Integral of a vector difference












1














Not sure how to deal with this problem:



$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{2}}right)$$



where $r_2$ and $r_1$ are vectors. The parameter $t$ represents time. $h$ is the specific angular momentum. The vector direction of the right side term, I suppose, would be $(r_2-r_1)$.



This is as far as I get. But I’m not sure if I’m headed in the right direction:



$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}left( overrightarrow {r_2}-overrightarrow {r_1}right) right)$$



$$int dfrac {doverrightarrow {h}}{dt} dt=int -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}r_2r_1sin theta dt$$



$$h=int dfrac {mr_2r_1sin theta }{left( left( r_{2}cos theta _{2}-r_{1}cos theta _{1}right) +left( r_{2}sin theta _{2}-r_{1}sin theta _{1}right) right) ^{frac {3}{2}}} dt$$



$$h=dfrac {mr_{2}r_{1}sin theta }{left( overrightarrow {r}_{2}-overrightarrow {r}_{1}right) ^{3}}t+C$$






calculus







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    1














    Not sure how to deal with this problem:



    $$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{2}}right)$$



    where $r_2$ and $r_1$ are vectors. The parameter $t$ represents time. $h$ is the specific angular momentum. The vector direction of the right side term, I suppose, would be $(r_2-r_1)$.



    This is as far as I get. But I’m not sure if I’m headed in the right direction:



    $$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}left( overrightarrow {r_2}-overrightarrow {r_1}right) right)$$



    $$int dfrac {doverrightarrow {h}}{dt} dt=int -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}r_2r_1sin theta dt$$



    $$h=int dfrac {mr_2r_1sin theta }{left( left( r_{2}cos theta _{2}-r_{1}cos theta _{1}right) +left( r_{2}sin theta _{2}-r_{1}sin theta _{1}right) right) ^{frac {3}{2}}} dt$$



    $$h=dfrac {mr_{2}r_{1}sin theta }{left( overrightarrow {r}_{2}-overrightarrow {r}_{1}right) ^{3}}t+C$$






    calculus







    share|cite|improve this question



























      1












      1








      1







      Not sure how to deal with this problem:



      $$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{2}}right)$$



      where $r_2$ and $r_1$ are vectors. The parameter $t$ represents time. $h$ is the specific angular momentum. The vector direction of the right side term, I suppose, would be $(r_2-r_1)$.



      This is as far as I get. But I’m not sure if I’m headed in the right direction:



      $$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}left( overrightarrow {r_2}-overrightarrow {r_1}right) right)$$



      $$int dfrac {doverrightarrow {h}}{dt} dt=int -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}r_2r_1sin theta dt$$



      $$h=int dfrac {mr_2r_1sin theta }{left( left( r_{2}cos theta _{2}-r_{1}cos theta _{1}right) +left( r_{2}sin theta _{2}-r_{1}sin theta _{1}right) right) ^{frac {3}{2}}} dt$$



      $$h=dfrac {mr_{2}r_{1}sin theta }{left( overrightarrow {r}_{2}-overrightarrow {r}_{1}right) ^{3}}t+C$$






      calculus







      share|cite|improve this question















      Not sure how to deal with this problem:



      $$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{2}}right)$$



      where $r_2$ and $r_1$ are vectors. The parameter $t$ represents time. $h$ is the specific angular momentum. The vector direction of the right side term, I suppose, would be $(r_2-r_1)$.



      This is as far as I get. But I’m not sure if I’m headed in the right direction:



      $$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}left( overrightarrow {r_2}-overrightarrow {r_1}right) right)$$



      $$int dfrac {doverrightarrow {h}}{dt} dt=int -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}r_2r_1sin theta dt$$



      $$h=int dfrac {mr_2r_1sin theta }{left( left( r_{2}cos theta _{2}-r_{1}cos theta _{1}right) +left( r_{2}sin theta _{2}-r_{1}sin theta _{1}right) right) ^{frac {3}{2}}} dt$$



      $$h=dfrac {mr_{2}r_{1}sin theta }{left( overrightarrow {r}_{2}-overrightarrow {r}_{1}right) ^{3}}t+C$$






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 23:38







      alexmarison

















      asked Nov 29 '18 at 3:17









      alexmarisonalexmarison

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