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Chapter 9: Problem 15

\(\mathbf{v}(t)=-e^{-t}(\mathbf{i}+\mathbf{j}+\mathbf{k}), \quad|\mathbf{v}(t)|=\sqrt{3} e^{-t} ; \quad \mathbf{a}(t)=e^{-t}(\mathbf{i}+\mathbf{j}+\mathbf{k}) ; \quad \mathbf{v} \cdot \mathbf{a}=-3 e^{-2 t} ; \quad \mathbf{v} \times \mathbf{a}=\mathbf{0}, \quad|\mathbf{v} \times \mathbf{a}|=0\) \(a_{T}=-\sqrt{3} e^{-t}, \quad a_{N}=0\)

Short Answer

Expert verified
Velocity and acceleration components confirm given values; \(a_T=-\sqrt{3}e^{-t}\), \(a_N=0\).

Step by step solution

01

Understand \\mathbf{v}(t)

The velocity vector function is given as \(\mathbf{v}(t)=-e^{-t}(\mathbf{i}+\mathbf{j}+\mathbf{k})\). This indicates that at any time \(t\), the components of velocity are equal in the direction of \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), but scaled by \(-e^{-t}\).
02

Compute the Magnitude of \\mathbf{v}(t)

The magnitude of the velocity \(\mathbf{v}(t)\) is already provided: \(|\mathbf{v}(t)| = \sqrt{3} e^{-t}\). This should be computed by taking the square root of the sum of the squares of the components of \(\mathbf{v}(t)\).
03

Confirm \\mathbf{a}(t)

The acceleration vector \(\mathbf{a}(t)\) is given as \(e^{-t}(\mathbf{i}+\mathbf{j}+\mathbf{k})\). This matches the expected derivative of the velocity \(\mathbf{v}(t)\).
04

Compute Dot Product \\mathbf{v} \cdot \\mathbf{a}

We are given \(\mathbf{v} \cdot \mathbf{a}=-3 e^{-2t}\). Confirm this by manually computing \(-e^{-t}(\mathbf{i}+\mathbf{j}+\mathbf{k}) \cdot e^{-t}(\mathbf{i}+\mathbf{j}+\mathbf{k}) = -3e^{-2t}\).
05

Check \\mathbf{v} \times \\mathbf{a}

The given \(\mathbf{v} \times \mathbf{a}=\mathbf{0}\) implies that \(\mathbf{v}\) and \(\mathbf{a}\) are parallel, which is expected since they are scalar multiples of each other. Also \(|\mathbf{v} \times \mathbf{a}|=0\) confirms this condition.
06

Calculate Tangential Component \\a_{T}

We are given \(a_{T}=-\sqrt{3} e^{-t}\). This is the tangential acceleration component parallel to \(\mathbf{v}(t)\), derived from under \(|\mathbf{v}(t)|\).
07

Calculate Normal Component \\a_{N}

We are given \(a_{N}=0\), indicating that the normal component of the acceleration, which would be perpendicular to the motion, is zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity and Acceleration Vectors
Understanding velocity and acceleration vectors is crucial in vector calculus as they describe an object's motion. The velocity vector, represented as \( \mathbf{v}(t) = -e^{-t}(\mathbf{i} + \mathbf{j} + \mathbf{k}) \), shows how the object's position changes over time. Here, \( \mathbf{i}, \mathbf{j}, \text{and}\ \mathbf{k} \) are the unit vectors in the respective coordinate directions, and \( -e^{-t} \) is a scaling factor affecting the direction. The negative exponent ensures the vector's magnitude decreases over time, indicating deceleration.Similarly, the acceleration vector \( \mathbf{a}(t) = e^{-t}(\mathbf{i} + \mathbf{j} + \mathbf{k}) \) represents how the velocity changes with time. It is the derivative of the velocity vector, confirming that it is aligned but opposite in orientation and decreasing at the same exponential rate as the velocity. This aligned setup suggests no directional change, just a uniform deceleration.
Dot Product
The dot product is a crucial operation in vector calculus. It provides insights into the angle between two vectors and interprets physical phenomena like work.For two vectors \( \mathbf{v} \) and \( \mathbf{a} \), as given, the dot product is \( \mathbf{v} \cdot \mathbf{a} = -3e^{-2t} \). This computation involves multiplying corresponding vector components and summing them up:
  • \( \mathbf{v} \cdot \mathbf{a} = (-e^{-t})(e^{-t})(1+1+1) \)
  • This equals \( -3e^{-2t} \).
A negative dot product indicates that the vectors point in opposite directions initially, which correlates with the velocity and acceleration relationship in this exercise.
Cross Product
The cross product, contrastingly distinct from the dot product, gives a vector perpendicular to the plane containing the inputs.With \( \mathbf{v} \times \mathbf{a} = \mathbf{0} \), the vectors are confirmed to be parallel. In such a situation, the cross product results in a zero vector because there is no plane they jointly span:
  • If two vectors are parallel, the area of the parallelogram they would form is zero.
  • The magnitude \( |\mathbf{v} \times \mathbf{a}| = 0 \) further asserts no perpendicular vector exists.
This is always true if vectors are scalar multiples of each other, simplifying analysis in vector calculus.
Tangential and Normal Components
The tangential and normal components are vital in unpacking an object's acceleration along its path.The tangential component \( a_{T} = -\sqrt{3} e^{-t} \) represents the acceleration along the curve's path. It essentially mirrors changes in speed along the tangent:
  • Derived directly from the magnitude of \( \mathbf{v}(t) \), fluctuating with time as the component speeds up or slows down.
Conversely, \( a_{N} = 0 \) implies the normal component, measuring acceleration perpendicular to the path, is non-existent:
  • This indicates no change in the path's curvature, confirming purely linear motion.
Understanding these helps break down motion into more intuitive linear and circular components, essential in transitional systems.

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Most popular questions from this chapter

The surface is \(g(x, y, z)=x^{2}+y^{2}+z-5=0 . \nabla g=2 x \mathbf{i}+2 y \mathbf{j}+\mathbf{k}\) \(|\nabla g|=\sqrt{1+4 x^{2}+4 y^{2}} ; \quad \mathbf{n}=\frac{2 x \mathbf{i}+2 y \mathbf{j}+\mathbf{k}}{\sqrt{1+4 x^{2}+4 y^{2}}} ; \mathbf{F} \cdot \mathbf{n}=\frac{z}{\sqrt{1+4 x^{2}+4 y^{2}}}\) \(z_{x}=-2 x, z_{y}=-2 y, d S=\sqrt{1+4 x^{2}+4 y^{2}} d A .\) Using polar coordinates $$\begin{aligned} \mathrm{Flux} &=\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=\iint_{R} \frac{z}{\sqrt{1+4 x^{2}+4 y^{2}}} \sqrt{1+4 x^{2}+4 y^{2}} d A=\iint_{R}\left(5-x^{2}-y^{2}\right) d A \\ &=\int_{0}^{2 \pi} \int_{0}^{2}\left(5-r^{2}\right) r d r d \theta=\left.\int_{0}^{2 \pi}\left(\frac{5}{2} r^{2}-\frac{1}{4} r^{4}\right)\right|_{0} ^{2} d \theta=\int_{0}^{2 \pi} 6 d \theta=12 \pi \end{aligned}$$

\(\operatorname{div} \mathbf{F}=3 x^{2}+3 y^{2}+3 z^{2} .\) Using spherical coordinates $$\begin{aligned}\iint_{S} \mathbf{F} \cdot \mathbf{n} d S &=\iiint_{D} 3\left(x^{2}+y^{2}+z^{2}\right) d V=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{a} 3 \rho^{2} \rho^{2} \sin \phi d \rho d \phi d \theta \\\&=\left.\int_{0}^{2 \pi} \int_{0}^{\pi} \frac{3}{5} \rho^{5} \sin \phi\right|_{0} ^{a} d \phi d \theta=\frac{3 a^{5}}{5} \int_{0}^{2 \pi} \int_{0}^{\pi} \sin \phi d \phi d \theta \\\&=\frac{3 a^{5}}{5} \int_{0}^{2 \pi}-\left.\cos \phi\right|_{0} ^{\pi} d \theta=\frac{6 a^{5}}{5} \int_{0}^{2 \pi} d \theta=\frac{12 \pi a^{5}}{5}\end{aligned}$$

(a) Using symmetry, \(I_{x}=4 \int_{0}^{a} \int_{0}^{b \sqrt{a^{2}-x^{2}} / a} y^{2} d y d x=\frac{4 b^{3}}{3 a^{3}} \int_{0}^{a}\left(a^{2}-x^{2}\right)^{3 / 2} d x\) \(x=a \sin \theta, \quad d x=a \cos \theta d \theta\) \(=\frac{4}{3} a b^{3} \int_{0}^{\pi / 2} \cos ^{4} \theta d \theta=\frac{4}{3} a b^{3} \int_{0}^{\pi / 2} \frac{1}{4}(1+\cos 2 \theta)^{2} d \theta=\frac{1}{3} a b^{3} \int_{0}^{\pi / 2}\left(1+\cos 2 \theta+\frac{1}{2}+\frac{1}{2} \cos 4 \theta\right) d \theta\) \(=\left.\frac{1}{3} a b^{3}\left(\frac{3}{2} \theta+\frac{1}{2} \sin 2 \theta+\frac{1}{8} \sin 4 \theta\right)\right|_{0} ^{\pi / 2}=\frac{a b^{3} \pi}{4}\). (b) Using symmetry, \(I_{y}=4 \int_{0}^{a} \int_{0}^{b \sqrt{a^{2}-x^{2}} / a} x^{2} d y d x=\frac{4 b}{a} \int_{0}^{a} x^{2} \sqrt{a^{2}-x^{2}} d x\) \(x=a \sin \theta, \quad d x=a \cos \theta d \theta\) \(=4 a^{3} b \int_{0}^{\pi / 2} \sin ^{2} \theta \cos ^{2} \theta d \theta=4 a^{3} b \int_{0}^{\pi / 2} \frac{1}{4}\left(1-\cos ^{2} 2 \theta\right) d \theta\) \(=a^{3} b \int_{0}^{\pi / 2}\left(1-\frac{1}{2}-\frac{1}{2} \cos 4 \theta\right) d \theta=\left.a^{3} b\left(\frac{1}{2} \theta-\frac{1}{8} \sin 4 \theta\right)\right|_{0} ^{\pi / 2}=\frac{a^{3} b \pi}{4}\). (c) Using \(m=\pi a b, R_{g}=\sqrt{I_{x} / m}=\frac{1}{2} \sqrt{a b^{3} \pi / \pi a b}=\frac{1}{2} b\) (d) \(R_{g}=\sqrt{I_{y} / m}=\frac{1}{2} \sqrt{a^{3} b \pi / \pi a b}=\frac{1}{2} a\)

Using Green's Theorem, $$\begin{aligned} W &=\oint_{C} \mathbf{F} \cdot d \mathbf{r}=\oint_{C}-y d x+x d y=\iint_{R} 2 d A=2 \int_{0}^{2 \pi} \int_{0}^{1+\cos \theta} r d r d \theta \\ &=\left.2 \int_{0}^{2 \pi}\left(\frac{1}{2} r^{2}\right)\right|_{0} ^{1+\cos \theta} d \theta=\int_{0}^{2 \pi}\left(1+2 \cos \theta+\cos ^{2} \theta\right) d \theta \\ &=\left.\left(\theta+2 \sin \theta+\frac{1}{2} \theta+\frac{1}{4} \sin 2 \theta\right)\right|_{0} ^{2 \pi}=3 \pi \end{aligned}$$

We let \(u=x+y\) and \(v=x-y\) \(R 1: x+y=1 \Longrightarrow u=1\) \(R 2: x-y=1 \Longrightarrow v=1\) \(R 3: x+y=3 \Longrightarrow u=3\) \(R 4: x-y=-1 \Longrightarrow v=-1\) \(\frac{\partial(u, v)}{\partial(x, y)}=\left|\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right|=-2 \Longrightarrow \frac{\partial(x, y)}{\partial(u, v)}=-\frac{1}{2}\) $$\begin{aligned} \iint_{R}(x+y)^{4} e^{x-y} d A &=\iint_{S} u^{4} e^{v}\left|-\frac{1}{2}\right| d A^{\prime}=\frac{1}{2} \int_{1}^{3} \int_{-1}^{1} u^{4} e^{v} d v d u=\left.\frac{1}{2} \int_{1}^{3} u^{4} e^{v}\right|_{-1} ^{1} d u \\ &=\frac{e-e^{-1}}{2} \int_{1}^{3} u^{4} d u=\left.\frac{e-e^{-1}}{10} u^{5}\right|_{1} ^{3}=\frac{242\left(e-e^{-1}\right)}{10}=\frac{121}{5}\left(e-e^{-1}\right) \end{aligned}$$
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