Answer
a) $-10$
b) $-3\sqrt{6}$
Work Step by Step
a) To Find: $\frac{dy}{dt} = \frac{dy}{dx}\times\frac{dx}{dt} = \frac{dy}{dx}\times(-5) \rightarrow①$
Implicitly differentiating the given equation $w.r.t.x$:
$\implies 2x+(2y\times\frac{dy}{dx}) = 2+4\times\frac{dy}{dx}$
Grouping,
$\implies (2x-2)-(4\times\frac{dy}{dx}-2y\times\frac{dy}{dx})$
$\implies 2(x-1) = 2\frac{dy}{dx}\times(2-y)$
$\implies \frac{dy}{dx} = \frac{x-1}{2-y}$
Substituting this in $①$, and substituting given values (x,y) = (3,1):
$\implies \frac{dy}{dt} = -5\times\frac{3-1}{2-1} = -10$
b) To Find: $\frac{dx}{dt} = \frac{dx}{dy}\times\frac{dy}{dt} = \frac{dx}{dy}\times6\rightarrow②$
$\implies \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{2-y}{x-1}$
Substituting this in ②, and substituting given values (x,y) = ($1+\sqrt{2}, 2+\sqrt{3}$):
$\implies \frac{dx}{dt} = 6\times\frac{2-2-\sqrt{3}}{1+\sqrt{2}-1} = \frac{-6\sqrt{3}}{\sqrt{2}}$
Rationalizing,
$\implies \frac{dx}{dt} = \frac{-6\sqrt{3}}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{-6\sqrt{6}}{2} = -3\sqrt{6}$
