Let $R(s,t) = G(u(s,t), v(s,t))$, where $G$, $u$, and $v$ are differentiable. What is $R_s(1,2)$ and...












1












$begingroup$


Here's everything that's given:



$u(1,2)$ = $5$



$u_s(1,2)$ = $4$



$u_t(1,2)$ = $-3$



$v(1,2)$ = $7$



$v_s(1,2)$ = $2$



$v_t(1,2)$ = $6$



$G_u(5,7)$ = $9$



$G_v(5,7)$ = $-2$



I would post my attempt, but I have no idea where to start. I know I have to split $R_s$ and $R_t$ into a bunch of partial derivatives involving the other variables, but I haven't found anything that works, nor do I have an answer in my book to check for an answer.






multivariable-calculus derivatives partial-derivative chain-rule







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$endgroup$

















    1












    $begingroup$


    Here's everything that's given:



    $u(1,2)$ = $5$



    $u_s(1,2)$ = $4$



    $u_t(1,2)$ = $-3$



    $v(1,2)$ = $7$



    $v_s(1,2)$ = $2$



    $v_t(1,2)$ = $6$



    $G_u(5,7)$ = $9$



    $G_v(5,7)$ = $-2$



    I would post my attempt, but I have no idea where to start. I know I have to split $R_s$ and $R_t$ into a bunch of partial derivatives involving the other variables, but I haven't found anything that works, nor do I have an answer in my book to check for an answer.






    multivariable-calculus derivatives partial-derivative chain-rule







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Here's everything that's given:



      $u(1,2)$ = $5$



      $u_s(1,2)$ = $4$



      $u_t(1,2)$ = $-3$



      $v(1,2)$ = $7$



      $v_s(1,2)$ = $2$



      $v_t(1,2)$ = $6$



      $G_u(5,7)$ = $9$



      $G_v(5,7)$ = $-2$



      I would post my attempt, but I have no idea where to start. I know I have to split $R_s$ and $R_t$ into a bunch of partial derivatives involving the other variables, but I haven't found anything that works, nor do I have an answer in my book to check for an answer.






      multivariable-calculus derivatives partial-derivative chain-rule







      share|cite|improve this question











      $endgroup$




      Here's everything that's given:



      $u(1,2)$ = $5$



      $u_s(1,2)$ = $4$



      $u_t(1,2)$ = $-3$



      $v(1,2)$ = $7$



      $v_s(1,2)$ = $2$



      $v_t(1,2)$ = $6$



      $G_u(5,7)$ = $9$



      $G_v(5,7)$ = $-2$



      I would post my attempt, but I have no idea where to start. I know I have to split $R_s$ and $R_t$ into a bunch of partial derivatives involving the other variables, but I haven't found anything that works, nor do I have an answer in my book to check for an answer.






      multivariable-calculus derivatives partial-derivative chain-rule




      multivariable-calculus derivatives partial-derivative chain-rule






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 10 '15 at 21:28









      JVV

      9981419




      9981419










      asked Nov 9 '15 at 7:12









      ChrisChris

      3691215




      3691215






















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          $begingroup$

          Since $G$ is a function of $u$ and $v$, and $u$ and $v$ are each functions of $s$ and $t$, we have: $$R_s=frac{partial G}{partial u}cdotfrac{partial u}{partial s} + frac{partial G}{partial v}cdotfrac{partial v}{partial s}, quad R_t=frac{partial G}{partial u}cdotfrac{partial u}{partial t} + frac{partial G}{partial v}cdotfrac{partial v}{partial t}.$$



          Now, at the point with coordinates $s=1,t=2$, we obtain: $$R_s(1,2)=G_u(5,7)cdot u_s(1,2)+ G_v(5,7)cdot v_s(1,2)=9cdot4-2cdot2=32.$$



          $$R_t(1,2)=G_u(5,7)cdot u_t(1,2)+ G_v(5,7)cdot v_t(1,2)=9cdot(-3)-2cdot6=-39.$$






          share|cite|improve this answer









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            $begingroup$

            Since $G$ is a function of $u$ and $v$, and $u$ and $v$ are each functions of $s$ and $t$, we have: $$R_s=frac{partial G}{partial u}cdotfrac{partial u}{partial s} + frac{partial G}{partial v}cdotfrac{partial v}{partial s}, quad R_t=frac{partial G}{partial u}cdotfrac{partial u}{partial t} + frac{partial G}{partial v}cdotfrac{partial v}{partial t}.$$



            Now, at the point with coordinates $s=1,t=2$, we obtain: $$R_s(1,2)=G_u(5,7)cdot u_s(1,2)+ G_v(5,7)cdot v_s(1,2)=9cdot4-2cdot2=32.$$



            $$R_t(1,2)=G_u(5,7)cdot u_t(1,2)+ G_v(5,7)cdot v_t(1,2)=9cdot(-3)-2cdot6=-39.$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Since $G$ is a function of $u$ and $v$, and $u$ and $v$ are each functions of $s$ and $t$, we have: $$R_s=frac{partial G}{partial u}cdotfrac{partial u}{partial s} + frac{partial G}{partial v}cdotfrac{partial v}{partial s}, quad R_t=frac{partial G}{partial u}cdotfrac{partial u}{partial t} + frac{partial G}{partial v}cdotfrac{partial v}{partial t}.$$



              Now, at the point with coordinates $s=1,t=2$, we obtain: $$R_s(1,2)=G_u(5,7)cdot u_s(1,2)+ G_v(5,7)cdot v_s(1,2)=9cdot4-2cdot2=32.$$



              $$R_t(1,2)=G_u(5,7)cdot u_t(1,2)+ G_v(5,7)cdot v_t(1,2)=9cdot(-3)-2cdot6=-39.$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Since $G$ is a function of $u$ and $v$, and $u$ and $v$ are each functions of $s$ and $t$, we have: $$R_s=frac{partial G}{partial u}cdotfrac{partial u}{partial s} + frac{partial G}{partial v}cdotfrac{partial v}{partial s}, quad R_t=frac{partial G}{partial u}cdotfrac{partial u}{partial t} + frac{partial G}{partial v}cdotfrac{partial v}{partial t}.$$



                Now, at the point with coordinates $s=1,t=2$, we obtain: $$R_s(1,2)=G_u(5,7)cdot u_s(1,2)+ G_v(5,7)cdot v_s(1,2)=9cdot4-2cdot2=32.$$



                $$R_t(1,2)=G_u(5,7)cdot u_t(1,2)+ G_v(5,7)cdot v_t(1,2)=9cdot(-3)-2cdot6=-39.$$






                share|cite|improve this answer









                $endgroup$



                Since $G$ is a function of $u$ and $v$, and $u$ and $v$ are each functions of $s$ and $t$, we have: $$R_s=frac{partial G}{partial u}cdotfrac{partial u}{partial s} + frac{partial G}{partial v}cdotfrac{partial v}{partial s}, quad R_t=frac{partial G}{partial u}cdotfrac{partial u}{partial t} + frac{partial G}{partial v}cdotfrac{partial v}{partial t}.$$



                Now, at the point with coordinates $s=1,t=2$, we obtain: $$R_s(1,2)=G_u(5,7)cdot u_s(1,2)+ G_v(5,7)cdot v_s(1,2)=9cdot4-2cdot2=32.$$



                $$R_t(1,2)=G_u(5,7)cdot u_t(1,2)+ G_v(5,7)cdot v_t(1,2)=9cdot(-3)-2cdot6=-39.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 10 '15 at 21:25









                JVVJVV

                9981419




                9981419






























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