Validity of Directional Derivatives for Unit Vectors

[ayao]

For directional derivatives:

Let \hat{u}=<a,b,c> be the direction.

Thus, \frac{∂\hat{u}}{∂x}=\frac{\sqrt{a^2+b^2+c^2}}{a} and so on. So,

\frac{∂x}{∂\hat{u}}=\frac{a}{\sqrt{a^2+b^2+c^2}}=a

Thus,

\frac{∂F}{∂\hat{u}}=\frac{∂F}{∂x}a+\frac{∂F}{∂y}b+\frac{∂F}{∂z}c=∇F \bullet \hat{u}.

 

[Mark44]

For directional derivatives:

Let \hat{u}=<a,b,c> be the direction.

Are a, b, and c constants? If so, the derivative of $\hat{u}$ would be the zero vector.

Thus, \frac{∂\hat{u}}{∂x}=\frac{\sqrt{a^2+b^2+c^2}}{a} and so on. So,

\frac{∂x}{∂\hat{u}}=\frac{a}{\sqrt{a^2+b^2+c^2}}=a

Thus,

\frac{∂F}{∂\hat{u}}=\frac{∂F}{∂x}a+\frac{∂F}{∂y}b+\frac{∂F}{∂z}c=∇F \bullet \hat{u}.

 

[ayao]

I probably should have defined it better; \hat{u} is a unit vector in the direction that we are trying to find the derivative in. The linear derivatives of \hat{u} are defined as derivatives of a line in that direction.