Curvature Formula (𝜅)

Curvature - Definition

Curvature of a Position Scalar-Valued Function

Assume we are given a position scalar-valued function:

• 𝑓(𝑡)

The curvature of 𝑓(𝑡) denoted as ||𝑑𝑇/𝑑𝑓|| is defined as:

• $𝜅=||\frac{dT}{df}||=\frac{||\frac{dT}{dt}||}{||\frac{df}{dt}||}$

where:

• 𝑇 is the unit tangent vector, defined as:
  • $T(t)=\frac{\frac{df}{dt}}{||\frac{df}{dt}||}$

Curvature of a Position Vector-Valued Function (2D)

Assume we are given a position 2D vector-valued function:

• $f(t)=\left[\begin{array}{c}x(t)\\ y(t)\end{array}\right]$

The curvature of 𝑓(𝑡) denoted as ||𝑑𝑇/𝑑𝑓|| is defined as:

• $𝜅=||\frac{dT}{df}||=\frac{||\frac{dT}{dt}||}{||\frac{df}{dt}||}=\frac{y^{\prime }(t)x^{\prime \prime }(t)-x^{\prime }(t)y^{\prime \prime }(t)}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}$

show work:

Click here to expand...

• $\frac{df}{dt}=f^{\prime }(t)=\left[\begin{array}{c}{x}^{\prime }(t)\\ y^{\prime }(t)\end{array}\right]$
• $||\frac{df}{dt}||=||f^{\prime }(t)||=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$
• $T(t)=\frac{1}{||\frac{df}{dt}||}\frac{df}{dt}=\frac{1}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t)}}\left[\begin{array}{c}{x}^{\prime }(t)\\ y^{\prime }(t)\end{array}\right]=\left[\begin{array}{c}\frac{x^{\prime }(t)}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}}\\ \frac{y^{\prime }(t)}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}}\end{array}\right]$

• $\frac{dT}{dt}=T^{\prime }(t)=\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}\left[\begin{array}{c}{y}^{\prime }(t{)}^2{x}^{\prime \prime }(t)-x^{\prime }(t)y^{\prime }(t)y^{\prime \prime }(t)\\ x^{\prime }(t{)}^2{y}^{\prime \prime }(t)-y^{\prime }(t)x^{\prime }(t)x^{\prime \prime }(t)\end{array}\right]$

show work

List indent undo

• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{y^{\prime }(t)x^{\prime \prime }(t)-x^{\prime }(t)y^{\prime \prime }(t)}{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$

show work

• $||\frac{dT}{dt}||=||T^{\prime }(t)||=||\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}\left[\begin{array}{c}{y}^{\prime }(t{)}^2{x}^{\prime \prime }(t)-x^{\prime }(t)y^{\prime }(t)y^{\prime \prime }(t)\\ x^{\prime }(t{)}^2{y}^{\prime \prime }(t)-y^{\prime }(t)x^{\prime }(t)x^{\prime \prime }(t)\end{array}\right]||$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}||\left[\begin{array}{c}{y}^{\prime }(t{)}^2{x}^{\prime \prime }(t)-x^{\prime }(t)y^{\prime }(t)y^{\prime \prime }(t)\\ x^{\prime }(t{)}^2{y}^{\prime \prime }(t)-y^{\prime }(t)x^{\prime }(t)x^{\prime \prime }(t)\end{array}\right]||$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}\sqrt{[y^{\prime }(t{)}^2{x}^{\prime \prime }(t)-x^{\prime }(t)y^{\prime }(t)y^{\prime \prime }(t){]}^2+[x^{\prime }(t{)}^2{y}^{\prime \prime }(t)-y^{\prime }(t)x^{\prime }(t)x^{\prime \prime }(t){]}^2}$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}\sqrt{y^{\prime }(t{)}^4{x}^{\prime \prime }(t{)}^2+x^{\prime }(t{)}^2{y}^{\prime }(t{)}^2{y}^{\prime \prime }(t{)}^2-2x^{\prime }(t)y^{\prime }(t{)}^3{x}^{\prime \prime }(t)y^{\prime \prime }(t)+x^{\prime }(t{)}^4{y}^{\prime \prime }(t{)}^2+y^{\prime }(t{)}^2{x}^{\prime }(t{)}^2{x}^{\prime \prime }(t{)}^2-2x^{\prime }(t{)}^3{y}^{\prime }(t)x^{\prime \prime }(t)y^{\prime \prime }(t)}$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}\sqrt{y^{\prime }(t{)}^4{x}^{\prime \prime }(t{)}^2+x^{\prime }(t{)}^2{y}^{\prime }(t{)}^2{y}^{\prime \prime }(t{)}^2+x^{\prime }(t{)}^4{y}^{\prime \prime }(t{)}^2+y^{\prime }(t{)}^2{x}^{\prime }(t{)}^2{x}^{\prime \prime }(t{)}^2-[x^{\prime }(t{)}^2+y^{\prime }(t{)}^2]\ast [2x^{\prime }(t)y^{\prime }(t)x^{\prime \prime }(t)y^{\prime \prime }(t)]}$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}\sqrt{[x^{\prime }(t{)}^2+y^{\prime }(t{)}^2]\ast [y^{\prime }(t{)}^2{x}^{\prime \prime }(t{)}^2]+[x^{\prime }(t{)}^2+y^{\prime }(t{)}^2]\ast [x^{\prime }(t{)}^2{y}^{\prime \prime }(t{)}^2]-[x^{\prime }(t{)}^2+y^{\prime }(t{)}^2]\ast [2x^{\prime }(t)y^{\prime }(t)x^{\prime \prime }(t)y^{\prime \prime }(t)]}$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}\sqrt{[x^{\prime }(t{)}^2+y^{\prime }(t{)}^2]\ast [y^{\prime }(t{)}^2{x}^{\prime \prime }(t{)}^2+x^{\prime }(t{)}^2{y}^{\prime \prime }(t{)}^2-2x^{\prime }(t)y^{\prime }(t)x^{\prime \prime }(t)y^{\prime \prime }(t)]}$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}\sqrt{y^{\prime }(t{)}^2{x}^{\prime \prime }(t{)}^2+x^{\prime }(t{)}^2{y}^{\prime \prime }(t{)}^2-2x^{\prime }(t)y^{\prime }(t)x^{\prime \prime }(t)y^{\prime \prime }(t)}$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}\sqrt{y^{\prime }(t{)}^2{x}^{\prime \prime }(t{)}^2+x^{\prime }(t{)}^2{y}^{\prime \prime }(t{)}^2-2x^{\prime }(t)y^{\prime }(t)x^{\prime \prime }(t)y^{\prime \prime }(t)}$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}\sqrt{[y^{\prime }(t)x^{\prime \prime }(t)-x^{\prime }(t)y^{\prime \prime }(t){]}^2}$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{1}{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}[y^{\prime }(t)x^{\prime \prime }(t)-x^{\prime }(t)y^{\prime \prime }(t)]$
• $||\frac{dT}{dt}||=||T^{\prime }(t)||=\frac{y^{\prime }(t)x^{\prime \prime }(t)-x^{\prime }(t)y^{\prime \prime }(t)}{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$

• $||\frac{dT}{df}||=\frac{||\frac{dT}{dt}||}{||\frac{df}{dt}||}=\frac{\frac{y^{\prime }(t)x^{\prime \prime }(t)-x^{\prime }(t)y^{\prime \prime }(t)}{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{1/2}}=\frac{y^{\prime }(t)x^{\prime \prime }(t)-x^{\prime }(t)y^{\prime \prime }(t)}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{3/2}}$

Curvature of a Position Vector-Valued Function (3D)

Assume we are given a position 3D vector-valued function:

• $f(t)=\left[\begin{array}{c}x(t)\\ y(t)\\ z(t)\end{array}\right]$

The curvature of 𝑓(𝑡) denoted as ||𝑑𝑇/𝑑??|| is defined as:

• $𝜅=||\frac{dT}{df}||=\frac{||\frac{dT}{dt}||}{||\frac{df}{dt}||}=\frac{||f^{\prime }⨯f^{\prime \prime }||}{||f^{\prime }|{|}^3}$

Curvature of a Position Vector-Valued Function (Generalized)

Assume we are given a position vector-valued function:

• $f(t)=\left[\begin{array}{c}x(t)\\ y(t)\\ z(t)\\ ⋮\end{array}\right]$

The curvature of 𝑓(𝑡) denoted as ||𝑑𝑇/𝑑??|| is defined as:

• $𝜅=\frac{f^{\prime }(t)\wedge f^{\prime \prime }(t)}{||f^{\prime }(t)|{|}^3}$

where:

• ∧ is the wedge product
• $f^{\prime }(t)=\left[\begin{array}{c}{x}^{\prime }(t)\\ y^{\prime }(t)\\ z^{\prime }(t)\\ ⋮\end{array}\right]$

Example

2D Curvature Example - Circle

from: https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/curvature/v/curvature-formula-part-2

Consider the following position vector-valued function 𝑓(𝑡):

• $f(t)=\left[\begin{array}{c}cos(t)r\\ sin(t)r\end{array}\right]$

The derivative of 𝑓(𝑡):

• $f^{\prime }(t)=\left[\begin{array}{c}-sin(t)r\\ cos(t)r\end{array}\right]$

The unit tangent function of 𝑓(𝑡):

• $T(t)=\frac{f^{\prime }(t)}{||f^{\prime }(t)||}$
  • $||f^{\prime }(t)||=||\left[\begin{array}{c}-sin(t)r\\ cos(t)r\end{array}\right]||$
  • $||f^{\prime }(t)||=\sqrt{si{n}^2(t)r^2+co{s}^2(t)r^2}$
  • $||f^{\prime }(t)||=r\sqrt{si{n}^2(t)+co{s}^2(t)}$
  • $||f^{\prime }(t)||=r$
• $T(t)=\frac{1}{r}{f}^{\prime }(t)$
• $T(t)=\frac{1}{r}\left[\begin{array}{c}-sin(t)r\\ cos(t)r\end{array}\right]$
• $T(t)=\left[\begin{array}{c}-sin(t)\\ cos(t)\end{array}\right]$

Next, find the derivative of 𝑇 over 𝑡:

• $\frac{dT}{dt}=\left[\begin{array}{c}-cos(t)\\ -sin(t)\end{array}\right]$

Next, solve for:

• $||\frac{dT}{df}||=\frac{||\frac{dT}{dt}||}{||\frac{df}{dt}||}$
  • $||\frac{dT}{dt}||=\sqrt{co{s}^2(t)+si{n}^2(t)}$
  • $||\frac{dT}{dt}||=1$
  • $||\frac{df}{dt}||=||f^{\prime }(t)||=r$
• $||\frac{dT}{df}||=\frac{1}{r}$

Resources

• Khan Academy
• YouTube’s Khan Academy’s Curvature videos