Q 66. - Calculus

Question

Prove Theorem when $f$ is a function of three variables. That is, show that if $\left(x_0,y_0,z_0\right)$ is a point in the domain of $f(x, y, z)$ at which the first-order partial derivatives of $f$ exist, and if $\mathbf{u}\in {\mathrm{ℝ}}^3$ is a unit vector for which the directional derivative $D_{\mathbf{u}}f\left(x_0,y_0,z_0\right)$ also exists, then$D_{\mathbf{u}}f\left(x_0,y_0,z_0\right)=\nabla f\left(x_0,y_0,z_0\right)·\mathbf{u}$.

Step-by-Step Solution

Answer

Find out the gradient to prove the above relation.

Step 1: Given Information

Let $f(x, y, z)$be a function of three variables defined on a open set containing

$\left(x_0,y_0,z_0\right)\text{ and let }\mathbf{u}=⟨\alpha ,\beta ,\gamma ⟩$ the point containing a unit vector

The direction derivative in $u$ direction is given by

$D_{\mathrm{u}}f\left(x_0,y_0,z_0\right)$

$D_{\mathrm{u}}f\left(x_0,y_0,z_0\right)=\underset{h\to 0}{\lim }\frac{f\left(x_0+\alpha h,y_0+\beta h,z_0+\gamma h\right)-f\left(x_0,y_0,z_0\right)}{h}$

limit also exists

The gradient of function is given by

$\nabla f\left(x_0,y_0,z_0\right)=f_x\left(x_0,y_0,z_0\right)\mathrm{i}+f_y\left(x_0,y_0,z_0\right)\mathrm{j}+f_z\left(x_0,y_0,z_0\right)\mathrm{k}$

As $\alpha ,\beta ,\gamma ,x_0,y_0\text{ and }{z}_0$ are constants

$F\left(h\right)=f\left(x_0+\alpha h,y_0+\beta h,z_0+\gamma h\right)$ is a single variable function

$\Rightarrow F\left(0\right)=f\left(x_0+\alpha ·0,y_0+\beta ·0,z_0+\gamma ·0\right)=f\left(x_0,y_0,z_0\right)$

Step 2: Simplification

From above equation $D_{\mathbf{u}}f\left(x_0,y_0,z_0\right)=\underset{h\to 0}{\lim }\frac{F\left(h\right)-F\left(0\right)}{h}=F^{\text{'}}\left(0\right)$

If $x=x_0+\alpha h,y=y_0+\beta h\text{ and }z=z_0+\gamma h$

By chain rule

$F^{\text{'}}\left(h\right)=\frac{\partial f}{\partial x}\frac{\partial x}{\partial h}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial h}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial h}$

$=f_x(x,y,z)\frac{\partial }{\partial h}\left(x_0+\alpha h\right)+f_y(x,y,z)\frac{\partial }{\partial h}\left(y_0+\beta h\right)+f_z(x,y,z)\frac{\partial }{\partial h}\left(z_0+\gamma h\right)$

$=f_x(x,y,z)(0+\alpha )+f_y(x,y,z)(0+\beta )+f_z(x,y,z)(0+\gamma )$

$=\alpha f_x(x,y,z)+\beta f_y(x,y,z)+\gamma f_z(x,y,z)$

If $h=0\text{ then }x=x_0, y=y_0\text{ and }z=z_0\text{. }$

$F^{\text{'}}\left(0\right)=\alpha f_x\left(x_0,y_0,z_0\right)+\beta f_y\left(x_0,y_0,z_0\right)+\gamma f_z\left(x_0,y_0,z_0\right)$

$D_{\mathrm{u}}f\left(x_0,y_0,z_0\right)=\alpha f_x\left(x_0,y_0,z_0\right)+\beta f_y\left(x_0,y_0,z_0\right)+\gamma f_z\left(x_0,y_0,z_0\right)$

$=\nabla f\left(x_0,y_0,z_0\right)·\mathbf{u}$

Hence proved