Cristofer Graves
2022-07-29
Answered

Consider the curve $a(t)-(2t,{t}^{2},{\mathrm{log}}_{}t)\text{on}I:t0$. Show that this curve passes through the points $p=(2,1,0)\text{and}q=(4,4,{\mathrm{log}}_{}2)$ and find its arc length between these points.

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iljovskint

Answered 2022-07-30
Author has **18** answers

Consider the equation of the curve

$a(t)=\u27e82t,{t}^{2}.\mathrm{log}t\u27e9,t\ge 0$

To show that the curve $a(t)=\u27e82t,{t}^{2}.\mathrm{log}t\u27e9$ 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 $\u27e8x(t),y(t),z(t)\u27e9$ gives

$x(t)=2t,y(t)={t}^{2},z(t)=\mathrm{log}t$

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

$a(t)=\u27e82t,{t}^{2}.\mathrm{log}t\u27e9,t\ge 0$

Find the arc lenght of curve.

From the point P = (2,1,0), we get

$(2t,{t}^{2}\mathrm{log}t)=(2,1,0)\Rightarrow 2t=2\Rightarrow t=1$

From the point Q = (4,4, log2), we get

$(2t,{t}^{2}\mathrm{log}t)=(4,4,\mathrm{log}2)\Rightarrow 2t=4\Rightarrow t=2$

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)

$a\le t\le bisL={\int}_{a}^{b}\sqrt{[{\frac{dx}{dt}}^{2}]+[{\frac{dy}{dt}}^{2}]+[{\frac{dz}{dt}}^{2}]dt}\phantom{\rule{0ex}{0ex}}$

Differentiate x and y with respect to t

$\frac{dx}{dt}=\frac{d}{dt}(2t)=2\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=\frac{d}{dt}({t}^{2})=2t\phantom{\rule{0ex}{0ex}}\frac{dz}{dt}=\frac{d}{dt}(\mathrm{log}t)=\frac{1}{t}$

The length of the curve is

$a\le t\le bisL={\int}_{1}^{2}\sqrt{[{\frac{dx}{dt}}^{2}]+[{\frac{dy}{dt}}^{2}]+[{\frac{dz}{dt}}^{2}]dt}\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{[2{]}^{2}+[2{]}^{2}=(\frac{1}{t}{)}^{2}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{4+4{t}^{2}+\frac{1}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{\frac{a{t}^{2}+4{t}^{4}+1}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{\frac{(2{t}^{2}+1{)}^{2}}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{\frac{2{t}^{2}+1}{t}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{2t+\frac{1}{t}}dt\phantom{\rule{0ex}{0ex}}=[{t}^{2}\mathrm{ln}t{]}_{1}^{2}\phantom{\rule{0ex}{0ex}}=[{2}^{2}+\mathrm{ln}2-{1}^{2}-\mathrm{ln}1]\phantom{\rule{0ex}{0ex}}=[4+\mathrm{ln}2-1-0]\phantom{\rule{0ex}{0ex}}=3+\mathrm{ln}2$

Hence the length of the curve is L=3+ln2

$a(t)=\u27e82t,{t}^{2}.\mathrm{log}t\u27e9,t\ge 0$

To show that the curve $a(t)=\u27e82t,{t}^{2}.\mathrm{log}t\u27e9$ 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 $\u27e8x(t),y(t),z(t)\u27e9$ gives

$x(t)=2t,y(t)={t}^{2},z(t)=\mathrm{log}t$

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

$a(t)=\u27e82t,{t}^{2}.\mathrm{log}t\u27e9,t\ge 0$

Find the arc lenght of curve.

From the point P = (2,1,0), we get

$(2t,{t}^{2}\mathrm{log}t)=(2,1,0)\Rightarrow 2t=2\Rightarrow t=1$

From the point Q = (4,4, log2), we get

$(2t,{t}^{2}\mathrm{log}t)=(4,4,\mathrm{log}2)\Rightarrow 2t=4\Rightarrow t=2$

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)

$a\le t\le bisL={\int}_{a}^{b}\sqrt{[{\frac{dx}{dt}}^{2}]+[{\frac{dy}{dt}}^{2}]+[{\frac{dz}{dt}}^{2}]dt}\phantom{\rule{0ex}{0ex}}$

Differentiate x and y with respect to t

$\frac{dx}{dt}=\frac{d}{dt}(2t)=2\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=\frac{d}{dt}({t}^{2})=2t\phantom{\rule{0ex}{0ex}}\frac{dz}{dt}=\frac{d}{dt}(\mathrm{log}t)=\frac{1}{t}$

The length of the curve is

$a\le t\le bisL={\int}_{1}^{2}\sqrt{[{\frac{dx}{dt}}^{2}]+[{\frac{dy}{dt}}^{2}]+[{\frac{dz}{dt}}^{2}]dt}\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{[2{]}^{2}+[2{]}^{2}=(\frac{1}{t}{)}^{2}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{4+4{t}^{2}+\frac{1}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{\frac{a{t}^{2}+4{t}^{4}+1}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{\frac{(2{t}^{2}+1{)}^{2}}{{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{\frac{2{t}^{2}+1}{t}}dt\phantom{\rule{0ex}{0ex}}={\int}_{1}^{2}\sqrt{2t+\frac{1}{t}}dt\phantom{\rule{0ex}{0ex}}=[{t}^{2}\mathrm{ln}t{]}_{1}^{2}\phantom{\rule{0ex}{0ex}}=[{2}^{2}+\mathrm{ln}2-{1}^{2}-\mathrm{ln}1]\phantom{\rule{0ex}{0ex}}=[4+\mathrm{ln}2-1-0]\phantom{\rule{0ex}{0ex}}=3+\mathrm{ln}2$

Hence the length of the curve is L=3+ln2

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