z=sqrt{x^2+y^2},x=e^2t,y=e^-2t2z/2t ਨੂੰ ਹੱਲ ਕਰੋ

$$z = \sqrt { x ^ { 2 } + y ^ { 2 } } , x = e ^ { 2 t } , y = e ^ { - 2 t } \frac { 2 z } { 2 t }$$

x, y, z ਲਈ ਹਲ ਕਰੋ

$$\left\{\begin{matrix}x=e^{2t}\text{, }y=-\frac{e^{2t}}{\sqrt{t^{2}e^{4t}-1}}\text{, }z=e^{4t}\sqrt{\frac{t^{2}}{t^{2}e^{4t}-1}}\text{, }&t^{2}e^{4t}\neq 1\text{ and }\frac{e^{4t}}{t^{2}e^{4t}-1}\geq 0\text{ and }t^{2}e^{4t}-1>0\text{ and }t<0\\x=e^{2t}\text{, }y=\frac{e^{2t}}{\sqrt{t^{2}e^{4t}-1}}\text{, }z=e^{4t}\sqrt{\frac{t^{2}}{t^{2}e^{4t}-1}}\text{, }&t^{2}e^{4t}\neq 1\text{ and }\frac{e^{4t}}{t^{2}e^{4t}-1}\geq 0\text{ and }t^{2}e^{4t}-1>0\text{ and }t>0\end{matrix}\right.$$