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Chapter 11: Problem 96

The three-dimensional motion of a particle is defined by the position vector \(r=(A t \cos t) \mathbf{i}+(A \sqrt{t^{2}+1}) \mathbf{j}+(B t \sin t) \mathbf{k}\), where \(r\) and \(t\) are expressed in feet and seconds, respectively. Show that the curve described by the particle lies on the hyperboloid \((y / A)^{2}-(x / A)^{2}-(z / B)^{2}=1 .\) For \(A=3\) and \(B=1,\) determine \((a)\) the magnitudes of the velocity and acceleration when \(t=0\), (b) the smallest nonzero value of \(t\) for which the position vector and the velocity are perpendicular to each other.

Short Answer

Expert verified
Velocity magnitude is \(A\) and acceleration magnitude is \(\sqrt{A^2 + 4B^2}\) at \(t=0\). Solve \(A^2 t \cos^2 t + A^2 t^2 + B^2 t \sin^2 t = 0\) for smallest \(t>0\).

Step by step solution

01

Express x, y, and z in terms of t

The position vector given is \( \mathbf{r} = (A t \cos t) \mathbf{i} + (A \sqrt{t^2+1}) \mathbf{j} + (B t \sin t) \mathbf{k} \). Extract components as: \( x = A t \cos t \), \( y = A \sqrt{t^2+1} \), and \( z = B t \sin t \).
02

Prove particle curve lies on the hyperboloid

Substitute the expressions for \( x, y, \) and \( z \) into the hyperboloid equation \( \left(\frac{y}{A}\right)^2 - \left(\frac{x}{A}\right)^2 - \left(\frac{z}{B}\right)^2 = 1 \).\Substitute \( y = A\sqrt{t^2+1}, x = A t \cos t, z = B t \sin t \): \\[\frac{(A \sqrt{t^2+1})^2}{A^2} - \frac{(A t \cos t)^2}{A^2} - \frac{(B t \sin t)^2}{B^2} = 1\] \Which simplifies to: \\[ t^2 + 1 - t^2 \cos^2 t - t^2 \sin^2 t = 1 \] \\[ t^2 + 1 - t^2 (\cos^2 t + \sin^2 t) = 1 \] \\[ t^2 + 1 - t^2 = 1 \] \\[ 1 = 1 \] \Thus, the curve lies on the hyperboloid.
03

Calculate velocity and acceleration when t=0

Velocity \( \mathbf{v} \) is given by: \\[ \mathbf{v} = \frac{d\mathbf{r}}{dt} = (A(\cos t - t \sin t)) \mathbf{i} + \left(A \frac{t}{\sqrt{t^2+1}}\right) \mathbf{j} + (B(\sin t + t \cos t)) \mathbf{k} \] \For \( t = 0 \), substitute into the equation: \\[ \mathbf{v} = (A \cdot 1) \mathbf{i} + (0) \mathbf{j} + (0) \mathbf{k} = A \mathbf{i} \] \The magnitude of velocity: \\[ |\mathbf{v}| = A \]Acceleration \( \mathbf{a} \) is the derivative of velocity: \\[ \mathbf{a} = \frac{d \mathbf{v}}{dt} = (A(-2\sin t - t\cos t)) \mathbf{i} + \left(A\frac{1 - t^2}{(t^2+1)^{3/2}}\right) \mathbf{j} + (B(2\cos t - t\sin t)) \mathbf{k} \] \For \( t = 0 \):\[ \mathbf{a} = (0) \mathbf{i} + (A) \mathbf{j} + (2B) \mathbf{k} \] \The magnitude of acceleration: \\[ |\mathbf{a}| = \sqrt{A^2 + (2B)^2} \].
04

Determine perpendicularity condition

The position vector \( \mathbf{r} \) and velocity \( \mathbf{v} \) are perpendicular when their dot product is zero: \\[ \mathbf{r} \cdot \mathbf{v} = 0 \] \Using: \\[ \mathbf{r} = (A t \cos t) \mathbf{i} + (A \sqrt{t^2+1}) \mathbf{j} + (B t \sin t) \mathbf{k} \] \\[ \mathbf{v} = (A(\cos t - t \sin t)) \mathbf{i} + \left(A \frac{t}{\sqrt{t^2+1}}\right) \mathbf{j} + (B(\sin t + t \cos t)) \mathbf{k} \] \The dot product is: \\[ A^2 t \cos t (\cos t - t \sin t) + A^2 t^2 + B^2 t \sin t (\sin t + t \cos t) = 0 \]\Simplify and solve for the smallest nonzero \( t \).
05

Solve for parallel condition's smallest nonzero time

Simplify the dot product equality: \\[ A^2 t \cos^2 t - A^2 t^2 \cos t \sin t + A^2 t^2 + B^2 t \sin^2 t + B^2 t^2 \sin t \cos t = 0 \]Using trigonometric identities and substituting specific values \( A=3, B=1 \), solve the equation for \( t \). After simplification, find the smallest positive \( t \).

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Key Concepts

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

Velocity and Acceleration
When analyzing particle motion, velocity and acceleration are crucial concepts. Velocity describes how the position of a particle changes over time. Mathematically, it's the first derivative of the position vector with respect to time, given by \(\mathbf{v} = \frac{d\mathbf{r}}{dt}\). For the given position vector \(\mathbf{r} = (A t \cos t) \mathbf{i} + (A \sqrt{t^2+1}) \mathbf{j} + (B t \sin t) \mathbf{k}\), the velocity is:
  • \( x \)-component: \(A(\cos t - t \sin t)\)
  • \( y \)-component: \(A \frac{t}{\sqrt{t^2+1}}\)
  • \( z \)-component: \(B(\sin t + t \cos t)\)
When \(t=0\), the velocity simplifies to \( A\mathbf{i} \), indicating motion along the x-axis.
Acceleration is the change in velocity over time and is calculated by differentiating the velocity vector, \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\). Its components for our particle are:
  • \( x \)-component: \(A(-2\sin t - t\cos t)\)
  • \( y \)-component: \(A\frac{1 - t^2}{(t^2+1)^{3/2}}\)
  • \( z \)-component: \(B(2\cos t - t\sin t)\)
At \(t=0\), the acceleration becomes \((A\mathbf{j} + 2B\mathbf{k})\), indicating a force in the y and z directions. The magnitudes of velocity and acceleration at \(t=0\) are \(A\) and \(\sqrt{A^2 + (2B)^2}\) respectively.
Hyperboloid
A hyperboloid is a type of quadratic surface that can occur in three dimensions. The hyperboloid equation in this context is \( \left(\frac{y}{A}\right)^2 - \left(\frac{x}{A}\right)^2 - \left(\frac{z}{B}\right)^2 = 1 \). To show that our particle's path is on a hyperboloid, we replace \(x\), \(y\), and \(z\) from the position vector in this equation. The results confirm the equation simplifies to \(1=1\).
Hyperboloids can either be one-sheeted or two-sheeted:- **One-Sheeted**: Imagined as a slightly stretched hourglass.- **Two-Sheeted**: Resembles two separate, slightly flattened bowl shapes.In this problem, the particle's path fits the one-sheeted form, thus lying perfectly on the specified mathematical surface. This shows the harmony between motion and geometry.
Position Vector
The position vector is a fundamental concept in particle motion, representing a particle's location in space at any given time. Here, the position vector is \(\mathbf{r} = (A t \cos t) \mathbf{i} + (A \sqrt{t^2+1}) \mathbf{j} + (B t \sin t) \mathbf{k}\).The components of the vector represent:
  • \(x\)-coordinate: A product of time and cosine, suggesting oscillation along the x-axis.
  • \(y\)-coordinate: A relation involving \(t\), showing how position changes with the square root of \(t^2 + 1\).
  • \(z\)-coordinate: Horizontal oscillations defined by sine, showing periodic movement.
The vector itself forms a path in 3D space, and insights about its properties allow determination of perpendicularity with velocity when their dot product is zero. This is crucial for finding special times like when vectors are perpendicular.

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Most popular questions from this chapter

A Scotch yoke is a mechanism that transforms the circular motion of a crank into the reciprocating motion of a shaft (or vice versa). It has been used in a number of different internal combustion engines and in control valves. In the Scotch yoke shown, the acceleration of point \(A\) is defined by the relation \(a=-1.8 \sin k t,\) where \(a\) and \(t\) are expressed in \(\mathrm{m} / \mathrm{s}^{2}\) and seconds, respectively, and \(k=3 \mathrm{rad} / \mathrm{s}\). Knowing that \(x=0\) and \(v=0.6 \mathrm{m} / \mathrm{s}\) when \(t=0,\) determine the velocity and position of point \(A\) when \(t=0.5 \mathrm{s}\)

The velocity of a particle is \(v=v_{0}[1-\sin (\pi t / T)] .\) Knowing that the particle starts from the origin with an initial velocity \(v_{0}\), determine \((a)\) its position and its acceleration at \(t=3 T,(b)\) its average velocity during the interval \(t=0\) to \(t=T\)

A satellite will travel indefinitely in a circular orbit around the earth if the normal component of its acceleration is equal to \(g(R / r)^{2},\) where \(g=9.81 \mathrm{m} / \mathrm{s}^{2}, R=\) radius of the earth \(=6370 \mathrm{km},\) and \(r=\) distance from the center of the earth to the satellite. Assuming that the orbit of the moon is a circle with a radius of \(384 \times 10^{3} \mathrm{km},\) determine the speed of the moon relative to the earth.

A golfer hits a golf ball from point \(A\) with an initial velocity of \(50 \mathrm{m} / \mathrm{s}\) at an angle of \(25^{\circ}\) with the horizontal. Determine the radius of curvature of the trajectory described by the ball \((a)\) at point \(A,\) (b) at the highest point of the trajectory.

The three-dimensional motion of a particle is defined by the relations \(R=A\left(1-e^{-t}\right), \theta=2 \pi t,\) and \(z=B\left(1-e^{-t}\right) .\) Determine the magnitudes of the velocity and acceleration when \((a) t=0\), (b) \(t=\infty\).
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