Consider the curve a ( t ) − ( 2 t , t 2 , log ⁡ t ) on I : t &gt; 0. Show that this curve passes through the points p = ( 2 , 1 , 0 ) and q = ( 4 , 4 , log ⁡ 2 ) and find its arc length between these points.

Question

Consider the curve   a  (  t  )  −  (  2  t  ,      t    2    ,      log            ⁡      t    )   on   I  :  t  &amp;gt;  0. Show that this curve passes through the points   p  =  (  2  ,  1  ,  0  )   and   q  =  (  4  ,  4  ,      log            ⁡      2    ) and find its arc length between these points.

iljovskint · Accepted Answer

Consider the equation of the curve  a  (  t  )  =  ⟨  2  t  ,      t    2    .  log  ⁡  t  ⟩  ,  t  ≥  0To show that the curve   a  (  t  )  =  ⟨  2  t  ,      t    2    .  log  ⁡  t  ⟩  is passingthrough the points P=(2,1,0) and Q=(4,4, log2)Compare the equation of the curve with the general curveequation   ⟨  x  (  t  )  ,  y  (  t  )  ,  z  (  t  )  ⟩ gives  x  (  t  )  =  2  t  ,  y  (  t  )  =      t    2    ,  z  (  t  )  =  log  ⁡  tNow substitute the point P=(2,1,0) in x, y, z gives  2  =  2  t  ,  1  =      t    2    ,  0  =  log  ⁡  tSolving these equations gives t=1Thus, for t=1, the curve passes throug the point PNow substitute the point Q=(4,4, log 2) in x,y,z gives  4  =  2  t  ,  4  =      t    2    ,  log  ⁡  2  =  log  ⁡  tSolving these equations gives t=2Thus, for t=2, the curve passes throught the point QHence, the curve passes through the point P = (2,1,0) and Q = (4,4, log 2)Consider the equation of the curve  a  (  t  )  =  ⟨  2  t  ,      t    2    .  log  ⁡  t  ⟩  ,  t  ≥  0Find the arc lenght of curve.From the point P = (2,1,0), we get  (  2  t  ,      t    2    log  ⁡  t  )  =  (  2  ,  1  ,  0  )  ⇒  2  t  =  2  ⇒  t  =  1From the point Q = (4,4, log2), we get  (  2  t  ,      t    2    log  ⁡  t  )  =  (  4  ,  4  ,  log  ⁡  2  )  ⇒  2  t  =  4  ⇒  t  =  2Thus, the limits of t are   1  ≤  t  ≤  2The length of the curve x=f(t), y=g(t), z=h(t)  a  ≤  t  ≤  b  i  s  L  =      ∫          a              b            [                            d          x                          d          t                    2        ]    +    [                            d          y                          d          t                    2        ]    +    [                            d          z                          d          t                    2        ]    d    t    Differentiate x and y with respect to t            d      x              d      t        =      d          d      t        (  2  t  )  =  2              d      y              d      t        =      d          d      t        (      t    2    )  =  2  t              d      z              d      t        =      d          d      t        (  log  ⁡  t  )  =      1    t  The length of the curve is  a  ≤  t  ≤  b  i  s  L  =      ∫          1              2            [                            d          x                          d          t                    2        ]    +    [                            d          y                          d          t                    2        ]    +    [                            d          z                          d          t                    2        ]    d    t      =      ∫          1              2            [    2          ]      2        +    [    2          ]      2        =    (          1      t              )      2        d  t    =      ∫          1              2            4    +    4          t      2        +          1              t        2              d  t    =      ∫          1              2                          a                  t          2                +        4                  t          4                +        1                    t        2              d  t    =      ∫          1              2                          (        2                  t          2                +        1                  )          2                            t        2              d  t    =      ∫          1              2                          2                  t          2                +        1            t        d  t    =      ∫          1              2            2    t    +          1      t        d  t    =  [      t    2    ln  ⁡  t      ]    1    2      =  [      2    2    +  ln  ⁡  2  −      1    2    −  ln  ⁡  1  ]    =  [  4  +  ln  ⁡  2  −  1  −  0  ]    =  3  +  ln  ⁡  2Hence the length of the curve is L=3+ln2

Consider the curve a(t) - (2t, t^2, log t) on I: t > 0. Show that this curve passes through the points p = (2,1,0) and q = (4,4,log2) and find its arc length between these points.

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