Mixing
Archimedean spiral and length
0
$begingroup$
The Archimedean spiral is given as $r=at,, a>0, , text{for}, t in [0, infty).$ I need to calculate the length of the first turn in the third quadrant. I have absolutely no idea how to compute that.
analysis differential-geometry manifolds
edited Jan 23 at 15:11
Bernard
122k741116
asked Jan 23 at 15:04
KingDingelingKingDingeling
1657
$endgroup$
add a comment |
0
$begingroup$
The Archimedean spiral is given as $r=at,, a>0, , text{for}, t in [0, infty).$ I need to calculate the length of the first turn in the third quadrant. I have absolutely no idea how to compute that.
analysis differential-geometry manifolds
edited Jan 23 at 15:11
Bernard
122k741116
asked Jan 23 at 15:04
KingDingelingKingDingeling
1657
$endgroup$
add a comment |
0
0
0
0
$begingroup$
The Archimedean spiral is given as $r=at,, a>0, , text{for}, t in [0, infty).$ I need to calculate the length of the first turn in the third quadrant. I have absolutely no idea how to compute that.
analysis differential-geometry manifolds
edited Jan 23 at 15:11
Bernard
122k741116
asked Jan 23 at 15:04
KingDingelingKingDingeling
1657
$endgroup$
The Archimedean spiral is given as $r=at,, a>0, , text{for}, t in [0, infty).$ I need to calculate the length of the first turn in the third quadrant. I have absolutely no idea how to compute that.
analysis differential-geometry manifolds
analysis differential-geometry manifolds
edited Jan 23 at 15:11
Bernard
122k741116
asked Jan 23 at 15:04
KingDingelingKingDingeling
1657
edited Jan 23 at 15:11
Bernard
122k741116
asked Jan 23 at 15:04
KingDingelingKingDingeling
1657
edited Jan 23 at 15:11
Bernard
122k741116
edited Jan 23 at 15:11
Bernard
122k741116
edited Jan 23 at 15:11
Bernard
122k741116
122k741116
asked Jan 23 at 15:04
KingDingelingKingDingeling
1657
asked Jan 23 at 15:04
KingDingelingKingDingeling
1657
asked Jan 23 at 15:04
KingDingelingKingDingeling
1657
1657
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Hint:
The formula for arc length in polar coordinates is
$$ell=int_pi^{tfrac{3pi}2}sqrt{r^2(t)+r'^2(t)},mathrm dt.$$
answered Jan 23 at 15:15
BernardBernard
122k741116
$endgroup$
$begingroup$
Thank you, I have the formula and also put in what I know. What irritates me is the "third quadrant" part...
$endgroup$
– KingDingeling
Jan 23 at 15:38
1
$begingroup$
Oh! I see. Look at my updated answer
$endgroup$
– Bernard
Jan 23 at 15:42
$begingroup$
@KingDingeling: You should put the work you've done into the question itself, so that people don't waste time duplicating your effort or telling you things that you already know. (Writing "I have set up the arc length integral [show it], but what irritates me is the 'third quadrant' part" is far more helpful to prospective answerers than "I have absolutely no idea how to compute that".)
$endgroup$
– Blue
Jan 23 at 15:49
1
$begingroup$
The radicand should be $(at)^2+a^2$.
$endgroup$
– Bernard
Jan 23 at 16:18
$begingroup$
Thank you. But what would be the formula for the first turn?
$endgroup$
– KingDingeling
Jan 23 at 16:47
|
show 1 more comment
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Hint:
The formula for arc length in polar coordinates is
$$ell=int_pi^{tfrac{3pi}2}sqrt{r^2(t)+r'^2(t)},mathrm dt.$$
answered Jan 23 at 15:15
BernardBernard
122k741116
$endgroup$
$begingroup$
Thank you, I have the formula and also put in what I know. What irritates me is the "third quadrant" part...
$endgroup$
– KingDingeling
Jan 23 at 15:38
1
$begingroup$
Oh! I see. Look at my updated answer
$endgroup$
– Bernard
Jan 23 at 15:42
$begingroup$
@KingDingeling: You should put the work you've done into the question itself, so that people don't waste time duplicating your effort or telling you things that you already know. (Writing "I have set up the arc length integral [show it], but what irritates me is the 'third quadrant' part" is far more helpful to prospective answerers than "I have absolutely no idea how to compute that".)
$endgroup$
– Blue
Jan 23 at 15:49
1
$begingroup$
The radicand should be $(at)^2+a^2$.
$endgroup$
– Bernard
Jan 23 at 16:18
$begingroup$
Thank you. But what would be the formula for the first turn?
$endgroup$
– KingDingeling
Jan 23 at 16:47
|
show 1 more comment
1
$begingroup$
Hint:
The formula for arc length in polar coordinates is
$$ell=int_pi^{tfrac{3pi}2}sqrt{r^2(t)+r'^2(t)},mathrm dt.$$
answered Jan 23 at 15:15
BernardBernard
122k741116
$endgroup$
$begingroup$
Thank you, I have the formula and also put in what I know. What irritates me is the "third quadrant" part...
$endgroup$
– KingDingeling
Jan 23 at 15:38
1
$begingroup$
Oh! I see. Look at my updated answer
$endgroup$
– Bernard
Jan 23 at 15:42
$begingroup$
@KingDingeling: You should put the work you've done into the question itself, so that people don't waste time duplicating your effort or telling you things that you already know. (Writing "I have set up the arc length integral [show it], but what irritates me is the 'third quadrant' part" is far more helpful to prospective answerers than "I have absolutely no idea how to compute that".)
$endgroup$
– Blue
Jan 23 at 15:49
1
$begingroup$
The radicand should be $(at)^2+a^2$.
$endgroup$
– Bernard
Jan 23 at 16:18
$begingroup$
Thank you. But what would be the formula for the first turn?
$endgroup$
– KingDingeling
Jan 23 at 16:47
|
show 1 more comment
1
1
1
$begingroup$
Hint:
The formula for arc length in polar coordinates is
$$ell=int_pi^{tfrac{3pi}2}sqrt{r^2(t)+r'^2(t)},mathrm dt.$$
answered Jan 23 at 15:15
BernardBernard
122k741116
$endgroup$
Hint:
The formula for arc length in polar coordinates is
$$ell=int_pi^{tfrac{3pi}2}sqrt{r^2(t)+r'^2(t)},mathrm dt.$$
answered Jan 23 at 15:15
BernardBernard
122k741116
edited Jan 23 at 15:40
answered Jan 23 at 15:15
BernardBernard
122k741116
answered Jan 23 at 15:15
BernardBernard
122k741116
answered Jan 23 at 15:15
BernardBernard
122k741116
122k741116
$begingroup$
Thank you, I have the formula and also put in what I know. What irritates me is the "third quadrant" part...
$endgroup$
– KingDingeling
Jan 23 at 15:38
1
$begingroup$
Oh! I see. Look at my updated answer
$endgroup$
– Bernard
Jan 23 at 15:42
$begingroup$
@KingDingeling: You should put the work you've done into the question itself, so that people don't waste time duplicating your effort or telling you things that you already know. (Writing "I have set up the arc length integral [show it], but what irritates me is the 'third quadrant' part" is far more helpful to prospective answerers than "I have absolutely no idea how to compute that".)
$endgroup$
– Blue
Jan 23 at 15:49
1
$begingroup$
The radicand should be $(at)^2+a^2$.
$endgroup$
– Bernard
Jan 23 at 16:18
$begingroup$
Thank you. But what would be the formula for the first turn?
$endgroup$
– KingDingeling
Jan 23 at 16:47
|
show 1 more comment
$begingroup$
Thank you, I have the formula and also put in what I know. What irritates me is the "third quadrant" part...
$endgroup$
– KingDingeling
Jan 23 at 15:38
1
$begingroup$
Oh! I see. Look at my updated answer
$endgroup$
– Bernard
Jan 23 at 15:42
$begingroup$
@KingDingeling: You should put the work you've done into the question itself, so that people don't waste time duplicating your effort or telling you things that you already know. (Writing "I have set up the arc length integral [show it], but what irritates me is the 'third quadrant' part" is far more helpful to prospective answerers than "I have absolutely no idea how to compute that".)
$endgroup$
– Blue
Jan 23 at 15:49
1
$begingroup$
The radicand should be $(at)^2+a^2$.
$endgroup$
– Bernard
Jan 23 at 16:18
$begingroup$
Thank you. But what would be the formula for the first turn?
$endgroup$
– KingDingeling
Jan 23 at 16:47
$begingroup$
Thank you, I have the formula and also put in what I know. What irritates me is the "third quadrant" part...
$endgroup$
– KingDingeling
Jan 23 at 15:38
$begingroup$
Thank you, I have the formula and also put in what I know. What irritates me is the "third quadrant" part...
$endgroup$
– KingDingeling
Jan 23 at 15:38
1
1
$begingroup$
Oh! I see. Look at my updated answer
$endgroup$
– Bernard
Jan 23 at 15:42
$begingroup$
Oh! I see. Look at my updated answer
$endgroup$
– Bernard
Jan 23 at 15:42
$begingroup$
@KingDingeling: You should put the work you've done into the question itself, so that people don't waste time duplicating your effort or telling you things that you already know. (Writing "I have set up the arc length integral [show it], but what irritates me is the 'third quadrant' part" is far more helpful to prospective answerers than "I have absolutely no idea how to compute that".)
$endgroup$
– Blue
Jan 23 at 15:49
$begingroup$
@KingDingeling: You should put the work you've done into the question itself, so that people don't waste time duplicating your effort or telling you things that you already know. (Writing "I have set up the arc length integral [show it], but what irritates me is the 'third quadrant' part" is far more helpful to prospective answerers than "I have absolutely no idea how to compute that".)
$endgroup$
– Blue
Jan 23 at 15:49
1
1
$begingroup$
The radicand should be $(at)^2+a^2$.
$endgroup$
– Bernard
Jan 23 at 16:18
$begingroup$
The radicand should be $(at)^2+a^2$.
$endgroup$
– Bernard
Jan 23 at 16:18
$begingroup$
Thank you. But what would be the formula for the first turn?
$endgroup$
– KingDingeling
Jan 23 at 16:47
$begingroup$
Thank you. But what would be the formula for the first turn?
$endgroup$
– KingDingeling
Jan 23 at 16:47
|
show 1 more comment
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