Q. 54. - Calculus

Question

Use Theorem 12.32 to find the indicated derivatives in Exercises

21–26. Express your answers as functions of a single variable

$w=\ln \left(\frac{x^2}{yz}\right)+\ln \left(\frac{z}{xy}\right)-\ln \left(\frac{x}{y^2}\right), P(3,5,8), \overrightarrow{v}=13\mathrm{i}+21\mathrm{j}+34\mathrm{k}$

Step-by-Step Solution

Answer

 The direction derivative of the given function at $P(3,5,8)$ in the direction $\overrightarrow{v}$

$D_{\overrightarrow{v}}w(x,y,z)=0$

Step 1: Given information

The function $w=\ln \left(\frac{x^2}{yz}\right)+\ln \left(\frac{z}{xy}\right)-\ln \left(\frac{x}{y^2}\right)\dots \dots .\left(1\right)$

And the given vector

$\overrightarrow{v}=13\mathrm{i}+21\mathrm{j}+34\mathrm{k}$

Step 2: The objective is to find the direct derivative of the given function at P ( 3 , 5 , 8 )   in the direction v →

The direction derivative of a function $f(x, y, z)$ in the direction of the unit vector $\overrightarrow{u}=ai+bj+cj$

$$\begin{gathered} D_u(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c \\ =\nabla f(x,y,z)·\overrightarrow{u} \end{gathered}$$

Rewrite function $\left(1\right)$ as follows:

$$\begin{gathered} w=\ln \left(\frac{x^2}{yz}\right)+\ln \left(\frac{z}{xy}\right)-\ln \left(\frac{x}{y^2}\right) \\ =\ln \left(\frac{x^2}{yz}\right)\left(\frac{z}{xy}\right)-\ln \left(\frac{x}{y^2}\right) \\ =\ln \left(\frac{x}{y^2}\right)\left(\frac{y^2}{x}\right) \\ =\ln 1 \\ =0 \\ Thus, \\ \nabla w=0 \end{gathered}$$

Hence, the directional derivative of the given function at $P(3,5,8)$ in the direction $\overrightarrow{v}$

$$\begin{gathered} D_{\overrightarrow{v}}w(x,y,z)=\nabla w(x,y,z)·\overrightarrow{v} \\ =0·\overrightarrow{v} \\ =0 \end{gathered}$$