Subjects vector calculus

Divergence Velocity F8A810

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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