Calculate path length of solution to differential equation
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Suppose I have a parameterized path $gamma:[0,infty)rightarrowmathbb C$ with initial condition $gamma(0)=1$ and which satisfies the following:
$$gamma'(t)=frac{i-1}{sqrt{2}}frac{gamma(t)}{|gamma(t)|}$$
In other words, $gamma(t)$ is always moving at constant unit speed towards its rotation in the plane by $pi/2$. I found computationally that
$$lim_{trightarrowinfty}gamma(t)=0$$

Questions:
Is there a formula for $gamma(t)$?
Does $gamma$ reach the origin in finite time? In other words, does there exist $T<infty$ such that $lim_{trightarrow T}gamma(t)=0$?
Edit: Without loss of generality set $s=1/sqrt{2}$ so that $gamma'(t)$ has unit magnitude. Also I realize that the path length $int_0^infty|gamma'(t)|dt$ must be infinite since $|gamma'(t)|$ is constant.
differential-equations
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up vote
2
down vote
favorite
Suppose I have a parameterized path $gamma:[0,infty)rightarrowmathbb C$ with initial condition $gamma(0)=1$ and which satisfies the following:
$$gamma'(t)=frac{i-1}{sqrt{2}}frac{gamma(t)}{|gamma(t)|}$$
In other words, $gamma(t)$ is always moving at constant unit speed towards its rotation in the plane by $pi/2$. I found computationally that
$$lim_{trightarrowinfty}gamma(t)=0$$

Questions:
Is there a formula for $gamma(t)$?
Does $gamma$ reach the origin in finite time? In other words, does there exist $T<infty$ such that $lim_{trightarrow T}gamma(t)=0$?
Edit: Without loss of generality set $s=1/sqrt{2}$ so that $gamma'(t)$ has unit magnitude. Also I realize that the path length $int_0^infty|gamma'(t)|dt$ must be infinite since $|gamma'(t)|$ is constant.
differential-equations
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose I have a parameterized path $gamma:[0,infty)rightarrowmathbb C$ with initial condition $gamma(0)=1$ and which satisfies the following:
$$gamma'(t)=frac{i-1}{sqrt{2}}frac{gamma(t)}{|gamma(t)|}$$
In other words, $gamma(t)$ is always moving at constant unit speed towards its rotation in the plane by $pi/2$. I found computationally that
$$lim_{trightarrowinfty}gamma(t)=0$$

Questions:
Is there a formula for $gamma(t)$?
Does $gamma$ reach the origin in finite time? In other words, does there exist $T<infty$ such that $lim_{trightarrow T}gamma(t)=0$?
Edit: Without loss of generality set $s=1/sqrt{2}$ so that $gamma'(t)$ has unit magnitude. Also I realize that the path length $int_0^infty|gamma'(t)|dt$ must be infinite since $|gamma'(t)|$ is constant.
differential-equations
Suppose I have a parameterized path $gamma:[0,infty)rightarrowmathbb C$ with initial condition $gamma(0)=1$ and which satisfies the following:
$$gamma'(t)=frac{i-1}{sqrt{2}}frac{gamma(t)}{|gamma(t)|}$$
In other words, $gamma(t)$ is always moving at constant unit speed towards its rotation in the plane by $pi/2$. I found computationally that
$$lim_{trightarrowinfty}gamma(t)=0$$

Questions:
Is there a formula for $gamma(t)$?
Does $gamma$ reach the origin in finite time? In other words, does there exist $T<infty$ such that $lim_{trightarrow T}gamma(t)=0$?
Edit: Without loss of generality set $s=1/sqrt{2}$ so that $gamma'(t)$ has unit magnitude. Also I realize that the path length $int_0^infty|gamma'(t)|dt$ must be infinite since $|gamma'(t)|$ is constant.
differential-equations
differential-equations
edited 17 hours ago
asked 22 hours ago
M. Nestor
71710
71710
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1 Answer
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It appears that this equation can be well-addressed via a transformation to polar coordinates; in polars, we write
$gamma(t) = r(t)e^{itheta(t)}; tag 1$
then
$dot gamma = dot r e^{itheta} + r i dot theta e^{itheta}; tag 2$
also,
$vert gamma vert = r, tag 3$
whence
$dfrac{gamma}{vert gamma vert} = e^{itheta}; tag 4$
assembling all this into the given equation
$dot gamma = s(i - 1) dfrac{gamma}{vert gamma vert} tag 5$
yields
$dot r e^{itheta} + r i dot theta e^{itheta} = s(i - 1) e^{itheta}; tag 6$
we divide out $e^{itheta}$:
$dot r + ridot theta = s(i - 1) = -s + is; tag 7$
equating real and imaginary parts,
$dot r = -s, tag 8$
$ridot theta = is; tag 9$
$rdot theta = s; tag{10}$
we may easily integrate (8) 'twixt $t_0$ and $t$, assuming $r(t_0) = r_0$:
$r - r_0 = -s(t - t_0), tag{11}$
$r = r_0 - s(t - t_0); tag{12}$
we observe that the coordinate constraint $r > 0$, along with the condition $s > 0$, implies that $t$ must remain less than $r_0/s + t_0$; we shall soon find that the polar solution circumvents this limit on $t$; inserting (12) into (10):
$(r_0 - s(t - t_0))dot theta = s; tag{13}$
$dot theta = dfrac{s}{r_0 - s(t - t_0)}, tag{14}$
which is also easily integrated 'twixt $t_0$ and $t$:
$theta - theta_0 = -ln(r_0 - s(t - t_0)) + ln r_0; tag{15}$
$theta = theta_0 - ln(r_0 - s(t - t_0)) + ln r_0; tag{16}$
using (12), we may express $r$ in term of $theta$:
$theta = theta_0 - ln r + ln r_0; tag{17}$
$e^theta = e^{theta_0} dfrac{r_0}{r}; tag{18}$
$r = r_0 e^{theta_0 - theta}; tag{19}$
the above presents the general solution for $t = t_0$, $r(t_0) = r_0$, $theta(t_0) = theta_0$; we specialize to the case at hand:
$t_0 = 0, ; gamma(0) = 1 + i; tag{20}$
$r_0 = vert gamma(0) vert = sqrt 2; tag{21}$
$theta_0 = dfrac{pi}{4}; tag{22}$
from (19):
$r = sqrt 2 e^{pi/4 - theta}; tag{23}$
from (12) and (16):
$r(t) = sqrt 2 - st; tag{24}$
$theta(t) = dfrac{pi}{4} - ln(sqrt 2 - s(t - t_0)) + ln sqrt 2. tag{25}$
The path length is perhaps easiest found by returning to the expressions for $r(t)$ and $theta(t)$, in terms of which
$vert dot gamma vert = sqrt{ (dot r)^2 + r^2 (dot theta)^2} = sqrt{s^2 + s^2} = sqrt 2 s; tag{26}$
since $t$ is restricted to $[0, r_0/s]$, the length of the curve is
$displaystyle int_0^{r_0/s} vert dot gamma vert ; dt = displaystyle int_0^{r_0/s} sqrt 2 s ; dt = sqrt 2 r_0. tag{27}$
1
Thanks! I see that $r=sqrt{2}exp(pi/4-theta)$ for $thetageqpi/4$ is a good polar expression for this curve. However this curve never reaches the origin, but $r(t)=sqrt{2}-st$ on line 24 seems to imply $r=0$ for $t=sqrt{2}/s$. Does the image of $tin[0,sqrt{2}/s)$ under $gamma$ contain the entire curve?
– M. Nestor
18 hours ago
@M.Nestor: well, a quick answer which I think adequate is that the polar coordinate system isn't really defined at $r = 0$, so we should really say $t < sqrt 2 / s$. I should probably edit that part of my answer.
– Robert Lewis
18 hours ago
1
@M.Nestor: you should be able to get answers to your edited question by taking $s = 1/sqrt 2$ in mine,
– Robert Lewis
18 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
It appears that this equation can be well-addressed via a transformation to polar coordinates; in polars, we write
$gamma(t) = r(t)e^{itheta(t)}; tag 1$
then
$dot gamma = dot r e^{itheta} + r i dot theta e^{itheta}; tag 2$
also,
$vert gamma vert = r, tag 3$
whence
$dfrac{gamma}{vert gamma vert} = e^{itheta}; tag 4$
assembling all this into the given equation
$dot gamma = s(i - 1) dfrac{gamma}{vert gamma vert} tag 5$
yields
$dot r e^{itheta} + r i dot theta e^{itheta} = s(i - 1) e^{itheta}; tag 6$
we divide out $e^{itheta}$:
$dot r + ridot theta = s(i - 1) = -s + is; tag 7$
equating real and imaginary parts,
$dot r = -s, tag 8$
$ridot theta = is; tag 9$
$rdot theta = s; tag{10}$
we may easily integrate (8) 'twixt $t_0$ and $t$, assuming $r(t_0) = r_0$:
$r - r_0 = -s(t - t_0), tag{11}$
$r = r_0 - s(t - t_0); tag{12}$
we observe that the coordinate constraint $r > 0$, along with the condition $s > 0$, implies that $t$ must remain less than $r_0/s + t_0$; we shall soon find that the polar solution circumvents this limit on $t$; inserting (12) into (10):
$(r_0 - s(t - t_0))dot theta = s; tag{13}$
$dot theta = dfrac{s}{r_0 - s(t - t_0)}, tag{14}$
which is also easily integrated 'twixt $t_0$ and $t$:
$theta - theta_0 = -ln(r_0 - s(t - t_0)) + ln r_0; tag{15}$
$theta = theta_0 - ln(r_0 - s(t - t_0)) + ln r_0; tag{16}$
using (12), we may express $r$ in term of $theta$:
$theta = theta_0 - ln r + ln r_0; tag{17}$
$e^theta = e^{theta_0} dfrac{r_0}{r}; tag{18}$
$r = r_0 e^{theta_0 - theta}; tag{19}$
the above presents the general solution for $t = t_0$, $r(t_0) = r_0$, $theta(t_0) = theta_0$; we specialize to the case at hand:
$t_0 = 0, ; gamma(0) = 1 + i; tag{20}$
$r_0 = vert gamma(0) vert = sqrt 2; tag{21}$
$theta_0 = dfrac{pi}{4}; tag{22}$
from (19):
$r = sqrt 2 e^{pi/4 - theta}; tag{23}$
from (12) and (16):
$r(t) = sqrt 2 - st; tag{24}$
$theta(t) = dfrac{pi}{4} - ln(sqrt 2 - s(t - t_0)) + ln sqrt 2. tag{25}$
The path length is perhaps easiest found by returning to the expressions for $r(t)$ and $theta(t)$, in terms of which
$vert dot gamma vert = sqrt{ (dot r)^2 + r^2 (dot theta)^2} = sqrt{s^2 + s^2} = sqrt 2 s; tag{26}$
since $t$ is restricted to $[0, r_0/s]$, the length of the curve is
$displaystyle int_0^{r_0/s} vert dot gamma vert ; dt = displaystyle int_0^{r_0/s} sqrt 2 s ; dt = sqrt 2 r_0. tag{27}$
1
Thanks! I see that $r=sqrt{2}exp(pi/4-theta)$ for $thetageqpi/4$ is a good polar expression for this curve. However this curve never reaches the origin, but $r(t)=sqrt{2}-st$ on line 24 seems to imply $r=0$ for $t=sqrt{2}/s$. Does the image of $tin[0,sqrt{2}/s)$ under $gamma$ contain the entire curve?
– M. Nestor
18 hours ago
@M.Nestor: well, a quick answer which I think adequate is that the polar coordinate system isn't really defined at $r = 0$, so we should really say $t < sqrt 2 / s$. I should probably edit that part of my answer.
– Robert Lewis
18 hours ago
1
@M.Nestor: you should be able to get answers to your edited question by taking $s = 1/sqrt 2$ in mine,
– Robert Lewis
18 hours ago
add a comment |
up vote
5
down vote
It appears that this equation can be well-addressed via a transformation to polar coordinates; in polars, we write
$gamma(t) = r(t)e^{itheta(t)}; tag 1$
then
$dot gamma = dot r e^{itheta} + r i dot theta e^{itheta}; tag 2$
also,
$vert gamma vert = r, tag 3$
whence
$dfrac{gamma}{vert gamma vert} = e^{itheta}; tag 4$
assembling all this into the given equation
$dot gamma = s(i - 1) dfrac{gamma}{vert gamma vert} tag 5$
yields
$dot r e^{itheta} + r i dot theta e^{itheta} = s(i - 1) e^{itheta}; tag 6$
we divide out $e^{itheta}$:
$dot r + ridot theta = s(i - 1) = -s + is; tag 7$
equating real and imaginary parts,
$dot r = -s, tag 8$
$ridot theta = is; tag 9$
$rdot theta = s; tag{10}$
we may easily integrate (8) 'twixt $t_0$ and $t$, assuming $r(t_0) = r_0$:
$r - r_0 = -s(t - t_0), tag{11}$
$r = r_0 - s(t - t_0); tag{12}$
we observe that the coordinate constraint $r > 0$, along with the condition $s > 0$, implies that $t$ must remain less than $r_0/s + t_0$; we shall soon find that the polar solution circumvents this limit on $t$; inserting (12) into (10):
$(r_0 - s(t - t_0))dot theta = s; tag{13}$
$dot theta = dfrac{s}{r_0 - s(t - t_0)}, tag{14}$
which is also easily integrated 'twixt $t_0$ and $t$:
$theta - theta_0 = -ln(r_0 - s(t - t_0)) + ln r_0; tag{15}$
$theta = theta_0 - ln(r_0 - s(t - t_0)) + ln r_0; tag{16}$
using (12), we may express $r$ in term of $theta$:
$theta = theta_0 - ln r + ln r_0; tag{17}$
$e^theta = e^{theta_0} dfrac{r_0}{r}; tag{18}$
$r = r_0 e^{theta_0 - theta}; tag{19}$
the above presents the general solution for $t = t_0$, $r(t_0) = r_0$, $theta(t_0) = theta_0$; we specialize to the case at hand:
$t_0 = 0, ; gamma(0) = 1 + i; tag{20}$
$r_0 = vert gamma(0) vert = sqrt 2; tag{21}$
$theta_0 = dfrac{pi}{4}; tag{22}$
from (19):
$r = sqrt 2 e^{pi/4 - theta}; tag{23}$
from (12) and (16):
$r(t) = sqrt 2 - st; tag{24}$
$theta(t) = dfrac{pi}{4} - ln(sqrt 2 - s(t - t_0)) + ln sqrt 2. tag{25}$
The path length is perhaps easiest found by returning to the expressions for $r(t)$ and $theta(t)$, in terms of which
$vert dot gamma vert = sqrt{ (dot r)^2 + r^2 (dot theta)^2} = sqrt{s^2 + s^2} = sqrt 2 s; tag{26}$
since $t$ is restricted to $[0, r_0/s]$, the length of the curve is
$displaystyle int_0^{r_0/s} vert dot gamma vert ; dt = displaystyle int_0^{r_0/s} sqrt 2 s ; dt = sqrt 2 r_0. tag{27}$
1
Thanks! I see that $r=sqrt{2}exp(pi/4-theta)$ for $thetageqpi/4$ is a good polar expression for this curve. However this curve never reaches the origin, but $r(t)=sqrt{2}-st$ on line 24 seems to imply $r=0$ for $t=sqrt{2}/s$. Does the image of $tin[0,sqrt{2}/s)$ under $gamma$ contain the entire curve?
– M. Nestor
18 hours ago
@M.Nestor: well, a quick answer which I think adequate is that the polar coordinate system isn't really defined at $r = 0$, so we should really say $t < sqrt 2 / s$. I should probably edit that part of my answer.
– Robert Lewis
18 hours ago
1
@M.Nestor: you should be able to get answers to your edited question by taking $s = 1/sqrt 2$ in mine,
– Robert Lewis
18 hours ago
add a comment |
up vote
5
down vote
up vote
5
down vote
It appears that this equation can be well-addressed via a transformation to polar coordinates; in polars, we write
$gamma(t) = r(t)e^{itheta(t)}; tag 1$
then
$dot gamma = dot r e^{itheta} + r i dot theta e^{itheta}; tag 2$
also,
$vert gamma vert = r, tag 3$
whence
$dfrac{gamma}{vert gamma vert} = e^{itheta}; tag 4$
assembling all this into the given equation
$dot gamma = s(i - 1) dfrac{gamma}{vert gamma vert} tag 5$
yields
$dot r e^{itheta} + r i dot theta e^{itheta} = s(i - 1) e^{itheta}; tag 6$
we divide out $e^{itheta}$:
$dot r + ridot theta = s(i - 1) = -s + is; tag 7$
equating real and imaginary parts,
$dot r = -s, tag 8$
$ridot theta = is; tag 9$
$rdot theta = s; tag{10}$
we may easily integrate (8) 'twixt $t_0$ and $t$, assuming $r(t_0) = r_0$:
$r - r_0 = -s(t - t_0), tag{11}$
$r = r_0 - s(t - t_0); tag{12}$
we observe that the coordinate constraint $r > 0$, along with the condition $s > 0$, implies that $t$ must remain less than $r_0/s + t_0$; we shall soon find that the polar solution circumvents this limit on $t$; inserting (12) into (10):
$(r_0 - s(t - t_0))dot theta = s; tag{13}$
$dot theta = dfrac{s}{r_0 - s(t - t_0)}, tag{14}$
which is also easily integrated 'twixt $t_0$ and $t$:
$theta - theta_0 = -ln(r_0 - s(t - t_0)) + ln r_0; tag{15}$
$theta = theta_0 - ln(r_0 - s(t - t_0)) + ln r_0; tag{16}$
using (12), we may express $r$ in term of $theta$:
$theta = theta_0 - ln r + ln r_0; tag{17}$
$e^theta = e^{theta_0} dfrac{r_0}{r}; tag{18}$
$r = r_0 e^{theta_0 - theta}; tag{19}$
the above presents the general solution for $t = t_0$, $r(t_0) = r_0$, $theta(t_0) = theta_0$; we specialize to the case at hand:
$t_0 = 0, ; gamma(0) = 1 + i; tag{20}$
$r_0 = vert gamma(0) vert = sqrt 2; tag{21}$
$theta_0 = dfrac{pi}{4}; tag{22}$
from (19):
$r = sqrt 2 e^{pi/4 - theta}; tag{23}$
from (12) and (16):
$r(t) = sqrt 2 - st; tag{24}$
$theta(t) = dfrac{pi}{4} - ln(sqrt 2 - s(t - t_0)) + ln sqrt 2. tag{25}$
The path length is perhaps easiest found by returning to the expressions for $r(t)$ and $theta(t)$, in terms of which
$vert dot gamma vert = sqrt{ (dot r)^2 + r^2 (dot theta)^2} = sqrt{s^2 + s^2} = sqrt 2 s; tag{26}$
since $t$ is restricted to $[0, r_0/s]$, the length of the curve is
$displaystyle int_0^{r_0/s} vert dot gamma vert ; dt = displaystyle int_0^{r_0/s} sqrt 2 s ; dt = sqrt 2 r_0. tag{27}$
It appears that this equation can be well-addressed via a transformation to polar coordinates; in polars, we write
$gamma(t) = r(t)e^{itheta(t)}; tag 1$
then
$dot gamma = dot r e^{itheta} + r i dot theta e^{itheta}; tag 2$
also,
$vert gamma vert = r, tag 3$
whence
$dfrac{gamma}{vert gamma vert} = e^{itheta}; tag 4$
assembling all this into the given equation
$dot gamma = s(i - 1) dfrac{gamma}{vert gamma vert} tag 5$
yields
$dot r e^{itheta} + r i dot theta e^{itheta} = s(i - 1) e^{itheta}; tag 6$
we divide out $e^{itheta}$:
$dot r + ridot theta = s(i - 1) = -s + is; tag 7$
equating real and imaginary parts,
$dot r = -s, tag 8$
$ridot theta = is; tag 9$
$rdot theta = s; tag{10}$
we may easily integrate (8) 'twixt $t_0$ and $t$, assuming $r(t_0) = r_0$:
$r - r_0 = -s(t - t_0), tag{11}$
$r = r_0 - s(t - t_0); tag{12}$
we observe that the coordinate constraint $r > 0$, along with the condition $s > 0$, implies that $t$ must remain less than $r_0/s + t_0$; we shall soon find that the polar solution circumvents this limit on $t$; inserting (12) into (10):
$(r_0 - s(t - t_0))dot theta = s; tag{13}$
$dot theta = dfrac{s}{r_0 - s(t - t_0)}, tag{14}$
which is also easily integrated 'twixt $t_0$ and $t$:
$theta - theta_0 = -ln(r_0 - s(t - t_0)) + ln r_0; tag{15}$
$theta = theta_0 - ln(r_0 - s(t - t_0)) + ln r_0; tag{16}$
using (12), we may express $r$ in term of $theta$:
$theta = theta_0 - ln r + ln r_0; tag{17}$
$e^theta = e^{theta_0} dfrac{r_0}{r}; tag{18}$
$r = r_0 e^{theta_0 - theta}; tag{19}$
the above presents the general solution for $t = t_0$, $r(t_0) = r_0$, $theta(t_0) = theta_0$; we specialize to the case at hand:
$t_0 = 0, ; gamma(0) = 1 + i; tag{20}$
$r_0 = vert gamma(0) vert = sqrt 2; tag{21}$
$theta_0 = dfrac{pi}{4}; tag{22}$
from (19):
$r = sqrt 2 e^{pi/4 - theta}; tag{23}$
from (12) and (16):
$r(t) = sqrt 2 - st; tag{24}$
$theta(t) = dfrac{pi}{4} - ln(sqrt 2 - s(t - t_0)) + ln sqrt 2. tag{25}$
The path length is perhaps easiest found by returning to the expressions for $r(t)$ and $theta(t)$, in terms of which
$vert dot gamma vert = sqrt{ (dot r)^2 + r^2 (dot theta)^2} = sqrt{s^2 + s^2} = sqrt 2 s; tag{26}$
since $t$ is restricted to $[0, r_0/s]$, the length of the curve is
$displaystyle int_0^{r_0/s} vert dot gamma vert ; dt = displaystyle int_0^{r_0/s} sqrt 2 s ; dt = sqrt 2 r_0. tag{27}$
edited 5 hours ago
answered 19 hours ago
Robert Lewis
42.4k22861
42.4k22861
1
Thanks! I see that $r=sqrt{2}exp(pi/4-theta)$ for $thetageqpi/4$ is a good polar expression for this curve. However this curve never reaches the origin, but $r(t)=sqrt{2}-st$ on line 24 seems to imply $r=0$ for $t=sqrt{2}/s$. Does the image of $tin[0,sqrt{2}/s)$ under $gamma$ contain the entire curve?
– M. Nestor
18 hours ago
@M.Nestor: well, a quick answer which I think adequate is that the polar coordinate system isn't really defined at $r = 0$, so we should really say $t < sqrt 2 / s$. I should probably edit that part of my answer.
– Robert Lewis
18 hours ago
1
@M.Nestor: you should be able to get answers to your edited question by taking $s = 1/sqrt 2$ in mine,
– Robert Lewis
18 hours ago
add a comment |
1
Thanks! I see that $r=sqrt{2}exp(pi/4-theta)$ for $thetageqpi/4$ is a good polar expression for this curve. However this curve never reaches the origin, but $r(t)=sqrt{2}-st$ on line 24 seems to imply $r=0$ for $t=sqrt{2}/s$. Does the image of $tin[0,sqrt{2}/s)$ under $gamma$ contain the entire curve?
– M. Nestor
18 hours ago
@M.Nestor: well, a quick answer which I think adequate is that the polar coordinate system isn't really defined at $r = 0$, so we should really say $t < sqrt 2 / s$. I should probably edit that part of my answer.
– Robert Lewis
18 hours ago
1
@M.Nestor: you should be able to get answers to your edited question by taking $s = 1/sqrt 2$ in mine,
– Robert Lewis
18 hours ago
1
1
Thanks! I see that $r=sqrt{2}exp(pi/4-theta)$ for $thetageqpi/4$ is a good polar expression for this curve. However this curve never reaches the origin, but $r(t)=sqrt{2}-st$ on line 24 seems to imply $r=0$ for $t=sqrt{2}/s$. Does the image of $tin[0,sqrt{2}/s)$ under $gamma$ contain the entire curve?
– M. Nestor
18 hours ago
Thanks! I see that $r=sqrt{2}exp(pi/4-theta)$ for $thetageqpi/4$ is a good polar expression for this curve. However this curve never reaches the origin, but $r(t)=sqrt{2}-st$ on line 24 seems to imply $r=0$ for $t=sqrt{2}/s$. Does the image of $tin[0,sqrt{2}/s)$ under $gamma$ contain the entire curve?
– M. Nestor
18 hours ago
@M.Nestor: well, a quick answer which I think adequate is that the polar coordinate system isn't really defined at $r = 0$, so we should really say $t < sqrt 2 / s$. I should probably edit that part of my answer.
– Robert Lewis
18 hours ago
@M.Nestor: well, a quick answer which I think adequate is that the polar coordinate system isn't really defined at $r = 0$, so we should really say $t < sqrt 2 / s$. I should probably edit that part of my answer.
– Robert Lewis
18 hours ago
1
1
@M.Nestor: you should be able to get answers to your edited question by taking $s = 1/sqrt 2$ in mine,
– Robert Lewis
18 hours ago
@M.Nestor: you should be able to get answers to your edited question by taking $s = 1/sqrt 2$ in mine,
– Robert Lewis
18 hours ago
add a comment |
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