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Chapter 12: Problem 40

Consider the following trajectories of moving objects. Find the tangential and normal components of the acceleration. $$\mathbf{r}(t)=\langle 20 \cos t, 20 \sin t, 30 t\rangle$$

Short Answer

Expert verified
Based on the given trajectory, the tangential component of acceleration is 0, while the normal component of acceleration is 20.

Step by step solution

01

Find the Velocity Vector Function \(\textbf{v}(t)\)

To find the velocity vector function, we need to differentiate \(\textbf{r}(t)\) with respect to time \(t\). So, \(\displaystyle \textbf{v}(t)=\dfrac{d\textbf{r}(t)}{dt}\). Differentiating \(\textbf{r}(t)=\left\langle 20\cos t,\ 20\sin t,\ 30t\right\rangle\), we get: $$ \textbf{v}(t) = \left\langle \frac{d}{dt}(20\cos t), \frac{d}{dt}(20\sin t), \frac{d}{dt}(30t)\right\rangle = \left\langle -20\sin t, 20\cos t, 30\right\rangle $$
02

Find the Acceleration Vector Function \(\textbf{a}(t)\)

To find the acceleration vector function, we differentiate the velocity vector function with respect to time \(t\). Hence, \(\displaystyle \textbf{a}(t)=\dfrac{d\textbf{v}(t)}{dt}\). Differentiating \(\textbf{v}(t) = \left\langle -20\sin t, 20\cos t, 30\right\rangle\), we get: $$ \textbf{a}(t) = \left\langle \frac{d}{dt}(-20\sin t), \frac{d}{dt}(20\cos t), \frac{d}{dt}(30) \right\rangle = \left\langle -20\cos t, -20\sin t, 0\right\rangle $$
03

Find the Unit Tangent Vector \(\textbf{T}(t)\)

The unit tangent vector \(\textbf{T}(t)\) is calculated by normalizing the velocity vector, which means dividing the velocity vector by its magnitude. First, find the magnitude of \(\textbf{v}(t)\): $$ \|\textbf{v}(t)\| = \sqrt{(-20\sin t)^2 + (20\cos t)^2 + 30^2} = \sqrt{400(\sin^2 t + \cos^2 t) + 900} = \sqrt{1300} $$ Now, we can find the unit tangent vector \(\textbf{T}(t)\): $$ \textbf{T}(t) = \frac{\textbf{v}(t)}{\|\textbf{v}(t)\|} = \left\langle \frac{-20\sin t}{\sqrt{1300}}, \frac{20\cos t}{\sqrt{1300}}, \frac{30}{\sqrt{1300}}\right\rangle $$
04

Find the Tangential Component of Acceleration

The tangential component of acceleration is calculated by projecting the acceleration vector \(\textbf{a}(t)\) onto the unit tangent vector \(\textbf{T}(t)\). We use the dot product formula to get the tangential component of acceleration: $$ a_T = \textbf{a}(t) \cdot \textbf{T}(t) = \left\langle -20\cos t, -20\sin t, 0\right\rangle \cdot \left\langle \frac{-20\sin t}{\sqrt{1300}}, \frac{20\cos t}{\sqrt{1300}}, \frac{30}{\sqrt{1300}}\right\rangle = 0 $$
05

Find the Normal Component of Acceleration

The normal component of acceleration can be found using the following formula: $$ a_N = \sqrt{\|\textbf{a}(t)\|^2 - a_T^2} $$ First, calculate the square of the magnitude of \(\textbf{a}(t)\): $$ \|\textbf{a}(t)\|^2 = (-20\cos t)^2 + (-20\sin t)^2 + 0^2 = 400(\cos^2 t + \sin^2 t) = 400 $$ Now, find the normal component of acceleration: $$ a_N = \sqrt{400 - 0^2} = 20 $$ #Conclusion# The tangential component of acceleration \(a_T\) is 0, and the normal component of acceleration \(a_N\) is 20.

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