Find the Arc length of the parametric curve
$begingroup$
$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = int_{0}^{2pi} sqrt{(6-6cost)^2+(6sint)^2}dt\
=int_{0}^{2pi} sqrt{36-72cost+36cos^2t+36sin^2t}dt \
=int_{0}^{2pi} 6 sqrt{1-2cost+cos^2t+sin^2t}dt\
=int_{0}^{2pi} 6sqrt{2-2cost}dt\
=int_{0}^{2pi} 6sqrt{2}sqrt{1-cost}dt\
= 6sqrt{2}int_{0}^{2pi}sqrt{1-cost}dt\
=6sqrt{2}int_{0}^{2pi}sqrt{1-cost}frac{sqrt{(1+cost)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sqrt{(1-cos^2t)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sint}{sqrt{(1+cost)}} dt\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6sqrt{2}int_{0}^{2pi}frac{1}{sqrt{u}}\
=-12sqrt{2}[u^frac{1}{2}]\
=0$$
But the answer is 48.
calculus
asked Dec 15 '18 at 16:41
EvianEvian
84
$endgroup$
add a comment |
$begingroup$
$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = int_{0}^{2pi} sqrt{(6-6cost)^2+(6sint)^2}dt\
=int_{0}^{2pi} sqrt{36-72cost+36cos^2t+36sin^2t}dt \
=int_{0}^{2pi} 6 sqrt{1-2cost+cos^2t+sin^2t}dt\
=int_{0}^{2pi} 6sqrt{2-2cost}dt\
=int_{0}^{2pi} 6sqrt{2}sqrt{1-cost}dt\
= 6sqrt{2}int_{0}^{2pi}sqrt{1-cost}dt\
=6sqrt{2}int_{0}^{2pi}sqrt{1-cost}frac{sqrt{(1+cost)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sqrt{(1-cos^2t)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sint}{sqrt{(1+cost)}} dt\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6sqrt{2}int_{0}^{2pi}frac{1}{sqrt{u}}\
=-12sqrt{2}[u^frac{1}{2}]\
=0$$
But the answer is 48.
calculus
asked Dec 15 '18 at 16:41
EvianEvian
84
$endgroup$
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
add a comment |
$begingroup$
$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = int_{0}^{2pi} sqrt{(6-6cost)^2+(6sint)^2}dt\
=int_{0}^{2pi} sqrt{36-72cost+36cos^2t+36sin^2t}dt \
=int_{0}^{2pi} 6 sqrt{1-2cost+cos^2t+sin^2t}dt\
=int_{0}^{2pi} 6sqrt{2-2cost}dt\
=int_{0}^{2pi} 6sqrt{2}sqrt{1-cost}dt\
= 6sqrt{2}int_{0}^{2pi}sqrt{1-cost}dt\
=6sqrt{2}int_{0}^{2pi}sqrt{1-cost}frac{sqrt{(1+cost)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sqrt{(1-cos^2t)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sint}{sqrt{(1+cost)}} dt\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6sqrt{2}int_{0}^{2pi}frac{1}{sqrt{u}}\
=-12sqrt{2}[u^frac{1}{2}]\
=0$$
But the answer is 48.
calculus
asked Dec 15 '18 at 16:41
EvianEvian
84
$endgroup$
$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = int_{0}^{2pi} sqrt{(6-6cost)^2+(6sint)^2}dt\
=int_{0}^{2pi} sqrt{36-72cost+36cos^2t+36sin^2t}dt \
=int_{0}^{2pi} 6 sqrt{1-2cost+cos^2t+sin^2t}dt\
=int_{0}^{2pi} 6sqrt{2-2cost}dt\
=int_{0}^{2pi} 6sqrt{2}sqrt{1-cost}dt\
= 6sqrt{2}int_{0}^{2pi}sqrt{1-cost}dt\
=6sqrt{2}int_{0}^{2pi}sqrt{1-cost}frac{sqrt{(1+cost)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sqrt{(1-cos^2t)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sint}{sqrt{(1+cost)}} dt\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6sqrt{2}int_{0}^{2pi}frac{1}{sqrt{u}}\
=-12sqrt{2}[u^frac{1}{2}]\
=0$$
But the answer is 48.
calculus
calculus
asked Dec 15 '18 at 16:41
EvianEvian
84
asked Dec 15 '18 at 16:41
EvianEvian
84
edited Dec 15 '18 at 16:49
Evian
asked Dec 15 '18 at 16:41
EvianEvian
84
asked Dec 15 '18 at 16:41
EvianEvian
84
asked Dec 15 '18 at 16:41
EvianEvian
84
84
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
add a comment |
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
1
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
add a comment |
1 Answer
1
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oldest
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$begingroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k673226
$endgroup$
add a comment |
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$begingroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k673226
$endgroup$
add a comment |
$begingroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k673226
$endgroup$
add a comment |
$begingroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k673226
$endgroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k673226
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k673226
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k673226
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k673226
128k673226
add a comment |
add a comment |
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$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47