Problem looks similar to T ( x , y , z ) = ( u , v ) u = x + y + z 2 v = cos ⁡ ( x ) + tan ⁡ ( y ) − z + 3vector r ( t ) = ( t ⋅ e , ( t 2 + 2 ) t , e − s 2 )At point t=0 what's the tangent vector of T(r(t))?Anybody who can help me out with this problem?

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Problem looks similar to  T  (  x  ,  y  ,  z  )  =  (  u  ,  v  )  u  =  x  +  y  +      z    2    v  =  cos  ⁡  (  x  )  +  tan  ⁡  (  y  )  −  z  +  3vector   r  (  t  )  =  (  t  ⋅  e  ,  (      t    2    +  2      )    t    ,      e          −              s        2              )At point t=0 what&#039;s the tangent vector of T(r(t))?Anybody who can help me out with this problem?

thenurssoullu · Accepted Answer

We have the functions   T  :            R        3    →            R        2   and   r  :      R    →            R        3  , so the composition is such that  T  ∘  r  :      R        ⟶    r              R        3        ⟶    T              R        2  r(0)=(0,1,1)  and the jacobian matrix of T in (0,1,1) is      J    T    (  0  ,  1  ,  1  )  =      (                            ∇          u          (          0          ,          1          ,          1          )                                      ∇          v          (          0          ,          0          ,          1          )                      )    =      (                            1                                  1                                  2          z                                      −          sin          ⁡          (          x          )                                                            1                                                cos                  2                                ⁡                (                x                )                                                                      −          1                      )                      |                    (      0      ,      1      ,      1      )        =      (                            1                                  1                                  2                                      0                                              cos                          −              2                                ⁡          (          1          )                                  −          1                      )    ,while the jacobian of r  in 0 is the vector       (                            e                                      log          ⁡          (          2          )                                      0                      )   , then for the rule of composition we get      J          T      ∘      r        (  0  )  =      (                            1                                  1                                  2                                      0                                              cos                          −              1                                ⁡          (          1          )                                  −          1                      )    ⋅      (                            e                                      log          ⁡          (          2          )                                      0                      )    =      (                            e          +          log          ⁡          (          2          )                                      log          ⁡          (          2          )                      cos                          −              2                                ⁡          (          1          )                      )

Marisol Rivers · Accepted Answer

Its an application of the Chain rule. Note that   r  (  0  )  =  (  0  ,  1  ,  1  ) and       r    ′    (  0  )  =  (  e  ,  ln  ⁡  2  ,  0  ). By the chain rule                    (*)                              ∂                      ∂            t                                                              |                                            t            =            0                          T        (        r        (        t        )        )        =        J        (        T                  )                      r            (            0            )                          ⋅                  r          ′                (        0        )            where   J  (  T      )          r      (      0      )       is the Jacobian of T evaluated at r(0). That is,                    J        (        T                  )                      r            (            0            )                                              =                              (                                                                                                      ∂                      u                                                              ∂                      x                                                                                                                                  ∂                      u                                                              ∂                      y                                                                                                                                  ∂                      u                                                              ∂                      z                                                                                                                                                              ∂                      v                                                              ∂                      x                                                                                                                                  ∂                      v                                                              ∂                      y                                                                                                                                  ∂                      v                                                              ∂                      z                                                                                            )                                                              |                                      r            (            0            )                                                                          =                              (                                                            1                                                  1                                                  2                  z                                                                              −                  sin                  ⁡                  x                                                                      1                                                                  cos                        2                                            ⁡                      y                                                                                        −                  1                                                      )                                (            0            ,            1            ,            1            )                                                                          =                  (                                                    1                                            1                                            2                                                                    0                                                              1                                                            cos                      2                                        ⁡                    1                                                                              −                1                                              )                    Multiplying the above matrix with the vector       r    ′    (  0  )  =  (  e  ,  ln  ⁡  2  ,  0  ) and using (∗) we obtain                              ∂                      ∂            t                                                              |                                            t            =            0                          T        (        r        (        t        )        )                            =                  (                                                    1                                            1                                            2                                                                    0                                                              1                                                            cos                      2                                        ⁡                    1                                                                              −                1                                              )                ⋅                  (                                                    e                                                                    ln                ⁡                2                                                                    0                                              )                                                                      =        (        e        +        ln        ⁡        2        ,                                       ln            ⁡            2                                              cos              2                        ⁡            1                          )            the desired tangent vector of   T  (  r  (  t  )  ) at t=0

The problem looks similar to T(x,y,z)=(u,v) u=x+y+z^2 v=cos(x)+tan(y)−z+3 vector r(t)=(t * e,(t^2+2)^t,e^(-s^2)) At point t=0 what's the tangent vector of T(r(t))?

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