Question: If $x=ucosv \;,\; y=usinv \;$ prove that $\dfrac{\partial(x,y)}{\partial(u,v)} \cdot \dfrac{\partial(u,v)}{\partial(x,y)} \;=\; 1$
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Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 6 M

Year : May 2013

 modified 3.0 years ago  • written 3.0 years ago by
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x$=ucosv \; \; \ldots (i) \; \; \; \; \; \; \; \; \; y \;=\; usinc \; \; \ldots (ii) \\ \; \\ \; \\ \therefore \dfrac{\partial x}{\partial u} \;=\; cosv \; \; \; \; \; \; \; \therefore \dfrac{\partial y}{\partial u} \;=\; sinv \\ \; \\ \; \\ \therefore \dfrac{\partial x}{\partial v} \;=\; -usinv \; \; \; \; \; \; \; \therefore \dfrac{\partial y}{\partial v} \;=\; ucosv \\ \; \\ \; \\ \therefore \dfrac{\partial(x,y)}{\partial(u,v)} \;=\; \left| \begin{array}{cc} \frac{\partial x }{\partial u} & \frac{\partial x }{\partial v} \\ \frac{\partial y }{\partial u} & \frac{\partial y }{\partial v} \end{array} \right| \;=\; \left| \begin{array}{ccc} cos v & -usinv \\ sin v & u cosv \end{array} \right| \;=\; u cos^2v - (-usin^v) \\ \; \\ \; \ \ = u cos^2v + usin^2v \\ \; \\ \; \\ = u(1) \;=\; u \\ \; \\ \; \\ \therefore \dfrac{\partial(x,y)}{\partial(u,v)} \;=\; u \; \; \; \ldots (iii) \\ \; \\ \; \\$

Equation(ii) divide by Equation(i)

$\therefore \dfrac{usinv}{ucosv} \;=\; \dfrac{y}{x} \; \; \; \; \; \; \therefore tanv \;=\; \dfrac{y}{x} \\ \; \\ \; \\ \therefore v= \;=\; tan^{-1} \bigg( \dfrac{y}{x} \bigg) \\ \; \\ \; \\ \therefore \dfrac{\partial v}{\partial x} \;=\; \dfrac{1}{1+ \dfrac{y^2}{x^2}} \dfrac{\partial }{\partial x} \bigg( \dfrac{y}{x} \bigg) \;=\; \dfrac{{x^2}}{ {x^2} + {y^2}} \bigg( \dfrac{-y}{x^2} \bigg) \\ \; \\ \; \\ \dfrac{{-y}}{ {x^2} + {y^2}} \\ \; \\ \; \\ \; \\ \dfrac{\partial v}{\partial y} \;=\; \dfrac{1}{1+ \dfrac{y^2}{x^2}} \dfrac{\partial }{\partial y} \bigg( \dfrac{y}{x} \bigg) \;=\; \dfrac{{x^2}}{ {x^2} + {y^2}} \bigg( \dfrac{1}{x} \bigg) \\ \; \\ \; \\ \dfrac{{x}}{ {x^2} + {y^2}} \\ \; \\ \; \\ \; \\$

Squaring and adding equation (i) and (ii)

$\therefore u^2 cos^2v \;=\; x^2 \; \; \; and \; \; \; u^2 sin^2v \;=\; y^2 \\ \; \\ \; \\ \therefore u^2 cos^2v + u^2 sin^2v \;=\; x^2+y^2 \\ \; \\ \; \\ u^2 \;=\; x^2+y^2 \\ \; \\ \; \\ u \;=\; \sqrt{ x^2+y^2 } \\ \; \\ \; \\ \; \\ \dfrac{\partial u}{\partial x} \;=\; \dfrac{1}{2 \sqrt{ x^2+y^2 } } \dfrac{\partial }{\partial x} (x^2+y^2) \;=\; \dfrac{2x}{2 \sqrt{ x^2+y^2 } } \;=\; \dfrac{x}{ \sqrt{ x^2+y^2 } } \\ \; \\ \; \\ \; \\ \dfrac{\partial u}{\partial y} \;=\; \dfrac{1}{2 \sqrt{ x^2+y^2 } } \dfrac{\partial }{\partial y} (x^2+y^2) \;=\; \dfrac{2y}{2 \sqrt{ x^2+y^2 } } \;=\; \dfrac{y}{ \sqrt{ x^2+y^2 } } \\ \; \\ \; \\ \therefore \dfrac{\partial(u,v)}{\partial(x,y)} \;=\; \left| \begin{array}{cc} \frac{\partial u }{\partial x} & \frac{\partial u }{\partial y} \\ \frac{\partial v }{\partial x} & \frac{\partial u }{\partial y} \end{array} \right| \;=\; \left| \begin{array}{ccc} \dfrac{x}{ \sqrt{ x^2+y^2 } } & \dfrac{y}{ \sqrt{ x^2+y^2 } } \\ \dfrac{-y}{ \sqrt{ x^2+y^2 } } & \dfrac{x}{ \sqrt{ x^2+y^2 } } \end{array} \right| \\ \; \\ \; \\ \; \\ \therefore \dfrac{\partial(u,v)}{\partial(x,y)} \;=\; \dfrac{x^2}{ (x^2+y^2)^{3/2} } + \dfrac{y^2}{ (x^2+y^2)^{3/2} } \\ \; \\ \; \\ \dfrac{1}{ (x^2+y^2)}^{1/2} \;=\; \dfrac{1}{ (u^2)}^{1/2} \\ \; \\ \; \\ \therefore \dfrac{\partial(u,v)}{\partial(x,y)} \;=\; \dfrac{1}{u} \; \; \; \; \ldots (iv) \\ \; \\ \; \\ \; \\$

From (iii) and (iv),

$\dfrac{\partial(x,y)}{\partial(u,v)} \cdot \dfrac{\partial(u,v)}{\partial(x,y)} \;=\; \dfrac{1}{u} \cdot u \;=\; 1 \\ \; \\ \; \\ \dfrac{\partial(x,y)}{\partial(u,v)} \cdot \dfrac{\partial(u,v)}{\partial(x,y)} \;=\; 1 \; \; \; \; Hence \; Proved.$