| 0.25 | 1.000 | 1.0625 | 4.00 | 4.25 |
| 0.50 | 2.000 | 2.000 | 4.00 | 3.00 |
| 0.75 | 3.000 | 2.4375 | 4.00 | 0.25 |
| 1 | 4 | 2 | 4 | -4 |

Example at $u=0.5$: $F_2'=1.5$, $F_3'=-0.25$, $F_4'=-0.25$, so $P'_y=1.5(2)+(-0.25)(4)+(-0.25)(-4)=3-1+1=3$ and $P'_x=4(1.5-0.25-0.25)=4$.

#### End conditions
At $u=0$: $F_1=1$, others 0, so $\mathbf P=(0,0)$ and $\mathbf P'=(4,4)$. At $u=1$: $F_2=1$, so $\mathbf P=(4,2)$ and $\mathbf P'=(4,-4)$. Both are satisfied.

**Answer: P(0.25) = (1.00, 1.0625), P(0.5) = (2.00, 2.00), P(0.75) = (3.00, 2.4375); tangents (4, 4.25), (4, 3), (4, 0.25).**

## Check yourself

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

1. With P0 = (0,0), which expression gives P(u)?
   A. F1(4,2) + F2(4,4)
   B. F2(4,2) + F3(4,4) + F4(4,-4)
   C. F3(4,2) + F4(4,4)
   D. F2(4,4) + F3(4,2)
   Answer: B. F1 multiplies P0 = 0, so P = F2 P1 + F3 P0' + F4 P1'. (Key point: Formula)

2. What is P(0.5)?
   A. (2.0, 1.5)
   B. (1.0, 1.0625)
   C. (2.0, 2.0)
   D. (3.0, 2.4375)
   Answer: C. F2 = 0.5, F3 = 0.125, F4 = -0.125 give x = 2, y = 2. (Key point: Midpoint)

3. What is P'(0.5)?
   A. (4, 4)
   B. (4, 0.25)
   C. (2, 3)
   D. (4, 3)
   Answer: D. F2' = 1.5, F3' = F4' = -0.25 give (4, 3). (Key point: Tangent at 0.5)

4. At u = 1 which blending function equals 1?
   A. F2
   B. F1
   C. F3
   D. F4
   Answer: A. At u = 1, F2 = 1 and the rest are 0, so P(1) = P1 = (4,2). (Key point: End condition)

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