Divergence Velocity F8A810
1. نبدأ بكتابة متجه السرعة المعطى:
$$\mathbf{V} = (6tx + z^2 y) \mathbf{i} + (3t + xy^2) \mathbf{j} + (xy - 2xyz - 6tz) \mathbf{k}$$
2. المطلوب هو حساب التباعد (divergence) للمتجه، وهو:
$$\nabla \cdot \mathbf{V} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}$$
حيث:
- $u = 6tx + z^2 y$
- $v = 3t + xy^2$
- $w = xy - 2xyz - 6tz$
3. نحسب كل مشتقة جزئية:
- $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(6tx + z^2 y) = 6t + 0 = 6t$
- $\frac{\partial v}{\partial y} = \frac{\partial}{\partial y}(3t + xy^2) = 0 + x \cdot 2y = 2xy$
- $\frac{\partial w}{\partial z} = \frac{\partial}{\partial z}(xy - 2xyz - 6tz) = 0 - 2xy - 6t = -2xy - 6t$
4. نجمع المشتقات الجزئية:
$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 6t + 2xy + (-2xy - 6t) = 6t + 2xy - 2xy - 6t = 0$$
5. إذن، التباعد يساوي صفر.
الجواب الصحيح هو: c. 0