Radius of curvature of $r=2 cdot e^3 phi$
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I have to find the curvature of $r=2 cdot e^3 phi$
$dfracleft[1+left(dfracd yd xright)^2right]^frac 3 2dfracd^2 yd x^2$
$x=rcdot cos(phi)=2 cdot e^3 phicdot cos(phi)$
$y=rcdot sin(phi)=2 cdot e^3 phicdot sin(phi)$
$dfracd xd phi=-2 e^3 phicdot (sin(phi) - 3 cos(phi))$
$dfracd yd phi=2 e^3 phicdot (3 sin(phi) + cos(phi))$
$dfracd yd x=dfrac2 e^3 phicdot (3 sin(phi) + cos(phi))-2 e^3 phicdot (sin(phi) - 3 cos(phi))
=dfrac-3 sin(phi) - cos(phi)sin(phi) - 3 cos(phi)$
$dfracd^2 yd phi^2=4 e^3 phicdot (4 sin(phi) + 3 cos(phi))$
$dfracd^2 yd ^2 x=dfracdfrac2 e^3 phicdot (4 sin(phi) - 3 cos(phi))50dfrac2 e^3 phicdot (sin(phi) + 3 cos(phi))10=dfrac4 sin(phi) - 3 cos(phi)5cdot (sin(phi) + 3 cos(phi))$
now using the radius of curvature formula i find an incredibly complex formula:
$rho=dfracleft[1+left(dfracd yd xright)^2right]^frac 3 2dfracd^2 yd x^2=dfracleft[1+left(dfrac3 sin(phi) - cos(phi)sin(phi) + 3 cos(phi)right)^2right]^frac 3 2dfrac4 sin(phi) - 3 cos(phi)5cdot (sin(phi) + 3 cos(phi))$
I can't find a way to write this in a more human way.
calculus derivatives
asked Aug 20 at 11:46
RubenDefour
918
add a comment |Â
up vote
3
down vote
favorite
I have to find the curvature of $r=2 cdot e^3 phi$
$dfracleft[1+left(dfracd yd xright)^2right]^frac 3 2dfracd^2 yd x^2$
$x=rcdot cos(phi)=2 cdot e^3 phicdot cos(phi)$
$y=rcdot sin(phi)=2 cdot e^3 phicdot sin(phi)$
$dfracd xd phi=-2 e^3 phicdot (sin(phi) - 3 cos(phi))$
$dfracd yd phi=2 e^3 phicdot (3 sin(phi) + cos(phi))$
$dfracd yd x=dfrac2 e^3 phicdot (3 sin(phi) + cos(phi))-2 e^3 phicdot (sin(phi) - 3 cos(phi))
=dfrac-3 sin(phi) - cos(phi)sin(phi) - 3 cos(phi)$
$dfracd^2 yd phi^2=4 e^3 phicdot (4 sin(phi) + 3 cos(phi))$
$dfracd^2 yd ^2 x=dfracdfrac2 e^3 phicdot (4 sin(phi) - 3 cos(phi))50dfrac2 e^3 phicdot (sin(phi) + 3 cos(phi))10=dfrac4 sin(phi) - 3 cos(phi)5cdot (sin(phi) + 3 cos(phi))$
now using the radius of curvature formula i find an incredibly complex formula:
$rho=dfracleft[1+left(dfracd yd xright)^2right]^frac 3 2dfracd^2 yd x^2=dfracleft[1+left(dfrac3 sin(phi) - cos(phi)sin(phi) + 3 cos(phi)right)^2right]^frac 3 2dfrac4 sin(phi) - 3 cos(phi)5cdot (sin(phi) + 3 cos(phi))$
I can't find a way to write this in a more human way.
calculus derivatives
asked Aug 20 at 11:46
RubenDefour
918
Have you tried using formula for polar coordinates?
– Rumpelstiltskin
Aug 20 at 11:53
Anyway there's a mistake you've done $fracdcosphidphi$ =-$ sinphi$ whereas you've written it as + $sinphi$ and same sign problem with $dsinphi/dphi$
– user187604
Aug 20 at 11:57
oh sorry i see my mistake here i used integrals in stead of derivatives
– RubenDefour
Aug 20 at 12:00
You can simplify your expression by taking things to the common fraction and cancelling the denominators and more. Additionally, you lost the exponential parts somehow, when you calculated the last expression. For better way, look at the answers about the polar coordinate calculation.
– orion
Aug 21 at 12:31
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have to find the curvature of $r=2 cdot e^3 phi$
$dfracleft[1+left(dfracd yd xright)^2right]^frac 3 2dfracd^2 yd x^2$
$x=rcdot cos(phi)=2 cdot e^3 phicdot cos(phi)$
$y=rcdot sin(phi)=2 cdot e^3 phicdot sin(phi)$
$dfracd xd phi=-2 e^3 phicdot (sin(phi) - 3 cos(phi))$
$dfracd yd phi=2 e^3 phicdot (3 sin(phi) + cos(phi))$
$dfracd yd x=dfrac2 e^3 phicdot (3 sin(phi) + cos(phi))-2 e^3 phicdot (sin(phi) - 3 cos(phi))
=dfrac-3 sin(phi) - cos(phi)sin(phi) - 3 cos(phi)$
$dfracd^2 yd phi^2=4 e^3 phicdot (4 sin(phi) + 3 cos(phi))$
$dfracd^2 yd ^2 x=dfracdfrac2 e^3 phicdot (4 sin(phi) - 3 cos(phi))50dfrac2 e^3 phicdot (sin(phi) + 3 cos(phi))10=dfrac4 sin(phi) - 3 cos(phi)5cdot (sin(phi) + 3 cos(phi))$
now using the radius of curvature formula i find an incredibly complex formula:
$rho=dfracleft[1+left(dfracd yd xright)^2right]^frac 3 2dfracd^2 yd x^2=dfracleft[1+left(dfrac3 sin(phi) - cos(phi)sin(phi) + 3 cos(phi)right)^2right]^frac 3 2dfrac4 sin(phi) - 3 cos(phi)5cdot (sin(phi) + 3 cos(phi))$
I can't find a way to write this in a more human way.
calculus derivatives
asked Aug 20 at 11:46
RubenDefour
918
I have to find the curvature of $r=2 cdot e^3 phi$
$dfracleft[1+left(dfracd yd xright)^2right]^frac 3 2dfracd^2 yd x^2$
$x=rcdot cos(phi)=2 cdot e^3 phicdot cos(phi)$
$y=rcdot sin(phi)=2 cdot e^3 phicdot sin(phi)$
$dfracd xd phi=-2 e^3 phicdot (sin(phi) - 3 cos(phi))$
$dfracd yd phi=2 e^3 phicdot (3 sin(phi) + cos(phi))$
$dfracd yd x=dfrac2 e^3 phicdot (3 sin(phi) + cos(phi))-2 e^3 phicdot (sin(phi) - 3 cos(phi))
=dfrac-3 sin(phi) - cos(phi)sin(phi) - 3 cos(phi)$
$dfracd^2 yd phi^2=4 e^3 phicdot (4 sin(phi) + 3 cos(phi))$
$dfracd^2 yd ^2 x=dfracdfrac2 e^3 phicdot (4 sin(phi) - 3 cos(phi))50dfrac2 e^3 phicdot (sin(phi) + 3 cos(phi))10=dfrac4 sin(phi) - 3 cos(phi)5cdot (sin(phi) + 3 cos(phi))$
now using the radius of curvature formula i find an incredibly complex formula:
$rho=dfracleft[1+left(dfracd yd xright)^2right]^frac 3 2dfracd^2 yd x^2=dfracleft[1+left(dfrac3 sin(phi) - cos(phi)sin(phi) + 3 cos(phi)right)^2right]^frac 3 2dfrac4 sin(phi) - 3 cos(phi)5cdot (sin(phi) + 3 cos(phi))$
I can't find a way to write this in a more human way.
calculus derivatives
asked Aug 20 at 11:46
RubenDefour
918
edited Aug 21 at 6:47
asked Aug 20 at 11:46
RubenDefour
918
asked Aug 20 at 11:46
RubenDefour
918
asked Aug 20 at 11:46
RubenDefour
918
918
Have you tried using formula for polar coordinates?
– Rumpelstiltskin
Aug 20 at 11:53
Anyway there's a mistake you've done $fracdcosphidphi$ =-$ sinphi$ whereas you've written it as + $sinphi$ and same sign problem with $dsinphi/dphi$
– user187604
Aug 20 at 11:57
oh sorry i see my mistake here i used integrals in stead of derivatives
– RubenDefour
Aug 20 at 12:00
You can simplify your expression by taking things to the common fraction and cancelling the denominators and more. Additionally, you lost the exponential parts somehow, when you calculated the last expression. For better way, look at the answers about the polar coordinate calculation.
– orion
Aug 21 at 12:31
add a comment |Â
Have you tried using formula for polar coordinates?
– Rumpelstiltskin
Aug 20 at 11:53
Anyway there's a mistake you've done $fracdcosphidphi$ =-$ sinphi$ whereas you've written it as + $sinphi$ and same sign problem with $dsinphi/dphi$
– user187604
Aug 20 at 11:57
oh sorry i see my mistake here i used integrals in stead of derivatives
– RubenDefour
Aug 20 at 12:00
You can simplify your expression by taking things to the common fraction and cancelling the denominators and more. Additionally, you lost the exponential parts somehow, when you calculated the last expression. For better way, look at the answers about the polar coordinate calculation.
– orion
Aug 21 at 12:31
Have you tried using formula for polar coordinates?
– Rumpelstiltskin
Aug 20 at 11:53
Have you tried using formula for polar coordinates?
– Rumpelstiltskin
Aug 20 at 11:53
Anyway there's a mistake you've done $fracdcosphidphi$ =-$ sinphi$ whereas you've written it as + $sinphi$ and same sign problem with $dsinphi/dphi$
– user187604
Aug 20 at 11:57
Anyway there's a mistake you've done $fracdcosphidphi$ =-$ sinphi$ whereas you've written it as + $sinphi$ and same sign problem with $dsinphi/dphi$
– user187604
Aug 20 at 11:57
oh sorry i see my mistake here i used integrals in stead of derivatives
– RubenDefour
Aug 20 at 12:00
oh sorry i see my mistake here i used integrals in stead of derivatives
– RubenDefour
Aug 20 at 12:00
You can simplify your expression by taking things to the common fraction and cancelling the denominators and more. Additionally, you lost the exponential parts somehow, when you calculated the last expression. For better way, look at the answers about the polar coordinate calculation.
– orion
Aug 21 at 12:31
You can simplify your expression by taking things to the common fraction and cancelling the denominators and more. Additionally, you lost the exponential parts somehow, when you calculated the last expression. For better way, look at the answers about the polar coordinate calculation.
– orion
Aug 21 at 12:31
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
It is easier to use the formula for curvature in terms of the function $r(theta)$:
If a curve is defined in polar coordinates as $r(theta)$, then its curvature is
$$
kappa(theta)
= frac(r^2 + r'^2)^3/2
$$
where here the prime refers to the differentiation with respect to $theta$.
The result is $e^-3theta/(2sqrt10)$.
edited Aug 21 at 19:05
Jendrik Stelzner
7,57221037
answered Aug 20 at 12:02
uniquesolution
8,251823
we can't use that formula on our exam
– RubenDefour
Aug 20 at 12:06
4
Please accept my deepest condolences.
– uniquesolution
Aug 20 at 12:06
Then show me where my mistake is please!
– Dr. Sonnhard Graubner
Aug 20 at 12:10
I cannot show you your mistake because you deleted it.
– uniquesolution
Aug 20 at 12:12
Yes since someone gave me a $-1$ i have checked it with WA!
– Dr. Sonnhard Graubner
Aug 20 at 12:13
add a comment |Â
up vote
0
down vote
The curve $gamma$ in question is a logarithmic spiral. From the selfsimilarity properties of these spirals it immediately follows that $rho(phi)=c r(phi)$ for some constant $c$. The following (human) computations will confirm this fact.
I shall use the parametric representation
$$gamma:quadphimapstobf z(phi):=2e^3phi(cosphi,sinphi) .$$One computes
$$eqalignbf z'(phi)&=2e^3phi(-sinphi+3cosphi, cosphi+3sinphi)cr &=2sqrt10e^3phibigl(cos(phi+alpha),sin(phi+alpha)bigr) ,cr$$
whereby $alpha:=arctan1over3$. It follows that the arc length $phimapsto s(phi)$ along $gamma$ satisfies
$$s'(phi)=bigl|bf z'(phi)bigr|=2sqrt10e^3phi ,$$ whereas the forward tangent direction is given by
$$theta(phi)=argbigl(bf z'(phi)bigr)=phi+alpha .$$
Now the curvature $kappa$ is the turning speed of the tangent with respect to arc length, hence
$$kappa(phi)=dthetaover ds=theta'(phi)over s'(phi)=1over2sqrt10e^-3phi ,$$
so that we finally obtain
$$rho(phi)=1overkappa(phi)=2sqrt10e^3phi .$$
answered Aug 21 at 12:24
Christian Blatter
165k7109310
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It is easier to use the formula for curvature in terms of the function $r(theta)$:
If a curve is defined in polar coordinates as $r(theta)$, then its curvature is
$$
kappa(theta)
= frac(r^2 + r'^2)^3/2
$$
where here the prime refers to the differentiation with respect to $theta$.
The result is $e^-3theta/(2sqrt10)$.
edited Aug 21 at 19:05
Jendrik Stelzner
7,57221037
answered Aug 20 at 12:02
uniquesolution
8,251823
we can't use that formula on our exam
– RubenDefour
Aug 20 at 12:06
4
Please accept my deepest condolences.
– uniquesolution
Aug 20 at 12:06
Then show me where my mistake is please!
– Dr. Sonnhard Graubner
Aug 20 at 12:10
I cannot show you your mistake because you deleted it.
– uniquesolution
Aug 20 at 12:12
Yes since someone gave me a $-1$ i have checked it with WA!
– Dr. Sonnhard Graubner
Aug 20 at 12:13
add a comment |Â
up vote
1
down vote
It is easier to use the formula for curvature in terms of the function $r(theta)$:
If a curve is defined in polar coordinates as $r(theta)$, then its curvature is
$$
kappa(theta)
= frac(r^2 + r'^2)^3/2
$$
where here the prime refers to the differentiation with respect to $theta$.
The result is $e^-3theta/(2sqrt10)$.
edited Aug 21 at 19:05
Jendrik Stelzner
7,57221037
answered Aug 20 at 12:02
uniquesolution
8,251823
we can't use that formula on our exam
– RubenDefour
Aug 20 at 12:06
4
Please accept my deepest condolences.
– uniquesolution
Aug 20 at 12:06
Then show me where my mistake is please!
– Dr. Sonnhard Graubner
Aug 20 at 12:10
I cannot show you your mistake because you deleted it.
– uniquesolution
Aug 20 at 12:12
Yes since someone gave me a $-1$ i have checked it with WA!
– Dr. Sonnhard Graubner
Aug 20 at 12:13
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is easier to use the formula for curvature in terms of the function $r(theta)$:
If a curve is defined in polar coordinates as $r(theta)$, then its curvature is
$$
kappa(theta)
= frac(r^2 + r'^2)^3/2
$$
where here the prime refers to the differentiation with respect to $theta$.
The result is $e^-3theta/(2sqrt10)$.
edited Aug 21 at 19:05
Jendrik Stelzner
7,57221037
answered Aug 20 at 12:02
uniquesolution
8,251823
It is easier to use the formula for curvature in terms of the function $r(theta)$:
If a curve is defined in polar coordinates as $r(theta)$, then its curvature is
$$
kappa(theta)
= frac(r^2 + r'^2)^3/2
$$
where here the prime refers to the differentiation with respect to $theta$.
The result is $e^-3theta/(2sqrt10)$.
edited Aug 21 at 19:05
Jendrik Stelzner
7,57221037
answered Aug 20 at 12:02
uniquesolution
8,251823
edited Aug 21 at 19:05
Jendrik Stelzner
7,57221037
edited Aug 21 at 19:05
Jendrik Stelzner
7,57221037
edited Aug 21 at 19:05
Jendrik Stelzner
7,57221037
7,57221037
answered Aug 20 at 12:02
uniquesolution
8,251823
answered Aug 20 at 12:02
uniquesolution
8,251823
answered Aug 20 at 12:02
uniquesolution
8,251823
8,251823
we can't use that formula on our exam
– RubenDefour
Aug 20 at 12:06
4
Please accept my deepest condolences.
– uniquesolution
Aug 20 at 12:06
Then show me where my mistake is please!
– Dr. Sonnhard Graubner
Aug 20 at 12:10
I cannot show you your mistake because you deleted it.
– uniquesolution
Aug 20 at 12:12
Yes since someone gave me a $-1$ i have checked it with WA!
– Dr. Sonnhard Graubner
Aug 20 at 12:13
add a comment |Â
we can't use that formula on our exam
– RubenDefour
Aug 20 at 12:06
4
Please accept my deepest condolences.
– uniquesolution
Aug 20 at 12:06
Then show me where my mistake is please!
– Dr. Sonnhard Graubner
Aug 20 at 12:10
I cannot show you your mistake because you deleted it.
– uniquesolution
Aug 20 at 12:12
Yes since someone gave me a $-1$ i have checked it with WA!
– Dr. Sonnhard Graubner
Aug 20 at 12:13
we can't use that formula on our exam
– RubenDefour
Aug 20 at 12:06
we can't use that formula on our exam
– RubenDefour
Aug 20 at 12:06
4
4
Please accept my deepest condolences.
– uniquesolution
Aug 20 at 12:06
Please accept my deepest condolences.
– uniquesolution
Aug 20 at 12:06
Then show me where my mistake is please!
– Dr. Sonnhard Graubner
Aug 20 at 12:10
Then show me where my mistake is please!
– Dr. Sonnhard Graubner
Aug 20 at 12:10
I cannot show you your mistake because you deleted it.
– uniquesolution
Aug 20 at 12:12
I cannot show you your mistake because you deleted it.
– uniquesolution
Aug 20 at 12:12
Yes since someone gave me a $-1$ i have checked it with WA!
– Dr. Sonnhard Graubner
Aug 20 at 12:13
Yes since someone gave me a $-1$ i have checked it with WA!
– Dr. Sonnhard Graubner
Aug 20 at 12:13
add a comment |Â
up vote
0
down vote
The curve $gamma$ in question is a logarithmic spiral. From the selfsimilarity properties of these spirals it immediately follows that $rho(phi)=c r(phi)$ for some constant $c$. The following (human) computations will confirm this fact.
I shall use the parametric representation
$$gamma:quadphimapstobf z(phi):=2e^3phi(cosphi,sinphi) .$$One computes
$$eqalignbf z'(phi)&=2e^3phi(-sinphi+3cosphi, cosphi+3sinphi)cr &=2sqrt10e^3phibigl(cos(phi+alpha),sin(phi+alpha)bigr) ,cr$$
whereby $alpha:=arctan1over3$. It follows that the arc length $phimapsto s(phi)$ along $gamma$ satisfies
$$s'(phi)=bigl|bf z'(phi)bigr|=2sqrt10e^3phi ,$$ whereas the forward tangent direction is given by
$$theta(phi)=argbigl(bf z'(phi)bigr)=phi+alpha .$$
Now the curvature $kappa$ is the turning speed of the tangent with respect to arc length, hence
$$kappa(phi)=dthetaover ds=theta'(phi)over s'(phi)=1over2sqrt10e^-3phi ,$$
so that we finally obtain
$$rho(phi)=1overkappa(phi)=2sqrt10e^3phi .$$
answered Aug 21 at 12:24
Christian Blatter
165k7109310
add a comment |Â
up vote
0
down vote
The curve $gamma$ in question is a logarithmic spiral. From the selfsimilarity properties of these spirals it immediately follows that $rho(phi)=c r(phi)$ for some constant $c$. The following (human) computations will confirm this fact.
I shall use the parametric representation
$$gamma:quadphimapstobf z(phi):=2e^3phi(cosphi,sinphi) .$$One computes
$$eqalignbf z'(phi)&=2e^3phi(-sinphi+3cosphi, cosphi+3sinphi)cr &=2sqrt10e^3phibigl(cos(phi+alpha),sin(phi+alpha)bigr) ,cr$$
whereby $alpha:=arctan1over3$. It follows that the arc length $phimapsto s(phi)$ along $gamma$ satisfies
$$s'(phi)=bigl|bf z'(phi)bigr|=2sqrt10e^3phi ,$$ whereas the forward tangent direction is given by
$$theta(phi)=argbigl(bf z'(phi)bigr)=phi+alpha .$$
Now the curvature $kappa$ is the turning speed of the tangent with respect to arc length, hence
$$kappa(phi)=dthetaover ds=theta'(phi)over s'(phi)=1over2sqrt10e^-3phi ,$$
so that we finally obtain
$$rho(phi)=1overkappa(phi)=2sqrt10e^3phi .$$
answered Aug 21 at 12:24
Christian Blatter
165k7109310
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The curve $gamma$ in question is a logarithmic spiral. From the selfsimilarity properties of these spirals it immediately follows that $rho(phi)=c r(phi)$ for some constant $c$. The following (human) computations will confirm this fact.
I shall use the parametric representation
$$gamma:quadphimapstobf z(phi):=2e^3phi(cosphi,sinphi) .$$One computes
$$eqalignbf z'(phi)&=2e^3phi(-sinphi+3cosphi, cosphi+3sinphi)cr &=2sqrt10e^3phibigl(cos(phi+alpha),sin(phi+alpha)bigr) ,cr$$
whereby $alpha:=arctan1over3$. It follows that the arc length $phimapsto s(phi)$ along $gamma$ satisfies
$$s'(phi)=bigl|bf z'(phi)bigr|=2sqrt10e^3phi ,$$ whereas the forward tangent direction is given by
$$theta(phi)=argbigl(bf z'(phi)bigr)=phi+alpha .$$
Now the curvature $kappa$ is the turning speed of the tangent with respect to arc length, hence
$$kappa(phi)=dthetaover ds=theta'(phi)over s'(phi)=1over2sqrt10e^-3phi ,$$
so that we finally obtain
$$rho(phi)=1overkappa(phi)=2sqrt10e^3phi .$$
answered Aug 21 at 12:24
Christian Blatter
165k7109310
The curve $gamma$ in question is a logarithmic spiral. From the selfsimilarity properties of these spirals it immediately follows that $rho(phi)=c r(phi)$ for some constant $c$. The following (human) computations will confirm this fact.
I shall use the parametric representation
$$gamma:quadphimapstobf z(phi):=2e^3phi(cosphi,sinphi) .$$One computes
$$eqalignbf z'(phi)&=2e^3phi(-sinphi+3cosphi, cosphi+3sinphi)cr &=2sqrt10e^3phibigl(cos(phi+alpha),sin(phi+alpha)bigr) ,cr$$
whereby $alpha:=arctan1over3$. It follows that the arc length $phimapsto s(phi)$ along $gamma$ satisfies
$$s'(phi)=bigl|bf z'(phi)bigr|=2sqrt10e^3phi ,$$ whereas the forward tangent direction is given by
$$theta(phi)=argbigl(bf z'(phi)bigr)=phi+alpha .$$
Now the curvature $kappa$ is the turning speed of the tangent with respect to arc length, hence
$$kappa(phi)=dthetaover ds=theta'(phi)over s'(phi)=1over2sqrt10e^-3phi ,$$
so that we finally obtain
$$rho(phi)=1overkappa(phi)=2sqrt10e^3phi .$$
answered Aug 21 at 12:24
Christian Blatter
165k7109310
answered Aug 21 at 12:24
Christian Blatter
165k7109310
answered Aug 21 at 12:24
Christian Blatter
165k7109310
answered Aug 21 at 12:24
Christian Blatter
165k7109310
165k7109310
add a comment |Â
add a comment |Â
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Have you tried using formula for polar coordinates?
– Rumpelstiltskin
Aug 20 at 11:53
Anyway there's a mistake you've done $fracdcosphidphi$ =-$ sinphi$ whereas you've written it as + $sinphi$ and same sign problem with $dsinphi/dphi$
– user187604
Aug 20 at 11:57
oh sorry i see my mistake here i used integrals in stead of derivatives
– RubenDefour
Aug 20 at 12:00
You can simplify your expression by taking things to the common fraction and cancelling the denominators and more. Additionally, you lost the exponential parts somehow, when you calculated the last expression. For better way, look at the answers about the polar coordinate calculation.
– orion
Aug 21 at 12:31