Mixing
How do I find the length of the polar spiral $r = acdot theta^2 - 1 + bcdot theta$?
-1
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How do I find the length of this polar spiral?
$$r = acdot theta^2 - 1 + bcdot theta,$$ where $a$ and $b$ are variables.
graphing-functions
edited Jan 21 at 13:00
Mohammad Zuhair Khan
1,6662625
asked Jan 21 at 12:04
user33915user33915
11
$endgroup$
add a comment |
-1
$begingroup$
How do I find the length of this polar spiral?
$$r = acdot theta^2 - 1 + bcdot theta,$$ where $a$ and $b$ are variables.
graphing-functions
edited Jan 21 at 13:00
Mohammad Zuhair Khan
1,6662625
asked Jan 21 at 12:04
user33915user33915
11
$endgroup$
2
$begingroup$
Do you know a formula for arc length using polar coordinates?
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:07
2
$begingroup$
khanacademy.org/math/integral-calculus/ic-adv-funcs/…
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 12:11
add a comment |
-1
-1
-1
$begingroup$
How do I find the length of this polar spiral?
$$r = acdot theta^2 - 1 + bcdot theta,$$ where $a$ and $b$ are variables.
graphing-functions
edited Jan 21 at 13:00
Mohammad Zuhair Khan
1,6662625
asked Jan 21 at 12:04
user33915user33915
11
$endgroup$
How do I find the length of this polar spiral?
$$r = acdot theta^2 - 1 + bcdot theta,$$ where $a$ and $b$ are variables.
graphing-functions
graphing-functions
edited Jan 21 at 13:00
Mohammad Zuhair Khan
1,6662625
asked Jan 21 at 12:04
user33915user33915
11
edited Jan 21 at 13:00
Mohammad Zuhair Khan
1,6662625
asked Jan 21 at 12:04
user33915user33915
11
edited Jan 21 at 13:00
Mohammad Zuhair Khan
1,6662625
edited Jan 21 at 13:00
Mohammad Zuhair Khan
1,6662625
edited Jan 21 at 13:00
Mohammad Zuhair Khan
1,6662625
1,6662625
asked Jan 21 at 12:04
user33915user33915
11
asked Jan 21 at 12:04
user33915user33915
11
asked Jan 21 at 12:04
user33915user33915
11
11
2
$begingroup$
Do you know a formula for arc length using polar coordinates?
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:07
2
$begingroup$
khanacademy.org/math/integral-calculus/ic-adv-funcs/…
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 12:11
add a comment |
2
$begingroup$
Do you know a formula for arc length using polar coordinates?
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:07
2
$begingroup$
khanacademy.org/math/integral-calculus/ic-adv-funcs/…
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 12:11
2
2
$begingroup$
Do you know a formula for arc length using polar coordinates?
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:07
$begingroup$
Do you know a formula for arc length using polar coordinates?
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:07
2
2
$begingroup$
khanacademy.org/math/integral-calculus/ic-adv-funcs/…
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 12:11
$begingroup$
khanacademy.org/math/integral-calculus/ic-adv-funcs/…
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 12:11
add a comment |
1 Answer
1
active
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1
$begingroup$
The formula tells us ( assuming $;0lethetale2pi;$ )
$$mathcal L=int_0^{2pi}sqrt{r^2+r_theta^2} ,dtheta=int_0^{2pi}sqrt{(atheta^2+btheta-1)^2+(2atheta+b)^2},dtheta$$
Not the nicest integral to see...but give it a try.
Pay attention to the fact that
$$r_theta^2=left(frac{dr}{dtheta}right)^2$$
answered Jan 21 at 12:11
DonAntonioDonAntonio
179k1494232
$endgroup$
$begingroup$
I've no idea. I haven't even tried to solve that integral: it looks nasty. Yet you can try to do it with Wolfram or some other program...
$endgroup$
– DonAntonio
Jan 23 at 12:50
$begingroup$
@user33915 I'm sorry but I still have no idea, yet this time it is because I can't understand what you mean: why the integral changes and all?? And how come you want to integrate with the variable ($,theta,$) being fixed?
$endgroup$
– DonAntonio
Jan 23 at 13:53
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
The formula tells us ( assuming $;0lethetale2pi;$ )
$$mathcal L=int_0^{2pi}sqrt{r^2+r_theta^2} ,dtheta=int_0^{2pi}sqrt{(atheta^2+btheta-1)^2+(2atheta+b)^2},dtheta$$
Not the nicest integral to see...but give it a try.
Pay attention to the fact that
$$r_theta^2=left(frac{dr}{dtheta}right)^2$$
answered Jan 21 at 12:11
DonAntonioDonAntonio
179k1494232
$endgroup$
$begingroup$
I've no idea. I haven't even tried to solve that integral: it looks nasty. Yet you can try to do it with Wolfram or some other program...
$endgroup$
– DonAntonio
Jan 23 at 12:50
$begingroup$
@user33915 I'm sorry but I still have no idea, yet this time it is because I can't understand what you mean: why the integral changes and all?? And how come you want to integrate with the variable ($,theta,$) being fixed?
$endgroup$
– DonAntonio
Jan 23 at 13:53
add a comment |
1
$begingroup$
The formula tells us ( assuming $;0lethetale2pi;$ )
$$mathcal L=int_0^{2pi}sqrt{r^2+r_theta^2} ,dtheta=int_0^{2pi}sqrt{(atheta^2+btheta-1)^2+(2atheta+b)^2},dtheta$$
Not the nicest integral to see...but give it a try.
Pay attention to the fact that
$$r_theta^2=left(frac{dr}{dtheta}right)^2$$
answered Jan 21 at 12:11
DonAntonioDonAntonio
179k1494232
$endgroup$
$begingroup$
I've no idea. I haven't even tried to solve that integral: it looks nasty. Yet you can try to do it with Wolfram or some other program...
$endgroup$
– DonAntonio
Jan 23 at 12:50
$begingroup$
@user33915 I'm sorry but I still have no idea, yet this time it is because I can't understand what you mean: why the integral changes and all?? And how come you want to integrate with the variable ($,theta,$) being fixed?
$endgroup$
– DonAntonio
Jan 23 at 13:53
add a comment |
1
1
1
$begingroup$
The formula tells us ( assuming $;0lethetale2pi;$ )
$$mathcal L=int_0^{2pi}sqrt{r^2+r_theta^2} ,dtheta=int_0^{2pi}sqrt{(atheta^2+btheta-1)^2+(2atheta+b)^2},dtheta$$
Not the nicest integral to see...but give it a try.
Pay attention to the fact that
$$r_theta^2=left(frac{dr}{dtheta}right)^2$$
answered Jan 21 at 12:11
DonAntonioDonAntonio
179k1494232
$endgroup$
The formula tells us ( assuming $;0lethetale2pi;$ )
$$mathcal L=int_0^{2pi}sqrt{r^2+r_theta^2} ,dtheta=int_0^{2pi}sqrt{(atheta^2+btheta-1)^2+(2atheta+b)^2},dtheta$$
Not the nicest integral to see...but give it a try.
Pay attention to the fact that
$$r_theta^2=left(frac{dr}{dtheta}right)^2$$
answered Jan 21 at 12:11
DonAntonioDonAntonio
179k1494232
answered Jan 21 at 12:11
DonAntonioDonAntonio
179k1494232
answered Jan 21 at 12:11
DonAntonioDonAntonio
179k1494232
answered Jan 21 at 12:11
DonAntonioDonAntonio
179k1494232
179k1494232
$begingroup$
I've no idea. I haven't even tried to solve that integral: it looks nasty. Yet you can try to do it with Wolfram or some other program...
$endgroup$
– DonAntonio
Jan 23 at 12:50
$begingroup$
@user33915 I'm sorry but I still have no idea, yet this time it is because I can't understand what you mean: why the integral changes and all?? And how come you want to integrate with the variable ($,theta,$) being fixed?
$endgroup$
– DonAntonio
Jan 23 at 13:53
add a comment |
$begingroup$
I've no idea. I haven't even tried to solve that integral: it looks nasty. Yet you can try to do it with Wolfram or some other program...
$endgroup$
– DonAntonio
Jan 23 at 12:50
$begingroup$
@user33915 I'm sorry but I still have no idea, yet this time it is because I can't understand what you mean: why the integral changes and all?? And how come you want to integrate with the variable ($,theta,$) being fixed?
$endgroup$
– DonAntonio
Jan 23 at 13:53
$begingroup$
I've no idea. I haven't even tried to solve that integral: it looks nasty. Yet you can try to do it with Wolfram or some other program...
$endgroup$
– DonAntonio
Jan 23 at 12:50
$begingroup$
I've no idea. I haven't even tried to solve that integral: it looks nasty. Yet you can try to do it with Wolfram or some other program...
$endgroup$
– DonAntonio
Jan 23 at 12:50
$begingroup$
@user33915 I'm sorry but I still have no idea, yet this time it is because I can't understand what you mean: why the integral changes and all?? And how come you want to integrate with the variable ($,theta,$) being fixed?
$endgroup$
– DonAntonio
Jan 23 at 13:53
$begingroup$
@user33915 I'm sorry but I still have no idea, yet this time it is because I can't understand what you mean: why the integral changes and all?? And how come you want to integrate with the variable ($,theta,$) being fixed?
$endgroup$
– DonAntonio
Jan 23 at 13:53
add a comment |
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2
$begingroup$
Do you know a formula for arc length using polar coordinates?
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:07
2
$begingroup$
khanacademy.org/math/integral-calculus/ic-adv-funcs/…
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 12:11