Suppose the position of a particle after t seconds is given by the following vector equation:...

Question:

Suppose the position of a particle after t seconds is given by the following vector equation: r(t) = (1 + cos 2 π t, sin2 π t)

At t = {eq}\frac{1}{4} {/eq} sec., compute each of the following vectors:

a) The velocity vector

b) The acceleration vector

c) The unit tangent vector

d) calculate the arc length for {eq}0 \leq t \leq \frac{1}{4} {/eq}

Consider the helical path described by the following vector equation:

(x, y, z) = {eq}(cos\frac{t}{2}, sin \frac{t}{2}, \frac{t}{2}) {/eq}

At t = π, the helix passes through the point {eq}(0, 1, \frac{\pi}{2}). {/eq}

a) Find a vector v that is tangent to the helix at {eq}(0, 1, \frac{\pi}{2}). {/eq}

b) Find the equation of the plane that passes through {eq}(0, 1, \frac{\pi}{2}). {/eq} and is perpendicular to {eq}\vec{v} {/eq}.

Velocity, Acceleration vectors:

Let {eq}p(t) {/eq} be a position vector then the velocity vector is defined as;

{eq}v(t) = \frac{{dp}}{{dt}} {/eq}

Let {eq}p(t) {/eq} be a position vector then acceleration vector is defined as;

{eq}a(t) = \frac{{{d^2}p}}{{d{t^2}}} {/eq}

Let {eq}p(t) {/eq} be a position vector then the speed of the particle is defined by the magnitude of the velocity vector. Such that;

The velocity vector is given by:

{eq}v(t) = \frac{{dp}}{{dt}} {/eq}

Unit Tangent Vector:

Let {eq}\textbf{r(t)} {/eq} be a differentiable vector valued function and {eq}\textbf{v(t)} {/eq} be the velocity vector. Then we define the unit tangent vector by: {eq}T(t) = \frac{{v(t)}}{{\left\| {v(t)} \right\|}} {/eq}

Unit Normal Vector:

Let {eq}\textbf{r(t)} {/eq} be a differentiable vector valued function and let {eq}\textbf{T(t)} {/eq} be the unit tangent vector. Then the unit normal vector {eq}\textbf{N(t)} {/eq} is defined by: {eq}N(t) = \frac{{T'(t)}}{{\left\| {T'(t)} \right\|}} {/eq}

Answer and Explanation:

Here, in this case, the given position vector is: {eq}\displaystyle {\text{ r(t) = }}\left( {{\text{1 + }}\cos 2\pi t{\text{, sin2}}\pi t} \right) {/eq}

Hence at the point {eq}\displaystyle t= \frac{1}{4} {/eq}

we have:

a) Velocity vector:

{eq}\eqalign{ {\left[ {v\left( t \right)} \right]_{t = \frac{1}{4}}} & = {\left[ {\frac{d}{{dt}}\left( {{\text{1 + }}\cos 2\pi t{\text{, sin2}}\pi t} \right)} \right]_{t = \frac{1}{4}}} \cr & = {\left( {\frac{d}{{dt}}\left( {{\text{1 + }}\cos 2\pi t} \right){\text{, }}\frac{d}{{dt}}{\text{sin2}}\pi t} \right)_{t = \frac{1}{4}}} \cr & = {\left( { - 2\pi \sin 2\pi t{\text{, }}2\pi {\text{cos2}}\pi t} \right)_{t = \frac{1}{4}}} \cr & = \left( { - 2\pi \sin \frac{\pi }{2}{\text{, }}2\pi {\text{cos}}\frac{\pi }{2}} \right) \cr & = \left( { - 2\pi ,0} \right) \cr} {/eq}

b) Acceleration Vector:

{eq}\eqalign{ {\left[ {a\left( t \right)} \right]_{t = \frac{1}{4}}} & = {\left[ {\frac{d}{{dt}}\left( { - 2\pi \sin 2\pi t{\text{, }}2\pi {\text{cos2}}\pi t} \right)} \right]_{t = \frac{1}{4}}} \cr & = {\left( { - 4{\pi ^2}\cos 2\pi t{\text{, - }}4{\pi ^2}{\text{sin2}}\pi t} \right)_{t = \frac{1}{4}}} \cr & = \left( { - 4{\pi ^2}\cos \frac{\pi }{2}{\text{, - }}4{\pi ^2}{\text{sin}}\frac{\pi }{2}} \right) \cr & = \left( {0{\text{, }}4{\pi ^2}} \right) \cr} {/eq}

c) Unit Tangent Vector:

{eq}\eqalign{ {\left[ {T\left( t \right)} \right]_{t = \frac{1}{4}}} & = {\left[ {\frac{{v\left( t \right)}}{{\left\| {v\left( t \right)} \right\|}}} \right]_{t = \frac{1}{4}}} \cr & = {\left[ {\frac{{\left( { - 2\pi \sin 2\pi t{\text{, }}2\pi {\text{cos2}}\pi t} \right)}}{{\left\| {\left( { - 2\pi \sin 2\pi t{\text{, }}2\pi {\text{cos2}}\pi t} \right)} \right\|}}} \right]_{t = \frac{1}{4}}} \cr & = {\left[ {\frac{{\left( { - 2\pi \sin 2\pi t{\text{, }}2\pi {\text{cos2}}\pi t} \right)}}{{2\pi }}} \right]_{t = \frac{1}{4}}} \cr & = {\left( { - \sin 2\pi t{\text{, cos2}}\pi t} \right)_{t = \frac{1}{4}}} \cr & = \left( { - \sin \frac{\pi }{2}{\text{, cos}}\frac{\pi }{2}} \right) \cr & = \left( { - 1{\text{, }}0} \right) \cr} {/eq}

(d) Arc length is given as:

{eq}\eqalign{ L& = \int\limits_{t = a}^b {\sqrt {f{{\left( t \right)}^2} + g{{\left( t \right)}^2}} dt} \cr & = \int\limits_{t = 0}^{\frac{1}{4}} {\sqrt {{{\left( {{\text{1 + }}\cos 2\pi t} \right)}^2} + {{\left( {\sin 2\pi t} \right)}^2}} dt} \cr & = \int\limits_{t = 0}^{\frac{1}{4}} {\sqrt {\left( {{\text{1 + 2}}\cos 2\pi t{\text{ + }}{{\cos }^2}2\pi t} \right) + \left( {{{\sin }^2}2\pi t} \right)} dt} \cr & = \int\limits_{t = 0}^{\frac{1}{4}} {\sqrt {\left( {{\text{2 + 2}}\cos 2\pi t} \right)} dt} \cr & = \sqrt 2 \int\limits_{t = 0}^{\frac{1}{4}} {\sqrt {\left( {{\text{1 + }}\cos 2\pi t} \right)} dt} \cr & = \sqrt 2 \int\limits_{t = 0}^{\frac{1}{4}} {\sqrt {\left( {2{{\cos }^2}\pi t} \right)} dt} \cr & = \sqrt 2 \int\limits_{t = 0}^{\frac{1}{4}} {\left( {2\cos \pi t} \right)dt} \cr & = \frac{2}{\pi } \cr} {/eq}

Here, in this case, the given position vector is: {eq}\displaystyle r\left( t \right) = \left( {cos\frac{t}{2},sin\frac{t}{2},\frac{t}{2}} \right) {/eq}

Unite Tangent vector is given as:

{eq}\eqalign{ {\left[ {T\left( t \right)} \right]_{\left( {x,y,z} \right) = \left( {0,1,\frac{\pi }{2}} \right)}}& = {\left[ {\frac{{v\left( t \right)}}{{\left\| {v\left( t \right)} \right\|}}} \right]_{\left( {x,y,z} \right) = \left( {0,1,\frac{\pi }{2}} \right)}} \cr & = {\left[ {\frac{{\frac{d}{{dt}}r\left( t \right)}}{{\left\| {\frac{d}{{dt}}r\left( t \right)} \right\|}}} \right]_{t = \pi }} \cr & = {\left[ {\frac{{\frac{d}{{dt}}\left( {cos\frac{t}{2},sin\frac{t}{2},\frac{t}{2}} \right)}}{{\left\| {\frac{d}{{dt}}\left( {cos\frac{t}{2},sin\frac{t}{2},\frac{t}{2}} \right)} \right\|}}} \right]_{t = \pi }} \cr & = {\left[ {\frac{{\left( { - \frac{1}{2}\sin \frac{t}{2},\frac{1}{2}\cos \frac{t}{2},\frac{1}{2}} \right)}}{{\left\| {\left( { - \frac{1}{2}\sin \frac{t}{2},\frac{1}{2}\cos \frac{t}{2},\frac{1}{2}} \right)} \right\|}}} \right]_{t = \pi }} \cr & = {\left[ {\frac{{\left( { - \frac{1}{2}\sin \frac{t}{2},\frac{1}{2}\cos \frac{t}{2},\frac{1}{2}} \right)}}{{\sqrt {\frac{1}{2}} }}} \right]_{t = \pi }} \cr & = \left[ {\frac{{\left( { - \frac{1}{2},0,\frac{1}{2}} \right)}}{{\sqrt {\frac{1}{2}} }}} \right] \cr & = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right) \cr} {/eq}

Unit Normal Vector is given as:

{eq}\displaystyle N\left( t \right) = \frac{{T\left( t \right)}}{{\left\| {T\left( t \right)} \right\|}} = \frac{{\left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)}}{{\left\| {\left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)} \right\|}} = \frac{{\left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)}}{1} = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right) {/eq}


Learn more about this topic:

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Vectors: Definition, Types & Examples

from Common Entrance Test (CET): Study Guide & Syllabus

Chapter 57 / Lesson 3
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