What's the PROOF of the Directional Derivative of a 3-variable function?

DuF(x,y,z)=(Fx,Fy,Fz).(U1,U2,U3) where U=(U1,U2,U3) is a unit vector.

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- Nov 9th 2009, 06:31 AMPhysicStuPROOF of the Directional Derivative
What's the PROOF of the Directional Derivative of a 3-variable function?

DuF(x,y,z)=(Fx,Fy,Fz).(U1,U2,U3) where U=(U1,U2,U3) is a unit vector. - Nov 9th 2009, 06:58 AMHallsofIvy
Assuming that F

**is**differentiable at the given point, $\displaystyle (x_0, y_0. z_0)$ then it can be approximated by the linear function $\displaystyle F_x(x_0,y_0,z_)(x- x_0)+ F_y(x_0,y_0,z_0)(x-y_0)+ F_z(x_0,y_0,z_0)(x- x_0)+ F(x_0,y_0,z_0)$. (That follows from the definition of "differentiable at a point")

The line through $\displaystyle (x_0, y_0, z_0)$, in the direction of $\displaystyle U= (U_1, U_2, U_3)$, is given by the parametric equations $\displaystyle x= x_0+ U_1t$ , $\displaystyle y= y_0+ U_2t$, $\displaystyle z= z_0+ U_2t$.

To find the derivative of F in the direction U, replace x, y, and z with those in the approximation for F, to get $\displaystyle F_x(U_1t)+ F_y(U_2t)+ F_z(U_3t)= (F_xU_1+ F_yU_2+ F_zU_3)t$, a linear function of t with derivative equal to its slope, $\displaystyle F_xU_1+ F_yU_2+ F_zU_3$.