---
title: "Curvature where tangent is parallel to x-axis"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:curvature-y-x2-x-minus-3"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/curvature-y-x2-x-minus-3
site: BCT Engineering
---

# Curvature where tangent is parallel to x-axis

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2071 Chaitra
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/curvature-y-x2-x-minus-3
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

Find the radius of curvature of the curve y = x²(x−3) at the points where the tangent is parallel to x-axis.

## Short answer

At horizontal tangents (x=0,2), y'=0 and y''=6x–6; ρ=1/|y''| = 1/6 at both (0,0) and (2,–4).

## Answer

Curve: y = x²(x − 3) = x³ − 3x².

**Points with horizontal tangent:** dy/dx = 3x² − 6x = 3x(x − 2) = 0 gives x = 0 or x = 2. The points are (0, 0) and (2, −4).

**Radius of curvature:** ρ = (1 + y₁²)^(3/2)/y₂, with y₂ = 6x − 6. At these points y₁ = 0, so ρ = 1/y₂.

- At (0, 0): y₂ = −6, so ρ = 1/(−6) = −1/6. Magnitude 1/6 (curve bends downward there, a local maximum).
- At (2, −4): y₂ = 6, so ρ = 1/6 (local minimum).

**Answer: the radius of curvature is 1/6 at both (0, 0) and (2, −4).**

## Check yourself

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

1. Points with tangent parallel to the x-axis are found from
   A. dy/dx = 0, giving x = 0 and x = 2
   B. d2y/dx2 = 0, giving x = 1
   C. y = 0, giving x = 0 and x = 3
   D. dx/dy = 0, giving x = 2 only
   Answer: A. A horizontal tangent means slope dy/dx = 3x^2 - 6x = 0. (Key point: Horizontal tangent)

2. At a point where y1 = 0, the formula rho = (1+y1^2)^(3/2)/y2 reduces to
   A. rho = y2
   B. rho = 1/y2
   C. rho = 0
   D. rho = 1/y2^2
   Answer: B. With y1 = 0 the numerator equals 1. (Key point: Simplification)

3. y2 for y = x^3 - 3x^2 is
   A. 3x^2 - 6x
   B. 6x - 3
   C. 6x - 6
   D. 6x
   Answer: C. Differentiating 3x^2 - 6x gives 6x - 6. (Key point: Second derivative)

4. The magnitude of the radius of curvature at both points is
   A. 6
   B. 1/3
   C. 1/12
   D. 1/6
   Answer: D. y2 = -6 at x=0 and 6 at x=2, so |rho| = 1/6 at both. (Key point: Result)

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