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Chapter 12: Problem 37

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

Short Answer

Expert verified
Answer: To find the tangential and normal components of the acceleration, you must perform several steps involving differentiation and vector operations. Firstly, calculate the velocity vector by differentiating the trajectory function. Next, calculate the acceleration vector by differentiating the velocity vector. Then, calculate the unit tangent vector by normalizing the velocity vector. The tangential component of acceleration can then be found by taking the dot product of the acceleration and unit tangent vectors. Finally, the normal component of acceleration can be obtained by first finding the cross product of the acceleration and unit tangent vectors and then calculating the magnitude of this resulting vector.

Step by step solution

01

Calculate the velocity vector

To calculate the velocity vector, take the derivative of the trajectory function \(\mathbf{r}(t)\) with respect to time \(t\). So, $$\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=\left\langle \frac{de^{t}\cos t}{dt},\frac{de^{t}\sin t}{dt},\frac{de^{t}}{dt}\right\rangle$$ Using the chain rule, we get: $$\mathbf{v}(t)=\left\langle e^t(\cos t - t\sin t), e^t(\sin t + t\cos t), e^t\right\rangle$$
02

Calculate the acceleration vector

To calculate the acceleration vector, take the derivative of the velocity vector \(\mathbf{v}(t)\) with respect to time \(t\). So, $$\mathbf{a}(t)=\frac{d\mathbf{v}}{dt}=\left\langle \frac{d(e^t(\cos t - t\sin t))}{dt},\frac{d(e^t(\sin t + t\cos t))}{dt},\frac{de^{t}}{dt}\right\rangle$$ Using the chain and product rules, we get: $$\mathbf{a}(t)=\left\langle e^t(-2\cos t-2t\sin t), e^t(2\sin t- 2t\cos t), e^t\right\rangle$$
03

Calculate the unit tangent vector

To calculate the unit tangent vector, normalize the velocity vector \(\mathbf{v}(t)\). We first need to find the magnitude of \(\mathbf{v}(t)\): $$|\mathbf{v}(t)|=\sqrt{(e^t(\cos t - t\sin t))^2+(e^t(\sin t + t\cos t))^2+(e^t)^2}$$ Then, normalize \(\mathbf{v}(t)\): $$\mathbf{T}(t)=\frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}=\left\langle \frac{e^t(\cos t - t\sin t)}{|\mathbf{v}(t)|}, \frac{e^t(\sin t + t\cos t)}{|\mathbf{v}(t)|}, \frac{e^t}{|\mathbf{v}(t)|}\right\rangle$$
04

Find the tangential component of the acceleration

To find the tangential component of the acceleration, take the dot product of the acceleration vector \(\mathbf{a}(t)\) and the unit tangent vector \(\mathbf{T}(t)\): $$a_{T}(t)=\mathbf{a}(t)\cdot\mathbf{T}(t)=\frac{e^t(-2\cos t-2t\sin t)e^t(\cos t - t\sin t)+ e^t(2\sin t- 2t\cos t)e^t(\sin t + t\cos t)+e^te^t}{|\mathbf{v}(t)|}$$
05

Find the normal component of the acceleration

To find the normal component of the acceleration, first calculate the cross product of the acceleration vector \(\mathbf{a}(t)\) and the unit tangent vector \(\mathbf{T}(t)\). Then calculate the magnitude of the resulting vector. \(\mathbf{A}=\mathbf{a}(t)\times\mathbf{T}(t)=\left\langle e^t(-2\sin(t)-2t\cos(t)),-e^t(-2\cos(t)+2t\sin(t)), 0 \right\rangle\) \(a_{N}(t)=|\mathbf{A}|=\sqrt{ (e^t(-2\sin(t)-2t\cos(t)))^2+(-e^t(-2\cos(t)+2t\sin(t)))^2} \) Now you have found the tangential and normal components of the acceleration for the given trajectory.

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