Q. 52. - Calculus

Question

Use Theorem 12.32 to find the indicated derivatives in Exercises

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

$z=\sqrt{x^2+y^2} ,P=(-3,4),v=(4,-3)$

Step-by-Step Solution

Answer

The required directional derivative of the function is $\nabla z·u=-0.96$

Step 1: Given information

Think about the following function.

$z=\sqrt{x^2+y^2}$

Step 2: The objective is to find the directional derivative of function at the point P = ( - 3 , 4 )  

$$\begin{gathered} \nabla z=i\frac{\partial z}{\partial x}+j\frac{\partial z}{\partial y} \\ z=\sqrt{x^2+y^2} \\ \frac{\partial z}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}} \\ \frac{\partial z}{\partial x}=\frac{x}{\sqrt{x^2+y^2}} \end{gathered}$$

Then,

$$\begin{gathered} \nabla z=i\frac{\partial z}{\partial x}+j\frac{\partial z}{\partial y} \\ \Rightarrow \nabla z=i\left(\frac{x}{\sqrt{x^2+y^2}}\right)+j\left(\frac{y}{\sqrt{x^2+y^2}}\right) \end{gathered}$$

Consider the vector below.

$v=4 i-3 j$

Along the vector$v=4 i-3 j$, there is a unit vector

$$\begin{gathered} u=\frac{4i-3j}{\sqrt{(4{)}^2+(-3{)}^2}} \\ \Rightarrow u=\frac{4i-3j}{\sqrt{16+9}} \\ \Rightarrow u=\frac{4i-3j}{5} \end{gathered}$$

Step 3: The directional derivatives of the function z in the direction

$$\begin{gathered} \nabla z·u=\left(i\left(\frac{x}{\sqrt{x^2+y^2}}\right)+j\left(\frac{y}{\sqrt{x^2+y^2}}\right)\right)·\left(\frac{4i-3j}{5}\right) \\ \nabla z·u=\frac{4x}{5\sqrt{x^2+y^2}}-\frac{3y}{5\sqrt{x^2+y^2}} \end{gathered}$$

At point $P=(-3,4)$

$$\begin{gathered} \nabla z·u=\frac{4(-3)}{5\sqrt{(-3{)}^2+(4{)}^2}}-\frac{3\left(4\right)}{\sqrt[5]{(-3{)}^2+(4{)}^2}} \\ \nabla z·u=-0.96 \end{gathered}$$

Hence, the directional derivative of the function is

$\nabla z·u=-0.96$