Answer to Question #270071 in Calculus for Festus

Question

If x(t+y) = t^2 + y^2. Show that (partial x/partial t - partial x/partial y)^2 = 4[1 - partial x/partial t - partial x/partial y]

Answer 1

$x(t+y) = t^2 + y^2\ \ \ \ \ \ \ \ (1)$

Differentiate both sides of the equation (1) with respect to $t$

$\dfrac{\partial x}{\partial t}(t+y)+x=2t$

Differentiate both sides of the equation (1) with respect to $y$

$\dfrac{\partial x}{\partial y}(t+y)+x=2y$

$\dfrac{\partial x}{\partial t}=\dfrac{2t-x}{t+y}$

$\dfrac{\partial x}{\partial y}=\dfrac{2y-x}{t+y}$

$\dfrac{\partial x}{\partial t}+\dfrac{\partial x}{\partial y}=\dfrac{2t+2y-2x}{t+y}$

$\dfrac{\partial x}{\partial t}-\dfrac{\partial x}{\partial y}=\dfrac{2t-2y}{t+y}$

$1-\dfrac{\partial x}{\partial t}-\dfrac{\partial x}{\partial y}=\dfrac{t+y-2t-2y+2x}{t+y}$

$=\dfrac{2x-(t+y)}{t+y}=\dfrac{2x(t+y)-(t+y)^2}{(t+y)^2}$

$=\dfrac{2(t^2+y^2)-t^2-2ty-y^2}{(t+y)^2}$

$=\dfrac{t^2+y^2-2ty}{(t+y)^2}=\dfrac{(t-y)^2}{(t+y)^2}$

Then

$(\dfrac{\partial x}{\partial t}-\dfrac{\partial x}{\partial y})^2=(\dfrac{2t-2y}{t+y})^2$

$=4\cdot\dfrac{(t-y)^2}{(t+y)^2}=4(1-\dfrac{\partial x}{\partial t}-\dfrac{\partial x}{\partial y})$

Therefore

$(\dfrac{\partial x}{\partial t}-\dfrac{\partial x}{\partial y})^2=4(1-\dfrac{\partial x}{\partial t}-\dfrac{\partial x}{\partial y})$