Curvature in Rectangular Coordinates

Mathematics · FE Reference Handbook section

Mathematics

5 formulas

10 exam-style examples

~55 min

All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

When it may be easier to differentiate the function with respect to y rather than x, the notation x′ will be used for the derivative.
The radius of curvature R at any point on a curve is defined as the absolute value of the reciprocal of the curvature K at that

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates

A deflected shape is modelled by f(x) = 1x³ − 6x² + 1x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 1x^{3} - 6x^{2} + 1x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 3x^{2} - 12x + 1$
Formula
$f″(x) = 6x - 12$
Set f″ = 0
$6x = 12$
Substituting
$x = 12/6 = 2.0000$
Ordinate
$f(2.000) = 1(8.0000) - 6(4.0000) + 1(2.0000) = -14.0000$
Slope
$f'(2.000) = -11.0000$

Answer:

Point of inflection at x = 2.000, f = -14.000, slope = -11.000

Why the other options are there

x = 2.000 (used f′ = 0 root)
x = 0.500 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 2

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (2)

A deflected shape is modelled by f(x) = 4x³ − 5x² + 3x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 4x^{3} - 5x^{2} + 3x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 12x^{2} - 10x + 3$
Formula
$f″(x) = 24x - 10$
Set f″ = 0
$24x = 10$
Substituting
$x = 10/24 = 0.4167$
Ordinate
$f(0.417) = 4(0.0723) - 5(0.1736) + 3(0.4167) = 0.6713$
Slope
$f'(0.417) = 0.9167$

Answer:

Point of inflection at x = 0.417, f = 0.671, slope = 0.917

Why the other options are there

x = 0.417 (used f′ = 0 root)
x = 2.400 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 3

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (3)

A deflected shape is modelled by f(x) = 4x³ − 4x² + 7x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 4x^{3} - 4x^{2} + 7x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 12x^{2} - 8x + 7$
Formula
$f″(x) = 24x - 8$
Set f″ = 0
$24x = 8$
Substituting
$x = 8/24 = 0.3333$
Ordinate
$f(0.333) = 4(0.0370) - 4(0.1111) + 7(0.3333) = 2.0370$
Slope
$f'(0.333) = 5.6667$

Answer:

Point of inflection at x = 0.333, f = 2.037, slope = 5.667

Why the other options are there

x = 0.333 (used f′ = 0 root)
x = 3.000 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 4

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (4)

A deflected shape is modelled by f(x) = 4x³ − 4x² + 5x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 4x^{3} - 4x^{2} + 5x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 12x^{2} - 8x + 5$
Formula
$f″(x) = 24x - 8$
Set f″ = 0
$24x = 8$
Substituting
$x = 8/24 = 0.3333$
Ordinate
$f(0.333) = 4(0.0370) - 4(0.1111) + 5(0.3333) = 1.3704$
Slope
$f'(0.333) = 3.6667$

Answer:

Point of inflection at x = 0.333, f = 1.370, slope = 3.667

Why the other options are there

x = 0.333 (used f′ = 0 root)
x = 3.000 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 5

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (5)

A deflected shape is modelled by f(x) = 2x³ − 4x² + 5x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 2x^{3} - 4x^{2} + 5x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 6x^{2} - 8x + 5$
Formula
$f″(x) = 12x - 8$
Set f″ = 0
$12x = 8$
Substituting
$x = 8/12 = 0.6667$
Ordinate
$f(0.667) = 2(0.2963) - 4(0.4444) + 5(0.6667) = 2.1481$
Slope
$f'(0.667) = 2.3333$

Answer:

Point of inflection at x = 0.667, f = 2.148, slope = 2.333

Why the other options are there

x = 0.667 (used f′ = 0 root)
x = 1.500 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 6

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (6)

A deflected shape is modelled by f(x) = 3x³ − 3x² + 4x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 3x^{3} - 3x^{2} + 4x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 9x^{2} - 6x + 4$
Formula
$f″(x) = 18x - 6$
Set f″ = 0
$18x = 6$
Substituting
$x = 6/18 = 0.3333$
Ordinate
$f(0.333) = 3(0.0370) - 3(0.1111) + 4(0.3333) = 1.1111$
Slope
$f'(0.333) = 3.0000$

Answer:

Point of inflection at x = 0.333, f = 1.111, slope = 3.000

Why the other options are there

x = 0.333 (used f′ = 0 root)
x = 3.000 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 7

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (7)

A deflected shape is modelled by f(x) = 3x³ − 9x² + 5x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 3x^{3} - 9x^{2} + 5x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 9x^{2} - 18x + 5$
Formula
$f″(x) = 18x - 18$
Set f″ = 0
$18x = 18$
Substituting
$x = 18/18 = 1.0000$
Ordinate
$f(1.000) = 3(1.0000) - 9(1.0000) + 5(1.0000) = -1.0000$
Slope
$f'(1.000) = -4.0000$

Answer:

Point of inflection at x = 1.000, f = -1.000, slope = -4.000

Why the other options are there

x = 1.000 (used f′ = 0 root)
x = 1.000 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 8

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (8)

A deflected shape is modelled by f(x) = 1x³ − 4x² + 3x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 1x^{3} - 4x^{2} + 3x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 3x^{2} - 8x + 3$
Formula
$f″(x) = 6x - 8$
Set f″ = 0
$6x = 8$
Substituting
$x = 8/6 = 1.3333$
Ordinate
$f(1.333) = 1(2.3704) - 4(1.7778) + 3(1.3333) = -0.7407$
Slope
$f'(1.333) = -2.3333$

Answer:

Point of inflection at x = 1.333, f = -0.741, slope = -2.333

Why the other options are there

x = 1.333 (used f′ = 0 root)
x = 0.750 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 9

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (9)

A deflected shape is modelled by f(x) = 1x³ − 4x² + 2x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 1x^{3} - 4x^{2} + 2x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 3x^{2} - 8x + 2$
Formula
$f″(x) = 6x - 8$
Set f″ = 0
$6x = 8$
Substituting
$x = 8/6 = 1.3333$
Ordinate
$f(1.333) = 1(2.3704) - 4(1.7778) + 2(1.3333) = -2.0741$
Slope
$f'(1.333) = -3.3333$

Answer:

Point of inflection at x = 1.333, f = -2.074, slope = -3.333

Why the other options are there

x = 1.333 (used f′ = 0 root)
x = 0.750 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates

Example 10

Locating a point of inflection from the second derivative — Curvature in Rectangular Coordinates (10)

A deflected shape is modelled by f(x) = 1x³ − 7x² + 1x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 1x^{3} - 7x^{2} + 1x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 3x^{2} - 14x + 1$
Formula
$f″(x) = 6x - 14$
Set f″ = 0
$6x = 14$
Substituting
$x = 14/6 = 2.3333$
Ordinate
$f(2.333) = 1(12.7037) - 7(5.4444) + 1(2.3333) = -23.0741$
Slope
$f'(2.333) = -15.3333$

Answer:

Point of inflection at x = 2.333, f = -23.074, slope = -15.333

Why the other options are there

x = 2.333 (used f′ = 0 root)
x = 0.429 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Curvature in Rectangular Coordinates