Find the exact length of the parametric curve: $x={e^t}+{e^{-t}}, y=5-2t,0le t le 3$.












1












$begingroup$


Find the exact length of the parametric curve: $$x={e^t}+{e^{-t}}, y=5-2t,0le t le 3$$
Solution:
$$frac{dx}{dt}=e^t -e^{-t}, frac{dy}{dt}=-2$$
$$int_{0}^{3} sqrt{(e^t-e^{-t})^2+(-2)^2}dt$$
$$= int_{0}^{3} sqrt{e^{2t}+e^{-2t}+2}dt$$
Let$$e^{2t}=u, e^{-2t}=frac{1}{u}$$
and then$$du=2u dt$$
Therefore, $$int sqrt{u+frac{1}{u}+2}left(frac{1}{2u}right)du$$
$$=int sqrt{frac{(u+1)^2}{u}}left(frac{1}{2u}right)du$$
$$=int frac{u+1}{2usqrt u}du$$
$$=int frac{1}{2sqrt u}+frac{1}{2usqrt u}du$$
$$=left[sqrt {e^{2t}} - frac{1}{sqrt {e^{2t}}}right]_{0}^{3}$$
$$=e^3-frac{1}{e^3}-1+1$$
$$=e^3+e^{-3}$$
Is my answer correct?






parametric







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  • $begingroup$
    Hint $(e^t+e^{-t})^2$ equals what?
    $endgroup$
    – AmateurMathPirate
    Dec 30 '18 at 13:25
















1












$begingroup$


Find the exact length of the parametric curve: $$x={e^t}+{e^{-t}}, y=5-2t,0le t le 3$$
Solution:
$$frac{dx}{dt}=e^t -e^{-t}, frac{dy}{dt}=-2$$
$$int_{0}^{3} sqrt{(e^t-e^{-t})^2+(-2)^2}dt$$
$$= int_{0}^{3} sqrt{e^{2t}+e^{-2t}+2}dt$$
Let$$e^{2t}=u, e^{-2t}=frac{1}{u}$$
and then$$du=2u dt$$
Therefore, $$int sqrt{u+frac{1}{u}+2}left(frac{1}{2u}right)du$$
$$=int sqrt{frac{(u+1)^2}{u}}left(frac{1}{2u}right)du$$
$$=int frac{u+1}{2usqrt u}du$$
$$=int frac{1}{2sqrt u}+frac{1}{2usqrt u}du$$
$$=left[sqrt {e^{2t}} - frac{1}{sqrt {e^{2t}}}right]_{0}^{3}$$
$$=e^3-frac{1}{e^3}-1+1$$
$$=e^3+e^{-3}$$
Is my answer correct?






parametric







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint $(e^t+e^{-t})^2$ equals what?
    $endgroup$
    – AmateurMathPirate
    Dec 30 '18 at 13:25














1












1








1





$begingroup$


Find the exact length of the parametric curve: $$x={e^t}+{e^{-t}}, y=5-2t,0le t le 3$$
Solution:
$$frac{dx}{dt}=e^t -e^{-t}, frac{dy}{dt}=-2$$
$$int_{0}^{3} sqrt{(e^t-e^{-t})^2+(-2)^2}dt$$
$$= int_{0}^{3} sqrt{e^{2t}+e^{-2t}+2}dt$$
Let$$e^{2t}=u, e^{-2t}=frac{1}{u}$$
and then$$du=2u dt$$
Therefore, $$int sqrt{u+frac{1}{u}+2}left(frac{1}{2u}right)du$$
$$=int sqrt{frac{(u+1)^2}{u}}left(frac{1}{2u}right)du$$
$$=int frac{u+1}{2usqrt u}du$$
$$=int frac{1}{2sqrt u}+frac{1}{2usqrt u}du$$
$$=left[sqrt {e^{2t}} - frac{1}{sqrt {e^{2t}}}right]_{0}^{3}$$
$$=e^3-frac{1}{e^3}-1+1$$
$$=e^3+e^{-3}$$
Is my answer correct?






parametric







share|cite|improve this question











$endgroup$




Find the exact length of the parametric curve: $$x={e^t}+{e^{-t}}, y=5-2t,0le t le 3$$
Solution:
$$frac{dx}{dt}=e^t -e^{-t}, frac{dy}{dt}=-2$$
$$int_{0}^{3} sqrt{(e^t-e^{-t})^2+(-2)^2}dt$$
$$= int_{0}^{3} sqrt{e^{2t}+e^{-2t}+2}dt$$
Let$$e^{2t}=u, e^{-2t}=frac{1}{u}$$
and then$$du=2u dt$$
Therefore, $$int sqrt{u+frac{1}{u}+2}left(frac{1}{2u}right)du$$
$$=int sqrt{frac{(u+1)^2}{u}}left(frac{1}{2u}right)du$$
$$=int frac{u+1}{2usqrt u}du$$
$$=int frac{1}{2sqrt u}+frac{1}{2usqrt u}du$$
$$=left[sqrt {e^{2t}} - frac{1}{sqrt {e^{2t}}}right]_{0}^{3}$$
$$=e^3-frac{1}{e^3}-1+1$$
$$=e^3+e^{-3}$$
Is my answer correct?






parametric




parametric






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share|cite|improve this question













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edited Dec 30 '18 at 13:31









Larry

2,53531131




2,53531131










asked Dec 30 '18 at 13:19









MaggieMaggie

888




888












  • $begingroup$
    Hint $(e^t+e^{-t})^2$ equals what?
    $endgroup$
    – AmateurMathPirate
    Dec 30 '18 at 13:25


















  • $begingroup$
    Hint $(e^t+e^{-t})^2$ equals what?
    $endgroup$
    – AmateurMathPirate
    Dec 30 '18 at 13:25
















$begingroup$
Hint $(e^t+e^{-t})^2$ equals what?
$endgroup$
– AmateurMathPirate
Dec 30 '18 at 13:25




$begingroup$
Hint $(e^t+e^{-t})^2$ equals what?
$endgroup$
– AmateurMathPirate
Dec 30 '18 at 13:25










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:



It's better to utilize $$e^{2t}+e^{-2t}+2=(e^t+e^{-t})^2$$



and $displaystyledfrac{d(e^{ax})}{dx}=ae^{ex}impliesint e^{ax} dx=dfrac{e^{ax}}a+K$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    OK, I see it. So, $$int_{0}^{3} ({e^t} +{e^{-t}})dt$$ $$=[e^t+e^{-t}]_{0}^{3} $$ $$=e^3 - e^{-3}$$
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:42





















4












$begingroup$

It is faster to use some hyperbolic trigonometry: as $x(t)=2cosh t$, we have $;x'(t)=2sinh t$, so
begin{align}
ell&=int_0^3sqrt{4(sinh^2t+1)},mathrm d t=2int_0^3cosh t,mathrm d t=2sinh tBiggmvert_0^3=2sinh 3=mathrm e^3-mathrm e^{-3}.
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This awesome! I forgot the formula $cosh^2 t - sinh^2 t =1$.
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:56











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Hint:



It's better to utilize $$e^{2t}+e^{-2t}+2=(e^t+e^{-t})^2$$



and $displaystyledfrac{d(e^{ax})}{dx}=ae^{ex}impliesint e^{ax} dx=dfrac{e^{ax}}a+K$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    OK, I see it. So, $$int_{0}^{3} ({e^t} +{e^{-t}})dt$$ $$=[e^t+e^{-t}]_{0}^{3} $$ $$=e^3 - e^{-3}$$
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:42


















1












$begingroup$

Hint:



It's better to utilize $$e^{2t}+e^{-2t}+2=(e^t+e^{-t})^2$$



and $displaystyledfrac{d(e^{ax})}{dx}=ae^{ex}impliesint e^{ax} dx=dfrac{e^{ax}}a+K$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    OK, I see it. So, $$int_{0}^{3} ({e^t} +{e^{-t}})dt$$ $$=[e^t+e^{-t}]_{0}^{3} $$ $$=e^3 - e^{-3}$$
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:42
















1












1








1





$begingroup$

Hint:



It's better to utilize $$e^{2t}+e^{-2t}+2=(e^t+e^{-t})^2$$



and $displaystyledfrac{d(e^{ax})}{dx}=ae^{ex}impliesint e^{ax} dx=dfrac{e^{ax}}a+K$






share|cite|improve this answer









$endgroup$



Hint:



It's better to utilize $$e^{2t}+e^{-2t}+2=(e^t+e^{-t})^2$$



and $displaystyledfrac{d(e^{ax})}{dx}=ae^{ex}impliesint e^{ax} dx=dfrac{e^{ax}}a+K$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 13:21









lab bhattacharjeelab bhattacharjee

227k15158277




227k15158277












  • $begingroup$
    OK, I see it. So, $$int_{0}^{3} ({e^t} +{e^{-t}})dt$$ $$=[e^t+e^{-t}]_{0}^{3} $$ $$=e^3 - e^{-3}$$
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:42




















  • $begingroup$
    OK, I see it. So, $$int_{0}^{3} ({e^t} +{e^{-t}})dt$$ $$=[e^t+e^{-t}]_{0}^{3} $$ $$=e^3 - e^{-3}$$
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:42


















$begingroup$
OK, I see it. So, $$int_{0}^{3} ({e^t} +{e^{-t}})dt$$ $$=[e^t+e^{-t}]_{0}^{3} $$ $$=e^3 - e^{-3}$$
$endgroup$
– Maggie
Dec 30 '18 at 13:42






$begingroup$
OK, I see it. So, $$int_{0}^{3} ({e^t} +{e^{-t}})dt$$ $$=[e^t+e^{-t}]_{0}^{3} $$ $$=e^3 - e^{-3}$$
$endgroup$
– Maggie
Dec 30 '18 at 13:42













4












$begingroup$

It is faster to use some hyperbolic trigonometry: as $x(t)=2cosh t$, we have $;x'(t)=2sinh t$, so
begin{align}
ell&=int_0^3sqrt{4(sinh^2t+1)},mathrm d t=2int_0^3cosh t,mathrm d t=2sinh tBiggmvert_0^3=2sinh 3=mathrm e^3-mathrm e^{-3}.
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This awesome! I forgot the formula $cosh^2 t - sinh^2 t =1$.
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:56
















4












$begingroup$

It is faster to use some hyperbolic trigonometry: as $x(t)=2cosh t$, we have $;x'(t)=2sinh t$, so
begin{align}
ell&=int_0^3sqrt{4(sinh^2t+1)},mathrm d t=2int_0^3cosh t,mathrm d t=2sinh tBiggmvert_0^3=2sinh 3=mathrm e^3-mathrm e^{-3}.
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This awesome! I forgot the formula $cosh^2 t - sinh^2 t =1$.
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:56














4












4








4





$begingroup$

It is faster to use some hyperbolic trigonometry: as $x(t)=2cosh t$, we have $;x'(t)=2sinh t$, so
begin{align}
ell&=int_0^3sqrt{4(sinh^2t+1)},mathrm d t=2int_0^3cosh t,mathrm d t=2sinh tBiggmvert_0^3=2sinh 3=mathrm e^3-mathrm e^{-3}.
end{align}






share|cite|improve this answer









$endgroup$



It is faster to use some hyperbolic trigonometry: as $x(t)=2cosh t$, we have $;x'(t)=2sinh t$, so
begin{align}
ell&=int_0^3sqrt{4(sinh^2t+1)},mathrm d t=2int_0^3cosh t,mathrm d t=2sinh tBiggmvert_0^3=2sinh 3=mathrm e^3-mathrm e^{-3}.
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 13:35









BernardBernard

123k741117




123k741117












  • $begingroup$
    This awesome! I forgot the formula $cosh^2 t - sinh^2 t =1$.
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:56


















  • $begingroup$
    This awesome! I forgot the formula $cosh^2 t - sinh^2 t =1$.
    $endgroup$
    – Maggie
    Dec 30 '18 at 13:56
















$begingroup$
This awesome! I forgot the formula $cosh^2 t - sinh^2 t =1$.
$endgroup$
– Maggie
Dec 30 '18 at 13:56




$begingroup$
This awesome! I forgot the formula $cosh^2 t - sinh^2 t =1$.
$endgroup$
– Maggie
Dec 30 '18 at 13:56


















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