Suppose we have the equation V = 1 3 π r 2 h. Find d r d h .[Chain Rule]We have d V d h = d V d r ⋅ d r d h { d V d h = 1 3 π r 2 d V d r = 2 3 π r h ⟹ 1 3 π r 2 = 2 3 π r h ⋅ d r d h ⟹ r = 2 h ⋅ d r d h ⟹ d r d h = r 2 h [Implicit Differentiation] d d h V = π 3 d d h ( r 2 h ) 0 = π 3 ( r 2 ⋅ d d h h + h d d h r 2 ) 0 = r 2 + 2 ⋅ h ⋅ r d r d h d r d h = − r 2 2 ⋅ h ⋅ r = − r 2 h Why does the solution using the chain rule method have a different sign compablack to the one using implicit differentiation? Did I make any mistake?

Question

Suppose we have the equation   V  =      1    3    π      r    2    h. Find             d      r              d      h      .[Chain Rule]We have             d      V              d      h        =            d      V              d      r        ⋅            d      r              d      h            {                                                      d              V                                      d              h                                =                      1            3                    π                      r            2                                                                          d              V                                      d              r                                =                      2            3                    π          r          h                            ⟹        1    3    π      r    2    =      2    3    π  r  h  ⋅            d      r              d      h          ⟹    r  =  2  h  ⋅            d      r              d      h          ⟹              d      r              d      h        =      r          2      h      [Implicit Differentiation]                              d                      d            h                          V                            =                  π          3                          d                      d            h                                                (                                    r          2                h                              )                                              0                            =                  π          3                                      (                                    r          2                ⋅                  d                      d            h                          h        +        h                  d                      d            h                                    r          2                                      )                                              0                            =                  r          2                +        2        ⋅        h        ⋅        r                              d            r                                d            h                                                                    d            r                                d            h                                              =        −                              r            2                                2            ⋅            h            ⋅            r                          =        −                  r                      2            h                              Why does the solution using the chain rule method have a different sign compablack to the one using implicit differentiation? Did I make any mistake?

Arjun Wright · Accepted Answer

There&#039;s a few things going on here. It&#039;s important to distinguish between V(r,h) the function and V the constant.First, your chain rule is wrong (for what you are asking). It should be:                    d        V                    d        h              =                    ∂        V        (        r        ,        h        )                    ∂        r                                d        r                    d        h              +                    ∂        V        (        r        ,        h        )                    ∂        h                                d        h                    d        h            Notice that the left hand side is V the constant, which when you did implicit differentiation you set equal to zero. Then we have                                                        d              V                                      d              h                                                          =                                            ∂              V              (              r              ,              h              )                                      ∂              r                                                                          d              r                                      d              h                                      +                                            ∂              V              (              r              ,              h              )                                      ∂              h                                                                          d              h                                      d              h                                                          0                            =        (                              2            3                          π        r        h        )                                            d              r                                      d              h                                      +                              1            3                          π                  r          2                (        1        )                                    ⟹                                                    d              r                                      d              h                                                          =        −                              r                          2              h                                          Now chain rule gives the same answer as implicit differentiation.To give a bit more intuiton.Your chain rule calculation that you originally had answeblack said &quot;How much do I have to change r by to get the same volume change if I changed h?&quot;Your implicit differentiation question said &quot;How much do I have to change r by to keep volume the same if I changed h?&quot;

sveiparnu · Accepted Answer

You are solving two different problems.In your first part, you have V a function of h and r where r and h are both independent variables and V is the dependent variable. That means your            d      r              d      h        =  0and            d      h              d      r        =  0In your second part, you keep   V constant in which case you have             d      V              d      h        =  0.. and             d      V              d      r        =  0. but you can solve for             d      h              d      r       or             d      r              d      h

Suppose we have the equation V=1/3 pir^2h. Find dr/dh. [Chain Rule] We have dV/dh=dV/dr x dr/dh. {dV/dh=1/3 pir^2 dV/dr=2/3 pirh=>1/3 pir^2=2/3 pirh x dr/dh=>r=2h x dr/dh=>dr/dh=r/2h

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New Questions in Calculus 2