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Length Of Curve $gamma(t)=(t cos t,tsin t)$
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Find the length of the curve $gamma(t)=(t cos t,tsin t)$, $tin[0,pi]$
So I first start of saying that the curve is in $C^1$ and therefore I can use the $L(gamma)=int_a^b|gamma'(t)|dt$
Is it right?
So
$gamma'(t)=( cos t-t sin t,sin t+tcos t)$
$|gamma'(t)|= sqrt(cos t-t sin t)^2+(sin t+tcos t)^2=\=sqrtcos^2t -2tcos t sin t+t^2sin ^2t+sin ^2t+2t cos t sin t+t^2cos ^2t=sqrt1+t^2$
So
$int_0^pisqrt1+t^2dt$
$t=tantheta, dt=sec theta d theta$
$int_0^pisqrt1+tan^2 t sec theta d theta=int_0^pisec^3 theta d theta=int_0^pisec theta cdotsec^2theta dtheta=int_0^pisec theta cdot(1+tan^2theta) dtheta$
$u=tan theta, du=sec theta dtheta$
$int_0^pi(1+u^2) du=u+fracu^33|_0^pi=tantheta+fractan^3theta3|_0^pi=sqrt1+t^2+(1+t^2)^frac32|_0^pi=\=sqrt1+pi^2+(1+pi^2)^frac32-2$
Which it wrong where is the mistake?
calculus integration
edited Jul 25 at 5:52
CiaPan
9,85311044
asked Jul 24 at 19:45
newhere
742310
add a comment |Â
up vote
2
down vote
favorite
Find the length of the curve $gamma(t)=(t cos t,tsin t)$, $tin[0,pi]$
So I first start of saying that the curve is in $C^1$ and therefore I can use the $L(gamma)=int_a^b|gamma'(t)|dt$
Is it right?
So
$gamma'(t)=( cos t-t sin t,sin t+tcos t)$
$|gamma'(t)|= sqrt(cos t-t sin t)^2+(sin t+tcos t)^2=\=sqrtcos^2t -2tcos t sin t+t^2sin ^2t+sin ^2t+2t cos t sin t+t^2cos ^2t=sqrt1+t^2$
So
$int_0^pisqrt1+t^2dt$
$t=tantheta, dt=sec theta d theta$
$int_0^pisqrt1+tan^2 t sec theta d theta=int_0^pisec^3 theta d theta=int_0^pisec theta cdotsec^2theta dtheta=int_0^pisec theta cdot(1+tan^2theta) dtheta$
$u=tan theta, du=sec theta dtheta$
$int_0^pi(1+u^2) du=u+fracu^33|_0^pi=tantheta+fractan^3theta3|_0^pi=sqrt1+t^2+(1+t^2)^frac32|_0^pi=\=sqrt1+pi^2+(1+pi^2)^frac32-2$
Which it wrong where is the mistake?
calculus integration
edited Jul 25 at 5:52
CiaPan
9,85311044
asked Jul 24 at 19:45
newhere
742310
1
When you make the change of variables from $t$ to $tan(theta )$, should the bounds of integration also change?
– gd1035
Jul 24 at 19:50
@gd1035 correct, but also the indefine integral does not come out right according to Wolfram
– newhere
Jul 24 at 20:41
1
$int sqrt1 + t^2dt$ could be looked up in a table of integrals like here: integral-table.com/downloads/single-page-integral-table.pdf
– gd1035
Jul 24 at 20:52
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find the length of the curve $gamma(t)=(t cos t,tsin t)$, $tin[0,pi]$
So I first start of saying that the curve is in $C^1$ and therefore I can use the $L(gamma)=int_a^b|gamma'(t)|dt$
Is it right?
So
$gamma'(t)=( cos t-t sin t,sin t+tcos t)$
$|gamma'(t)|= sqrt(cos t-t sin t)^2+(sin t+tcos t)^2=\=sqrtcos^2t -2tcos t sin t+t^2sin ^2t+sin ^2t+2t cos t sin t+t^2cos ^2t=sqrt1+t^2$
So
$int_0^pisqrt1+t^2dt$
$t=tantheta, dt=sec theta d theta$
$int_0^pisqrt1+tan^2 t sec theta d theta=int_0^pisec^3 theta d theta=int_0^pisec theta cdotsec^2theta dtheta=int_0^pisec theta cdot(1+tan^2theta) dtheta$
$u=tan theta, du=sec theta dtheta$
$int_0^pi(1+u^2) du=u+fracu^33|_0^pi=tantheta+fractan^3theta3|_0^pi=sqrt1+t^2+(1+t^2)^frac32|_0^pi=\=sqrt1+pi^2+(1+pi^2)^frac32-2$
Which it wrong where is the mistake?
calculus integration
edited Jul 25 at 5:52
CiaPan
9,85311044
asked Jul 24 at 19:45
newhere
742310
Find the length of the curve $gamma(t)=(t cos t,tsin t)$, $tin[0,pi]$
So I first start of saying that the curve is in $C^1$ and therefore I can use the $L(gamma)=int_a^b|gamma'(t)|dt$
Is it right?
So
$gamma'(t)=( cos t-t sin t,sin t+tcos t)$
$|gamma'(t)|= sqrt(cos t-t sin t)^2+(sin t+tcos t)^2=\=sqrtcos^2t -2tcos t sin t+t^2sin ^2t+sin ^2t+2t cos t sin t+t^2cos ^2t=sqrt1+t^2$
So
$int_0^pisqrt1+t^2dt$
$t=tantheta, dt=sec theta d theta$
$int_0^pisqrt1+tan^2 t sec theta d theta=int_0^pisec^3 theta d theta=int_0^pisec theta cdotsec^2theta dtheta=int_0^pisec theta cdot(1+tan^2theta) dtheta$
$u=tan theta, du=sec theta dtheta$
$int_0^pi(1+u^2) du=u+fracu^33|_0^pi=tantheta+fractan^3theta3|_0^pi=sqrt1+t^2+(1+t^2)^frac32|_0^pi=\=sqrt1+pi^2+(1+pi^2)^frac32-2$
Which it wrong where is the mistake?
calculus integration
edited Jul 25 at 5:52
CiaPan
9,85311044
asked Jul 24 at 19:45
newhere
742310
edited Jul 25 at 5:52
CiaPan
9,85311044
edited Jul 25 at 5:52
CiaPan
9,85311044
edited Jul 25 at 5:52
CiaPan
9,85311044
9,85311044
asked Jul 24 at 19:45
newhere
742310
asked Jul 24 at 19:45
newhere
742310
asked Jul 24 at 19:45
newhere
742310
742310
1
When you make the change of variables from $t$ to $tan(theta )$, should the bounds of integration also change?
– gd1035
Jul 24 at 19:50
@gd1035 correct, but also the indefine integral does not come out right according to Wolfram
– newhere
Jul 24 at 20:41
1
$int sqrt1 + t^2dt$ could be looked up in a table of integrals like here: integral-table.com/downloads/single-page-integral-table.pdf
– gd1035
Jul 24 at 20:52
add a comment |Â
1
When you make the change of variables from $t$ to $tan(theta )$, should the bounds of integration also change?
– gd1035
Jul 24 at 19:50
@gd1035 correct, but also the indefine integral does not come out right according to Wolfram
– newhere
Jul 24 at 20:41
1
$int sqrt1 + t^2dt$ could be looked up in a table of integrals like here: integral-table.com/downloads/single-page-integral-table.pdf
– gd1035
Jul 24 at 20:52
1
1
When you make the change of variables from $t$ to $tan(theta )$, should the bounds of integration also change?
– gd1035
Jul 24 at 19:50
When you make the change of variables from $t$ to $tan(theta )$, should the bounds of integration also change?
– gd1035
Jul 24 at 19:50
@gd1035 correct, but also the indefine integral does not come out right according to Wolfram
– newhere
Jul 24 at 20:41
@gd1035 correct, but also the indefine integral does not come out right according to Wolfram
– newhere
Jul 24 at 20:41
1
1
$int sqrt1 + t^2dt$ could be looked up in a table of integrals like here: integral-table.com/downloads/single-page-integral-table.pdf
– gd1035
Jul 24 at 20:52
$int sqrt1 + t^2dt$ could be looked up in a table of integrals like here: integral-table.com/downloads/single-page-integral-table.pdf
– gd1035
Jul 24 at 20:52
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Your integral is $frac12[tsqrt1+t^2+ln(t+sqrt1+t^2)]_0^pi=fracpisqrt1+pi^2+ln(pi+sqrt1+pi^2)2$. Your substitution $t=tantheta$ would also finish the problem (see here), but the upper limit must become $arctanpi$.
answered Jul 25 at 5:46
J.G.
13.1k11424
add a comment |Â
up vote
1
down vote
After the substitution $t=tan theta$ you must change the limits of integration and, more important, you have
$$
dt=sec^2theta dtheta
$$
(and the same for
$
u=tan theta qquad du=sec^2 theta d theta
quad$
that returns the integration to the starting point: $int_0^pisqrt1+u^2du$
answered Jul 24 at 20:43
Emilio Novati
50.2k43170
1
So I need to try different method then the $u$ substitution
– newhere
Jul 24 at 21:29
add a comment |Â
up vote
1
down vote
What about hyperbolic functions?:
$$t=sinh ximplies dt=cosh x,dximpliesint_0^pisqrt1+t^2,dt=int_0^textarcsinhpisqrt1+sinh^2xcdotcosh xdx=$$
$$=int_0^textarcsinhpicosh^2x,dx=left.frac12left(x+sinh xcosh xright)right|_0^textarcsinhpi=frac12left(textarcsinhpi+pisqrt1+pi^2right)$$
answered Jul 25 at 14:18
DonAntonio
172k1483217
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your integral is $frac12[tsqrt1+t^2+ln(t+sqrt1+t^2)]_0^pi=fracpisqrt1+pi^2+ln(pi+sqrt1+pi^2)2$. Your substitution $t=tantheta$ would also finish the problem (see here), but the upper limit must become $arctanpi$.
answered Jul 25 at 5:46
J.G.
13.1k11424
add a comment |Â
up vote
2
down vote
accepted
Your integral is $frac12[tsqrt1+t^2+ln(t+sqrt1+t^2)]_0^pi=fracpisqrt1+pi^2+ln(pi+sqrt1+pi^2)2$. Your substitution $t=tantheta$ would also finish the problem (see here), but the upper limit must become $arctanpi$.
answered Jul 25 at 5:46
J.G.
13.1k11424
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your integral is $frac12[tsqrt1+t^2+ln(t+sqrt1+t^2)]_0^pi=fracpisqrt1+pi^2+ln(pi+sqrt1+pi^2)2$. Your substitution $t=tantheta$ would also finish the problem (see here), but the upper limit must become $arctanpi$.
answered Jul 25 at 5:46
J.G.
13.1k11424
Your integral is $frac12[tsqrt1+t^2+ln(t+sqrt1+t^2)]_0^pi=fracpisqrt1+pi^2+ln(pi+sqrt1+pi^2)2$. Your substitution $t=tantheta$ would also finish the problem (see here), but the upper limit must become $arctanpi$.
answered Jul 25 at 5:46
J.G.
13.1k11424
answered Jul 25 at 5:46
J.G.
13.1k11424
answered Jul 25 at 5:46
J.G.
13.1k11424
answered Jul 25 at 5:46
J.G.
13.1k11424
13.1k11424
add a comment |Â
add a comment |Â
up vote
1
down vote
After the substitution $t=tan theta$ you must change the limits of integration and, more important, you have
$$
dt=sec^2theta dtheta
$$
(and the same for
$
u=tan theta qquad du=sec^2 theta d theta
quad$
that returns the integration to the starting point: $int_0^pisqrt1+u^2du$
answered Jul 24 at 20:43
Emilio Novati
50.2k43170
1
So I need to try different method then the $u$ substitution
– newhere
Jul 24 at 21:29
add a comment |Â
up vote
1
down vote
After the substitution $t=tan theta$ you must change the limits of integration and, more important, you have
$$
dt=sec^2theta dtheta
$$
(and the same for
$
u=tan theta qquad du=sec^2 theta d theta
quad$
that returns the integration to the starting point: $int_0^pisqrt1+u^2du$
answered Jul 24 at 20:43
Emilio Novati
50.2k43170
1
So I need to try different method then the $u$ substitution
– newhere
Jul 24 at 21:29
add a comment |Â
up vote
1
down vote
up vote
1
down vote
After the substitution $t=tan theta$ you must change the limits of integration and, more important, you have
$$
dt=sec^2theta dtheta
$$
(and the same for
$
u=tan theta qquad du=sec^2 theta d theta
quad$
that returns the integration to the starting point: $int_0^pisqrt1+u^2du$
answered Jul 24 at 20:43
Emilio Novati
50.2k43170
After the substitution $t=tan theta$ you must change the limits of integration and, more important, you have
$$
dt=sec^2theta dtheta
$$
(and the same for
$
u=tan theta qquad du=sec^2 theta d theta
quad$
that returns the integration to the starting point: $int_0^pisqrt1+u^2du$
answered Jul 24 at 20:43
Emilio Novati
50.2k43170
edited Jul 24 at 20:51
answered Jul 24 at 20:43
Emilio Novati
50.2k43170
answered Jul 24 at 20:43
Emilio Novati
50.2k43170
answered Jul 24 at 20:43
Emilio Novati
50.2k43170
50.2k43170
1
So I need to try different method then the $u$ substitution
– newhere
Jul 24 at 21:29
add a comment |Â
1
So I need to try different method then the $u$ substitution
– newhere
Jul 24 at 21:29
1
1
So I need to try different method then the $u$ substitution
– newhere
Jul 24 at 21:29
So I need to try different method then the $u$ substitution
– newhere
Jul 24 at 21:29
add a comment |Â
up vote
1
down vote
What about hyperbolic functions?:
$$t=sinh ximplies dt=cosh x,dximpliesint_0^pisqrt1+t^2,dt=int_0^textarcsinhpisqrt1+sinh^2xcdotcosh xdx=$$
$$=int_0^textarcsinhpicosh^2x,dx=left.frac12left(x+sinh xcosh xright)right|_0^textarcsinhpi=frac12left(textarcsinhpi+pisqrt1+pi^2right)$$
answered Jul 25 at 14:18
DonAntonio
172k1483217
add a comment |Â
up vote
1
down vote
What about hyperbolic functions?:
$$t=sinh ximplies dt=cosh x,dximpliesint_0^pisqrt1+t^2,dt=int_0^textarcsinhpisqrt1+sinh^2xcdotcosh xdx=$$
$$=int_0^textarcsinhpicosh^2x,dx=left.frac12left(x+sinh xcosh xright)right|_0^textarcsinhpi=frac12left(textarcsinhpi+pisqrt1+pi^2right)$$
answered Jul 25 at 14:18
DonAntonio
172k1483217
add a comment |Â
up vote
1
down vote
up vote
1
down vote
What about hyperbolic functions?:
$$t=sinh ximplies dt=cosh x,dximpliesint_0^pisqrt1+t^2,dt=int_0^textarcsinhpisqrt1+sinh^2xcdotcosh xdx=$$
$$=int_0^textarcsinhpicosh^2x,dx=left.frac12left(x+sinh xcosh xright)right|_0^textarcsinhpi=frac12left(textarcsinhpi+pisqrt1+pi^2right)$$
answered Jul 25 at 14:18
DonAntonio
172k1483217
What about hyperbolic functions?:
$$t=sinh ximplies dt=cosh x,dximpliesint_0^pisqrt1+t^2,dt=int_0^textarcsinhpisqrt1+sinh^2xcdotcosh xdx=$$
$$=int_0^textarcsinhpicosh^2x,dx=left.frac12left(x+sinh xcosh xright)right|_0^textarcsinhpi=frac12left(textarcsinhpi+pisqrt1+pi^2right)$$
answered Jul 25 at 14:18
DonAntonio
172k1483217
answered Jul 25 at 14:18
DonAntonio
172k1483217
answered Jul 25 at 14:18
DonAntonio
172k1483217
answered Jul 25 at 14:18
DonAntonio
172k1483217
172k1483217
add a comment |Â
add a comment |Â
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1
When you make the change of variables from $t$ to $tan(theta )$, should the bounds of integration also change?
– gd1035
Jul 24 at 19:50
@gd1035 correct, but also the indefine integral does not come out right according to Wolfram
– newhere
Jul 24 at 20:41
1
$int sqrt1 + t^2dt$ could be looked up in a table of integrals like here: integral-table.com/downloads/single-page-integral-table.pdf
– gd1035
Jul 24 at 20:52