• <blockquote id="a6oys"></blockquote>
  • <tt id="a6oys"></tt>
  • Paul's Online Notes
    Paul's Online Notes
    Home / Calculus II / 3-Dimensional Space / Arc Length with Vector Functions
    Show General Notice Show Mobile Notice Show All Notes Hide All Notes
    General Notice

    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    Paul
    August 27, 2020

    Mobile Notice
    You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

    Section 6-9 : Arc Length with Vector Functions

    In this section we’ll recast an old formula into terms of vector functions. We want to determine the length of a vector function,

    \[\vec r\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \]

    on the interval \(a \le t \le b\).

    We actually already know how to do this. Recall that we can write the vector function into the parametric form,

    \[x = f\left( t \right)\hspace{0.25in}y = g\left( t \right)\hspace{0.25in}z = h\left( t \right)\]

    Also, recall that with two dimensional parametric curves the arc length is given by,

    \[L = \int_{{\,a}}^{{\,b}}{{\sqrt {{{\left[ {f'\left( t \right)} \right]}^2} + {{\left[ {g'\left( t \right)} \right]}^2}} \,dt}}\]

    There is a natural extension of this to three dimensions. So, the length of the curve \(\vec r\left( t \right)\) on the interval \(a \le t \le b\) is,

    \[L = \int_{{\,a}}^{{\,b}}{{\sqrt {{{\left[ {f'\left( t \right)} \right]}^2} + {{\left[ {g'\left( t \right)} \right]}^2} + {{\left[ {h'\left( t \right)} \right]}^2}} \,dt}}\]

    There is a nice simplification that we can make for this. Notice that the integrand (the function we’re integrating) is nothing more than the magnitude of the tangent vector,

    \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left[ {f'\left( t \right)} \right]}^2} + {{\left[ {g'\left( t \right)} \right]}^2} + {{\left[ {h'\left( t \right)} \right]}^2}} \]

    Therefore, the arc length can be written as,

    \[L = \int_{{\,a}}^{{\,b}}{{\left\| {\vec r'\left( t \right)} \right\|\,dt}}\]
    Let’s work a quick example of this.

    Example 1 Determine the length of the curve \(\vec r\left( t \right) = \left\langle {2t,3\sin \left( {2t} \right),3\cos \left( {2t} \right)} \right\rangle \) on the interval \(0 \le t \le 2\pi \).
    Show Solution

    We will first need the tangent vector and its magnitude.

    \[\begin{align*}\vec r'\left( t \right) & = \left\langle {2,6\cos \left( {2t} \right), - 6\sin \left( {2t} \right)} \right\rangle \\ \left\| {\vec r'\left( t \right)} \right\| & = \sqrt {4 + 36{{\cos }^2}\left( {2t} \right) + 36{{\sin }^2}\left( {2t} \right)} = \sqrt {4 + 36} = 2\sqrt {10} \end{align*}\]

    The length is then,

    \[\begin{align*}L & = \int_{{\,a}}^{{\,b}}{{\left\| {\vec r'\left( t \right)} \right\|\,dt}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{2\sqrt {10} \,dt}}\\ & = 4\pi \sqrt {10} \end{align*}\]

    We need to take a quick look at another concept here. We define the arc length function as,

    \[s\left( t \right) = \int_{{\,0}}^{{\,t}}{{\left\| {\vec r'\left( u \right)} \right\|\,du}}\]

    Before we look at why this might be important let’s work a quick example.

    Example 2 Determine the arc length function for \(\vec r\left( t \right) = \left\langle {2t,3\sin \left( {2t} \right),3\cos \left( {2t} \right)} \right\rangle \).
    Show Solution

    From the previous example we know that,

    \[\left\| {\vec r'\left( t \right)} \right\| = 2\sqrt {10} \]

    The arc length function is then,

    \[s\left( t \right) = \int_{{\,0}}^{{\,t}}{{2\sqrt {10} \,du}} = \left( {2\sqrt {10} \,u} \right)_0^t = 2\sqrt {10} \,t\]

    Okay, just why would we want to do this? Well let’s take the result of the example above and solve it for \(t\).

    \[t = \frac{s}{{2\sqrt {10} }}\]

    Now, taking this and plugging it into the original vector function and we can reparametrize the function into the form, \(\vec r\left( {t\left( s \right)} \right)\). For our function this is,

    \[\vec r\left( {t\left( s \right)} \right) = \left\langle {\frac{s}{{\sqrt {10} }},3\sin \left( {\frac{s}{{\sqrt {10} }}} \right),3\cos \left( {\frac{s}{{\sqrt {10} }}} \right)} \right\rangle \]

    So, why would we want to do this? Well with the reparameterization we can now tell where we are on the curve after we’ve traveled a distance of \(s\) along the curve. Note as well that we will start the measurement of distance from where we are at \(t = 0\).

    Example 3 Where on the curve \(\vec r\left( t \right) = \left\langle {2t,3\sin \left( {2t} \right),3\cos \left( {2t} \right)} \right\rangle \) are we after traveling for a distance of \(\displaystyle \frac{{\pi \sqrt {10} }}{3}\)?
    Show Solution

    To determine this we need the reparameterization, which we have from above.

    \[\vec r\left( {t\left( s \right)} \right) = \left\langle {\frac{s}{{\sqrt {10} }},3\sin \left( {\frac{s}{{\sqrt {10} }}} \right),3\cos \left( {\frac{s}{{\sqrt {10} }}} \right)} \right\rangle \]

    Then, to determine where we are all that we need to do is plug in \(s = \frac{{\pi \sqrt {10} }}{3}\) into this and we’ll get our location.

    \[\vec r\left( {t\left( {\frac{{\pi \sqrt {10} }}{3}} \right)} \right) = \left\langle {\frac{\pi }{3},3\sin \left( {\frac{\pi }{3}} \right),3\cos \left( {\frac{\pi }{3}} \right)} \right\rangle = \left\langle {\frac{\pi }{3},\frac{{3\sqrt 3 }}{2},\frac{3}{2}} \right\rangle \]

    So, after traveling a distance of \(\frac{{\pi \sqrt {10} }}{3}\) along the curve we are at the point \(\left( {\frac{\pi }{3},\frac{{3\sqrt 3 }}{2},\frac{3}{2}} \right)\).

    亚洲欧美中文日韩视频 - 视频 - 在线观看 - 影视资讯 - av网 <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <文本链> <文本链> <文本链> <文本链> <文本链> <文本链>