Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 6 M

Year : Dec 2012

Let, $ f_1 \;=\;u+e^{-v}sinu-x=0 \; \& \; f_2 \;=\; v+e^{-u} cosu-y=0 \\ \; \\ \; \\ Now, \\ \dfrac{\partial u}{\partial y} \;=\; \dfrac{ \dfrac{\partial(f_1,f_2)}{ \partial(y,v) } }{ \dfrac{\partial(f_1,f_2)}{ \partial(u,v) } } \\ \; \\ \; \\ \dfrac{\partial(f_1,f_2)}{ \partial(y,v)} \;=\; \left| \begin{array}{cc} \frac{\partial f_1 }{\partial y} & \frac{\partial f_1 }{\partial v} \\ \frac{\partial f_2 }{\partial y} & \frac{\partial f_2 }{\partial v} \end{array} \right| \;=\; \left| \begin{array}{cc} 0 & -e^{-v} sinu \\ -1 & 1-e^{-v} cosu \end{array} \right| \\ \; \\ \; \\ = 0-(-1)(-e^{-v} sinu ) \;=\; -e^{-v} sinu \\ \; \\ \; \\ \dfrac{\partial(f_1,f_2)}{ \partial(u,v)} \;=\; \left| \begin{array}{cc} \frac{\partial f_1 }{\partial u} & \frac{\partial f_1 }{\partial v} \\ \frac{\partial f_2 }{\partial u} & \frac{\partial f_2 }{\partial v} \end{array} \right| \;=\; \left| \begin{array}{cc} 1+e^{-v}cosu & -e^{-v} sinu \\ 1- e^{-u} sinu -e^{-v} cosu & 1 \end{array} \right| \\ \; \\ \; \\ = 0-(-1)(-e^{-v} sinu ) \;=\; -e^{-v} sinu \\ \; \\ \; \\ = 1+e^{-v}cosu - e^{-(u+v)} sin^2u - e^{-(u+v)} sinu \; cosu \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial y} \;=\; \dfrac{ -e^{-v} sinu } { 1+e^{-v}cosu - e^{-(u+v)} sin^2u - e^{-(u+v)} sinu \; cosu } \\ \; \\ \; \\ = \dfrac{ -e^{-v} sinu } { 1+e^{-v}cosu - e^{-(u+v)} (sinu) (sinu + cosu) } \\ \; \\ \; \\ Now, \\ \dfrac{\partial v}{\partial x} \;=\; -\dfrac{ \dfrac{\partial(f_1,f_2)}{ \partial(u,x) } }{ \dfrac{\partial(f_1,f_2)}{ \partial(u,v) } } \\ \; \\ \; \\ \dfrac{\partial(f_1,f_2)}{ \partial(u,x) } \;=\; \left| \begin{array}{cc} \frac{\partial f_1 }{\partial u} & \frac{\partial f_1 }{\partial x} \\ \frac{\partial f_2 }{\partial u} & \frac{\partial f_2 }{\partial x} \end{array} \right| \; = \; \left| \begin{array}{cc} 1+e^{-v} cosu & -1 \\ -e^{-v} sinu & 0 \end{array} \right| \\ \; \\ \; \\ = -e^{-v} sinu \\ \; \\ \; \\ \dfrac{\partial(f_1,f_2)}{ \partial(u,v) }\;=\; 1+e^{-v}cosu - e^{-(u+v)} (sinu) (sinu + cosu) \\ \; \\ \; \\ \therefore \dfrac{\partial v}{\partial x} \;=\; \dfrac{ -e^{-v} sinu } { 1+e^{-v}cosu - e^{-(u+v)} (sinu) (sinu + cosu) } \\ \; \\ \; \\ \therefore \dfrac{\partial v}{\partial x} \;=\; \dfrac{\partial u}{\partial y} \\ \; \\ \; \\ \; \\ $

Note: Actually, there seems a printing mistake in the question.

When referred in textbook(H.K.Dass), y was equal to $ v+e^{-v}cosu$ and not $ v+e^{-v}cosu$ . In that case, answer comes
$ \dfrac{\partial v}{\partial x} \;=\; \dfrac{\partial u}{\partial y} \;=\; \dfrac{e^{-v} sinu}{1-e^{-2v}} $

However, steps remain the same as above.