 # Consider the curve a(t) - (2t, t^2, log t) on I: t > 0. Show that this curve passes through the points p = (2,1,0) and q = (4,4,log2) and find its arc length between these points. Cristofer Graves 2022-07-29 Answered
Consider the curve . Show that this curve passes through the points and find its arc length between these points.
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Consider the equation of the curve
$a\left(t\right)=⟨2t,{t}^{2}.\mathrm{log}t⟩,t\ge 0$
To show that the curve $a\left(t\right)=⟨2t,{t}^{2}.\mathrm{log}t⟩$ is passing
through the points P=(2,1,0) and Q=(4,4, log2)
Compare the equation of the curve with the general curve
equation $⟨x\left(t\right),y\left(t\right),z\left(t\right)⟩$ gives
$x\left(t\right)=2t,y\left(t\right)={t}^{2},z\left(t\right)=\mathrm{log}t$
Now substitute the point P=(2,1,0) in x, y, z gives
$2=2t,1={t}^{2},0=\mathrm{log}t$
Solving these equations gives t=1
Thus, for t=1, the curve passes throug the point P
Now substitute the point Q=(4,4, log 2) in x,y,z gives
$4=2t,4={t}^{2},\mathrm{log}2=\mathrm{log}t$
Solving these equations gives t=2
Thus, for t=2, the curve passes throught the point Q
Hence, the curve passes through the point P = (2,1,0) and Q = (4,4, log 2)
Consider the equation of the curve
$a\left(t\right)=⟨2t,{t}^{2}.\mathrm{log}t⟩,t\ge 0$
Find the arc lenght of curve.
From the point P = (2,1,0), we get
$\left(2t,{t}^{2}\mathrm{log}t\right)=\left(2,1,0\right)⇒2t=2⇒t=1$
From the point Q = (4,4, log2), we get
$\left(2t,{t}^{2}\mathrm{log}t\right)=\left(4,4,\mathrm{log}2\right)⇒2t=4⇒t=2$
Thus, the limits of t are $1\le t\le 2$
The length of the curve x=f(t), y=g(t), z=h(t)
$a\le t\le bisL={\int }_{a}^{b}\sqrt{\left[{\frac{dx}{dt}}^{2}\right]+\left[{\frac{dy}{dt}}^{2}\right]+\left[{\frac{dz}{dt}}^{2}\right]dt}\phantom{\rule{0ex}{0ex}}$
Differentiate x and y with respect to t
$\frac{dx}{dt}=\frac{d}{dt}\left(2t\right)=2\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=\frac{d}{dt}\left({t}^{2}\right)=2t\phantom{\rule{0ex}{0ex}}\frac{dz}{dt}=\frac{d}{dt}\left(\mathrm{log}t\right)=\frac{1}{t}$
The length of the curve is
$a\le t\le bisL={\int }_{1}^{2}\sqrt{\left[{\frac{dx}{dt}}^{2}\right]+\left[{\frac{dy}{dt}}^{2}\right]+\left[{\frac{dz}{dt}}^{2}\right]dt}\phantom{\rule{0ex}{0ex}}={\int }_{1}^{2}\sqrt{\left[2{\right]}^{2}+\left[2{\right]}^{2}=\left(\frac{1}{t}{\right)}^{2}}dt\phantom{\rule{0ex}{0ex}}={\int }_{1}^{2}\sqrt{4+4{t}^{2}+\frac{1}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int }_{1}^{2}\sqrt{\frac{a{t}^{2}+4{t}^{4}+1}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int }_{1}^{2}\sqrt{\frac{\left(2{t}^{2}+1{\right)}^{2}}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int }_{1}^{2}\sqrt{\frac{2{t}^{2}+1}{t}}dt\phantom{\rule{0ex}{0ex}}={\int }_{1}^{2}\sqrt{2t+\frac{1}{t}}dt\phantom{\rule{0ex}{0ex}}=\left[{t}^{2}\mathrm{ln}t{\right]}_{1}^{2}\phantom{\rule{0ex}{0ex}}=\left[{2}^{2}+\mathrm{ln}2-{1}^{2}-\mathrm{ln}1\right]\phantom{\rule{0ex}{0ex}}=\left[4+\mathrm{ln}2-1-0\right]\phantom{\rule{0ex}{0ex}}=3+\mathrm{ln}2$
Hence the length of the curve is L=3+ln2