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Let $ r = \sqrt {x^2+y^2}$ and $\theta= tan^{-1}(y/x)$ be the usual polar/rectangular relationships. Furthermore, define $u(r(x,y),\theta(x,y)) = -sech^2(r)tanh(r)sin(\theta)$ and $v(r(x,y),\theta(x,y)) = sech^2(r)tanh(r)cos(\theta)$.

Show that $\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = sech^2(r)[sech^2(r)-2tanh^2(r)+\frac{tanh(r)}{r}] $

How would I go about doing this? I know I would have to use the chain rule, but how exactly would I go about finding $\frac{\partial v}{\partial x}$ and $\frac{\partial u}{\partial y}$?

Thanks for the help.

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2 Answers 2

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Hint $$x=r\cos\theta\quad,\quad y=r \sin\theta$$ $$\frac{\partial v}{\partial r}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial r}$$ $$-2{{\operatorname{sech}}^{2}}\,r\,{{\tanh }^{2}}r\cos \theta +{{\operatorname{sech}}^{4}}\,r\cos \theta=\cos \theta\frac{\partial v}{\partial x} +\sin \theta\frac{\partial v}{\partial y}\quad(1)$$ $$\frac{\partial v}{\partial \theta }=\frac{\partial v}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial v}{\partial y}\frac{\partial y}{\partial \theta }$$ $$-{{\operatorname{sech}}^{2}}r\tanh \,r\,\sin \theta =-r\sin \theta \frac{\partial v}{\partial x}+r\cos \theta \frac{\partial v}{\partial y}\quad(2)$$ we have $$\left\{ \begin{align} & \quad\,\,\,\cos \theta \frac{\partial v}{\partial x}\,\,+\,\,\,\,\,\sin \theta \frac{\partial v}{\partial y}=-2{{\operatorname{sech}}^{2}}r\,{{\tanh }^{2}}r\cos \theta +{{\operatorname{sech}}^{4}}r\cos \theta \\ & -r\sin \theta \frac{\partial v}{\partial x}\,+r\cos \theta \frac{\partial v}{\partial y}=-{{\operatorname{sech}}^{2}}r\tanh r\sin \theta \\ \end{align} \right.$$ $$\frac{\partial v}{\partial x}=?$$ Similarly $$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$$ $$2{{\operatorname{sech}}^{2}}\,r\,{{\tanh }^{2}}r\sin \theta -{{\operatorname{sech}}^{4}}\,r\sin \theta=\cos \theta\frac{\partial u}{\partial x} +\sin \theta\frac{\partial u}{\partial y}\quad(3)$$ $$\frac{\partial u}{\partial \theta }=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta }$$ $${{\operatorname{sech}}^{2}}r\tanh \,r\,\cos \theta =-r\sin \theta \frac{\partial u}{\partial x}+r\cos \theta \frac{\partial u}{\partial y}\quad(4)$$ we have $$\left\{ \begin{align} &\quad\,\,\,\cos \theta \frac{\partial u}{\partial x}+\,\,\,\,\sin \theta \frac{\partial u}{\partial y}=2{{\operatorname{sech}}^{2}}r{{\tanh }^{2}}r\sin \theta -{{\operatorname{sech}}^{4}}r\sin \theta \\ & -r\sin \theta \frac{\partial u}{\partial x}+r\cos \theta \frac{\partial u}{\partial y}={{\operatorname{sech}}^{2}}r\tanh r\cos \theta \\ \end{align} \right.$$ $$\frac{\partial u}{\partial x}=?$$

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Use the chain rule: $$\dfrac{\partial v}{\partial x} = \dfrac{\partial v}{\partial r}\dfrac{\partial r}{\partial x} + \dfrac{\partial v}{\partial \theta}\dfrac{\partial \theta}{\partial x}$$ $$\dfrac{\partial u}{\partial y} = \dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial y} + \dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial y}$$ for example: $$\dfrac{\partial v}{\partial r}=\left( \text{sech}^4r-2\text{tanh}^2(r)\,\text{sech}^2r\right)\cos\theta$$ ...etc

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