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ayao
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For directional derivatives:
Let [itex]\hat{u}=<a,b,c>[/itex] be the direction.
Thus, [itex]\frac{∂\hat{u}}{∂x}=\frac{\sqrt{a^2+b^2+c^2}}{a}[/itex] and so on. So,
[itex]\frac{∂x}{∂\hat{u}}=\frac{a}{\sqrt{a^2+b^2+c^2}}=a[/itex]
Thus,
[itex]\frac{∂F}{∂\hat{u}}=\frac{∂F}{∂x}a+\frac{∂F}{∂y}b+\frac{∂F}{∂z}c=∇F \bullet \hat{u}[/itex].
Let [itex]\hat{u}=<a,b,c>[/itex] be the direction.
Thus, [itex]\frac{∂\hat{u}}{∂x}=\frac{\sqrt{a^2+b^2+c^2}}{a}[/itex] and so on. So,
[itex]\frac{∂x}{∂\hat{u}}=\frac{a}{\sqrt{a^2+b^2+c^2}}=a[/itex]
Thus,
[itex]\frac{∂F}{∂\hat{u}}=\frac{∂F}{∂x}a+\frac{∂F}{∂y}b+\frac{∂F}{∂z}c=∇F \bullet \hat{u}[/itex].
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