What is wrong with my solution?
$$ S=f(x, y)=\left\{\begin{array}{ll} \displaystyle \frac{x y}{x^{2}+y^{2}} & ((x, y) \neq(0,0)) \\ 0 & ((x, y)=(0,0)) \end{array}\right. $$
$x=2t, y=3t$
- $\displaystyle\left(\frac{d S}{dt}\right)_{t=0}$
My solution:
When $t \neq 0$, $$S = \frac{2t \times 3t}{(2t)^2 + (3t)^2} = \frac{6}{13}$$
$$\displaystyle\left(\frac{d S}{dt}\right)_{t=0} = 0$$
When $t = 0 $, $$S=0$$ $$\displaystyle\left(\frac{d S}{dt}\right)_{t=0} = 0$$
The answer: $$\displaystyle\left(\frac{d S}{dt}\right)_{t=0} = 0$$
- When $t=0$, how $\displaystyle\frac{\partial S}{\partial x} \frac{d x}{d t}+\frac{\partial S}{\partial y} \frac{d y}{d t}$ would be?
My solution:
When $t \neq 0$, $$\begin{eqnarray} \displaystyle\frac{\partial S}{\partial x} \frac{d x}{d t}+\frac{\partial S}{\partial y} \frac{d y}{d t} &=& \frac{y\left(x^{2}+y^{2}\right)-x y \times 2 x}{\left(x^{2}+y^{2}\right)^{2}} \times 2+\frac{x\left(x^{2}+y^{2}\right)-xy \times 2y}{\left(x^{2}+y\right)^{2}} \times 3 \\ &=& \frac{2 y\left(y^{2}-x^{2}\right)}{\left(x+y^{2}\right)^{2}}+\frac{3 x\left(x^{2}-y^{2}\right)}{\left( x^2+y^{2}\right)^{2}} \\ &=& \frac{3 x^{3}-2 x^{2} y-3 x y^{2}+2 y^{3}}{\left(x^{2}+y^{2}\right)^{2}} \\ &=& 0 \end{eqnarray}$$
$$\displaystyle\frac{\partial S}{\partial x} \frac{d x}{d t}+\frac{\partial S}{\partial y} \frac{d y}{d t} = 0$$
When $t = 0 $, $$S=0$$ $$\displaystyle\frac{\partial S}{\partial x} \frac{d x}{d t}+\frac{\partial S}{\partial y} \frac{d y}{d t} = 0$$
The answer: $$\displaystyle\frac{\partial S}{\partial x} \frac{d x}{d t}+\frac{\partial S}{\partial y} \frac{d y}{d t} = 0$$