Tangential and Normal Components of Acceleration: Calculus 3 project by Caroline Ting
Overview
This Project has been completed as part of a standard 10 weeks Calculus 3 asynchronous online course with optional WebEx sessions during Summer 2021 Semester at MassBay Community College, Wellesley Hills, MA.
Tangential and Normal Components of Acceleration
Introduction
We can find an acceleration vector when given a position vector. Similar to velocty, it can be represented in two dimensions, the unit tangent vector T(t), and the unit normal vector N(t), and have coefficients called the tangential component of acceleration and the normal component of acceleration repectively. The tangential component is denoted by \(\alpha_T\) and the normal component by \(\alpha_N\) .
Formulas
The coefficients of T(t) and N(t) are \(\alpha_T\) and \(\alpha_N\), but there are also a few seperate formulas for these two components:
\(\displaystyle\alpha_{{T}}={\mathbf{{a}}}\cdot{\mathbf{{T}}}={\frac{{\vec{{v}}\cdot\vec{{a}}}}{{{\left|{\left|{v}\right|}\right|}}}}\)
\(\displaystyle\alpha_{{N}}={\mathbf{{a}}}\cdot{\mathbf{{N}}}=\frac{{\left|{\left|\vec{{v}}\times\vec{{a}}\right|}\right|}}{{{\left|{\left|{v}\right|}\right|}}}=\sqrt{{{\left|{\left|\alpha\right|}\right|}^{2}-{\alpha_{{T}}^{{2}}}}}\)
We also arrive at the formula:
\(\displaystyle{\mathbf{{{a}}}}{\left({t}\right)}=\alpha_{{T}}{\mathbf{{T}}}{\left({t}\right)}+\alpha_{{N}}{\mathbf{{N}}}{\left({t}\right)}\)
This is also a great video to show what the components represent:
Tangential and Normal components of Acceleration | Multi-variable Calculus
Duration: 10:48
Visual representation
Example 1
A particle moves in a path defined by the vector-valued function \(\displaystyle{\mathbf{{r}}}{\left({t}\right)}={t}^{2}{\mathbf{{i}}}+{\left({2}{t}-{3}\right)}{\mathbf{{j}}}+{\left({3}{t}^{2}-{3}{t}\right)}{\mathbf{{k}}}\) where t measures time in seconds and distance is measured in feet. Find \(\alpha_T\) and \(\alpha_N\) at time t=2.
In order to find the \(\alpha_N\) and \(\alpha_T\), we will first need to calcuate v(t) and a(t). We can then use these findings to continue with the formulas.
\(\mathbf{v}(t)=\mathbf{r'}(t)=2t~\mathbf{i}+2~\mathbf{j}+(6t-3)~\mathbf{k} \\ \mathbf{a}(t)=\mathbf{v'}(t)=2~\mathbf{i}+6~\mathbf{k}\\ \alpha_T=\dfrac{\mathbf{v}\cdot \mathbf{a}}{||\mathbf{v}||}\\ =\dfrac{(2t~\mathbf{i}+2~\mathbf{j}+(6t-3)\mathbf{k})\cdot(2~\mathbf{i}+6~\mathbf{k})}{||2t~\mathbf{i}+2~\mathbf{j}+(6t-3)\mathbf{k}||}\\ =\dfrac{4t+6(6t-3)}{\sqrt{(2t)^2+2^2+(6t-3)^2}} \\ =\dfrac{40t -18}{\sqrt{40t^2-36t+13}} \\ \alpha_T(2)=\dfrac{40(2) -18}{\sqrt{40(2)^2-36(2)+13}} \\ =\dfrac{80-18}{\sqrt{160-72+13}}\\ =\dfrac{62}{\sqrt{101}}\)
\(\alpha_N=\sqrt{||\mathbf{a}||^2-\alpha_T^2} \\ =\sqrt{||2~\mathbf{i}+6~\mathbf{k}||^2-\Big(\dfrac{40t-18}{\sqrt{40t^2-36t+13}}\Big)^2}\\ =\sqrt{4+36-\dfrac{(40t-18)^2}{40t^2-36t+13}}\\ =\sqrt{\dfrac{196}{40t^2-36t+13}}\\ =\dfrac{14}{\sqrt{40t^2-36t+13}}\\ \alpha_N(2)=\dfrac{14}{\sqrt{40(2)^2-36(2)+13}}\\ =\dfrac{14}{\sqrt{160-72+13}}\\ =\dfrac{14}{\sqrt{101}}\)
Example 2
For the curve defined by \(\displaystyle{\mathbf{{r}}}{\left({t}\right)}=<{e}^{{-{t}}} \cos{{\left({t}\right)}},{e}^{t} \sin{{\left({t}\right)}}>\) . Find the unit tangent vector, unit normal vector, normal acceleration, and tangential acceleration at \(t=\dfrac{\pi}{2}\)
\(\mathbf{T}(t)=\dfrac{\mathbf{r'}(t)}{||\mathbf{r'}(t)||} =\dfrac{<-\cos(t)e^{-t}-\sin(t)e^{-t}, \sin(t)e^{-t}+\cos(t)e^{-t}>}{\sqrt{(-\cos(t)e^{-t}-\sin(t)e^{-t})^2+(\sin(t)e^{-t}+\cos(t)e^{-t})^2}}\\ \mathbf{T}\Big(\dfrac{\pi}{2}\Big) =\dfrac{<-\cos(\pi/2)e^{-\pi/2}-\sin(\pi/2)e^{-\pi/2}, \sin(\pi/2)e^{-\pi/2}+\cos(\pi/2)e^{-\pi/2}>}{\sqrt{(-\cos(\pi/2)e^{-\pi/2}-\sin(\pi/2)e^{-\pi/2})^2+(\sin(\pi/2)e^{-\pi/2}+\cos(\pi/2)e^{-\pi/2})^2}}\\ =<-0.043173624930332,0.99906758435572>\)
Because this vector is in a plane,
\(\text{if}~~ \mathbf{T}(t)=<x(t),y(t)>~~\text{then} \\ \mathbf{N}(t)=<y(t),-x(t)>\\ \text{So}~~\mathbf{N}(t)=<0.99906758435572,0.043173624930332>\)
\(\mathbf{v}(t)= <-\cos(t)e^{-t}-\sin(t)e^{-t}, \sin(t)e^{-t}+\cos(t)e^{-t}> \\ \mathbf{a}(t)= <2\sin(t)e^{-t},2\cos(t)e^{-t}>\)
\(\alpha_T=\mathbf{a}\cdot \mathbf{T}\\ =<2\sin(\pi/2)e^{-\pi/2}, 2\cos(\pi/2)e^{-\pi/2}>\cdot <-0.043173624930332,0.99906758435572>\\ =-0.017949829720122 \\ \alpha_N=\mathbf{a}\cdot \mathbf{N}\\ =<2\sin(\pi/2)e^{-\pi/2}, 2\cos(\pi/2)e^{-\pi/2}>\cdot <0.99906758435572,0.043173624930332>\\ =0.41537149236329\)
Example 3
This is also a great video walking through how to calculate the Tangential and Normal Components of Acceleration.
Conclusion
The tangential and normal compoents of acceleration are essential to visualize and describe the direction of acceleration. They can be helpful to calculate the components of acceleration at one point in time, even if the path of the particle is not uniform. Convertiting to a tangent-normal coordinate system is useful when faced with motion questions.
To further understand how we arrived at the formulas, this wiki on curvature would help.
References
Benjamin Woodruff, K. R. B. S. Multivariable Calculus. Tangential And Normal Components. https://faculty.valpo.edu/calculus3ibl/section-29.html Accessed 15 July 2021
bullcleo1. (2011, January 23). Determining the Tangential and Normal Components of Acceleration. YouTube. YouTube. https://www.youtube.com/watch?v=vuHvWYgAY3c Accessed 15 July 2021
Libretexts. (2020, December 21). 2.6: Tangential and Normal Components of Acceleration. Mathematics LibreTexts. Libretexts. https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Vector_Calculus/2%3A_Vector-Valued_Functions_and_Motion_in_Space/2.6%3A_Tangential_and_Normal_Components_of_Acceleration Accessed 15 July 2021
Libretexts. (2020, November 10). 12.5: Tangential and Normal Components of Acceleration. Mathematics LibreTexts. Libretexts. https://math.libretexts.org/Bookshelves/Calculus/Map%3A_University_Calculus_(Hass_et_al)/12%3A_Vector-Valued_Functions_and_Motion_in_Space/12.5%3A_Tangential_and_Normal_Components_of_Acceleration Accessed 15 July 2021
Tangential and Normal components of Acceleration | Multi-variable Calculus. (2020, May 17) YouTube. https://www.youtube.com/watch?v=LQNC7klG0F8 Accessed 15 July 2021