Mixing
Explanation on the steps of this total arc-length solution
$begingroup$
Can someone please explain to me what rules have been used to calculate ${bf{dot{x}}}$ and $|{bf{dot{x}}}|$ in the definition for total arc-length of this problem. I've tried to calculate this by expanding everything out, but get lost in the workings. Has the identity $cos^2t+sin^2t=1$ been used to eliminate $cos$ and $sin$?
Example: Calculate the total length of the following curve.
The logarithmic spiral, $x=e^{−ωt}$ $(Rcos2πt, Rsin2πt)$ with ω > 0 and $R > 0$ constants and I = [0,∞).
Solution:
The total length is
begin{align}
L &=int_0^∞
Re^{−ωt}(ω^2 + 4π^2)^{1/2} dt\
&=
frac{R}{ω} (ω^2 + 4π^2)^{1/2}.
end{align}
parametric arc-length
$endgroup$
add a comment |
$begingroup$
Can someone please explain to me what rules have been used to calculate ${bf{dot{x}}}$ and $|{bf{dot{x}}}|$ in the definition for total arc-length of this problem. I've tried to calculate this by expanding everything out, but get lost in the workings. Has the identity $cos^2t+sin^2t=1$ been used to eliminate $cos$ and $sin$?
Example: Calculate the total length of the following curve.
The logarithmic spiral, $x=e^{−ωt}$ $(Rcos2πt, Rsin2πt)$ with ω > 0 and $R > 0$ constants and I = [0,∞).
Solution:
The total length is
begin{align}
L &=int_0^∞
Re^{−ωt}(ω^2 + 4π^2)^{1/2} dt\
&=
frac{R}{ω} (ω^2 + 4π^2)^{1/2}.
end{align}
parametric arc-length
$endgroup$
add a comment |
$begingroup$
Can someone please explain to me what rules have been used to calculate ${bf{dot{x}}}$ and $|{bf{dot{x}}}|$ in the definition for total arc-length of this problem. I've tried to calculate this by expanding everything out, but get lost in the workings. Has the identity $cos^2t+sin^2t=1$ been used to eliminate $cos$ and $sin$?
Example: Calculate the total length of the following curve.
The logarithmic spiral, $x=e^{−ωt}$ $(Rcos2πt, Rsin2πt)$ with ω > 0 and $R > 0$ constants and I = [0,∞).
Solution:
The total length is
begin{align}
L &=int_0^∞
Re^{−ωt}(ω^2 + 4π^2)^{1/2} dt\
&=
frac{R}{ω} (ω^2 + 4π^2)^{1/2}.
end{align}
parametric arc-length
$endgroup$
Can someone please explain to me what rules have been used to calculate ${bf{dot{x}}}$ and $|{bf{dot{x}}}|$ in the definition for total arc-length of this problem. I've tried to calculate this by expanding everything out, but get lost in the workings. Has the identity $cos^2t+sin^2t=1$ been used to eliminate $cos$ and $sin$?
Example: Calculate the total length of the following curve.
The logarithmic spiral, $x=e^{−ωt}$ $(Rcos2πt, Rsin2πt)$ with ω > 0 and $R > 0$ constants and I = [0,∞).
Solution:
The total length is
begin{align}
L &=int_0^∞
Re^{−ωt}(ω^2 + 4π^2)^{1/2} dt\
&=
frac{R}{ω} (ω^2 + 4π^2)^{1/2}.
end{align}
parametric arc-length
parametric arc-length
edited Jan 11 at 1:49
asked Jan 11 at 1:43
user606466
add a comment |
add a comment |
1 Answer
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$begingroup$
$L = int |frac {dmathbf x}{dt}| dt$
$frac {d}{dt} mathbf x = -omega e^{-omega t}(Rcos 2pi t,Rsin 2pi t) + e^{-omega t}(-2pi Rsin 2pi t,2pi Rcos 2pi t)$
The two terms are orthogonal to one another.
$|frac {dmathbf x}{dt}|^2 = |-omega R e^{-omega t}(cos 2pi t,sin 2pi t)|^2 + |2pi Re^{-omega t}(-sin 2pi t,cos 2pi t)|^2$
$|frac {dmathbf x}{dt}| =Re^{-omega t}sqrt{omega^2+(2pi)^2} $
answered Jan 11 at 1:56
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
Thank you. That cleared my confusion :)
$endgroup$
– user606466
Jan 11 at 2:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$L = int |frac {dmathbf x}{dt}| dt$
$frac {d}{dt} mathbf x = -omega e^{-omega t}(Rcos 2pi t,Rsin 2pi t) + e^{-omega t}(-2pi Rsin 2pi t,2pi Rcos 2pi t)$
The two terms are orthogonal to one another.
$|frac {dmathbf x}{dt}|^2 = |-omega R e^{-omega t}(cos 2pi t,sin 2pi t)|^2 + |2pi Re^{-omega t}(-sin 2pi t,cos 2pi t)|^2$
$|frac {dmathbf x}{dt}| =Re^{-omega t}sqrt{omega^2+(2pi)^2} $
answered Jan 11 at 1:56
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
Thank you. That cleared my confusion :)
$endgroup$
– user606466
Jan 11 at 2:10
add a comment |
$begingroup$
$L = int |frac {dmathbf x}{dt}| dt$
$frac {d}{dt} mathbf x = -omega e^{-omega t}(Rcos 2pi t,Rsin 2pi t) + e^{-omega t}(-2pi Rsin 2pi t,2pi Rcos 2pi t)$
The two terms are orthogonal to one another.
$|frac {dmathbf x}{dt}|^2 = |-omega R e^{-omega t}(cos 2pi t,sin 2pi t)|^2 + |2pi Re^{-omega t}(-sin 2pi t,cos 2pi t)|^2$
$|frac {dmathbf x}{dt}| =Re^{-omega t}sqrt{omega^2+(2pi)^2} $
answered Jan 11 at 1:56
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
Thank you. That cleared my confusion :)
$endgroup$
– user606466
Jan 11 at 2:10
add a comment |
$begingroup$
$L = int |frac {dmathbf x}{dt}| dt$
$frac {d}{dt} mathbf x = -omega e^{-omega t}(Rcos 2pi t,Rsin 2pi t) + e^{-omega t}(-2pi Rsin 2pi t,2pi Rcos 2pi t)$
The two terms are orthogonal to one another.
$|frac {dmathbf x}{dt}|^2 = |-omega R e^{-omega t}(cos 2pi t,sin 2pi t)|^2 + |2pi Re^{-omega t}(-sin 2pi t,cos 2pi t)|^2$
$|frac {dmathbf x}{dt}| =Re^{-omega t}sqrt{omega^2+(2pi)^2} $
answered Jan 11 at 1:56
Doug MDoug M
44.8k31854
$endgroup$
$L = int |frac {dmathbf x}{dt}| dt$
$frac {d}{dt} mathbf x = -omega e^{-omega t}(Rcos 2pi t,Rsin 2pi t) + e^{-omega t}(-2pi Rsin 2pi t,2pi Rcos 2pi t)$
The two terms are orthogonal to one another.
$|frac {dmathbf x}{dt}|^2 = |-omega R e^{-omega t}(cos 2pi t,sin 2pi t)|^2 + |2pi Re^{-omega t}(-sin 2pi t,cos 2pi t)|^2$
$|frac {dmathbf x}{dt}| =Re^{-omega t}sqrt{omega^2+(2pi)^2} $
answered Jan 11 at 1:56
Doug MDoug M
44.8k31854
answered Jan 11 at 1:56
Doug MDoug M
44.8k31854
answered Jan 11 at 1:56
Doug MDoug M
44.8k31854
answered Jan 11 at 1:56
Doug MDoug M
44.8k31854
44.8k31854
$begingroup$
Thank you. That cleared my confusion :)
$endgroup$
– user606466
Jan 11 at 2:10
add a comment |
$begingroup$
Thank you. That cleared my confusion :)
$endgroup$
– user606466
Jan 11 at 2:10
$begingroup$
Thank you. That cleared my confusion :)
$endgroup$
– user606466
Jan 11 at 2:10
$begingroup$
Thank you. That cleared my confusion :)
$endgroup$
– user606466
Jan 11 at 2:10
add a comment |
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