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{\bf Question}

Find $\displaystyle \frac {dz}{dt}$, when $z = (x^2 +
y^2)^{\frac{1}{2}}, \, x = (t - 1)^2, \, y = 2(t - 1)$

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{\bf Answer}

\begin{eqnarray*} \frac{dz}{dt} & = & \frac{\partial z}{\partial
x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial
y}\frac{\partial y}{\partial t}; \hspace{.2in} \begin{array}{c} x
= (t - 1)^2 \\ y = 2(t - 1) \end{array} \\ \frac{dx}{dt} & = &
2(t-1); \hspace{.2in} \frac{dy}{dt} = 2 \\ \frac{\partial
z}{\partial x} & = & \frac{1}{2}(x^2 + y^2)^{-\frac{1}{2}} \times
2x = \frac{x}{(x^2 + y^2)^{\frac{1}{2}}} \\ \frac{\partial
z}{\partial y} & = & \frac{1}{2}(x^2 + y^2)^{-\frac{1}{2}} \times
2y = \frac{y}{(x^2 + y^2)^{\frac{1}{2}}} \\ \frac{dz}{dt} & = &
\frac{1}{(x^2 + y^2)^{-\frac{1}{2}}} \{ 2x(t-1) + 2y \} \\ & = &
\frac{2[(t-1)^3 + 2(t-1)]}{[(t-1)^4 + 4(t-1)^2]^\frac{1}{2}} \\ &
= & \frac{2(t-1)([t-1]^2 + 2)}{(t-1)[(t-1)^2 + 4]^{\frac{1}{2}}}
\\ & = & \frac{2[(t-1)^2 + 2]}{[(t-1)^2 + 4]^{\frac{1}{2}}}
\end{eqnarray*}



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