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Chapter 11: 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
The tangential component of the acceleration is 0, and the normal component of the acceleration is 20.

Step by step solution

01

Find the velocity vector

Differentiate the position vector \(\mathbf{r}(t)\) with respect to time to get the velocity vector \(\mathbf{v}(t)\): $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle-20\sin t, 20\cos t, 30\rangle$$
02

Find the acceleration vector

Differentiate the velocity vector \(\mathbf{v}(t)\) with respect to time to get the acceleration vector \(\mathbf{a}(t)\): $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \langle-20\cos t, -20\sin t, 0\rangle$$
03

Find the unit tangent vector

To find the unit tangent vector \(\mathbf{T}(t)\), first find the magnitude of the velocity vector: $$||\mathbf{v}(t)|| = \sqrt{(-20\sin t)^2 + (20\cos t)^2 + 30^2} = \sqrt{400} = 20$$ Now, divide the velocity vector by its magnitude to get the unit tangent vector: $$\mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||} = \langle -\frac{\sin t}{\sqrt{2}},\frac{\cos t}{\sqrt{2}},\frac{3\sqrt{2}}{2}\rangle$$
04

Find the tangential component of the acceleration

The tangential component of the acceleration is the dot product of the acceleration vector and the unit tangent vector: $$a_T(t) = \mathbf{a}(t) \cdot\mathbf{T}(t) = (-20\cos t)(-\frac{\sin t}{\sqrt{2}}) + (-20\sin t)(\frac{\cos t}{\sqrt{2}}) + (0)(\frac{3\sqrt{2}}{2}) = 0$$
05

Find the normal component of the acceleration

The normal component of the acceleration is the vector difference between the total acceleration and the tangential component: $$\mathbf{a}_N = \mathbf{a}(t) - a_T(t)\mathbf{T}(t) = \langle -20\cos t, -20\sin t, 0\rangle - 0\langle -\frac{\sin t}{\sqrt{2}},\frac{\cos t}{\sqrt{2}},\frac{3\sqrt{2}}{2}\rangle = \langle -20\cos t, -20\sin t, 0\rangle$$ The magnitude of the normal component of the acceleration is: $$||\mathbf{a}_N|| = \sqrt{(-20\cos t)^2 + (-20\sin t)^2 + 0^2} = 20$$ So, the tangential component of the acceleration is \(0\), and the normal component of the acceleration is \(20\).

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Most popular questions from this chapter

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$\begin{aligned}\mathbf{r}(t)=&\left(\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{i}+\left(-\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{j} \\\&+\left(\frac{1}{\sqrt{3}} \sin t\right) \mathbf{k} \end{aligned}$$

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Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=e^{3 t} \mathbf{i}+\frac{1}{1+t^{2}} \mathbf{j}-\frac{1}{\sqrt{2 t}} \mathbf{k}$$

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0} .\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}\). $$\mathbf{r}(t)=\left\langle e^{t}, e^{2 t}, e^{3 t}\right\rangle ; t_{0}=0$$
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