Q. 53. - Calculus

Question

Use Theorem 12.32 to find the indicated derivatives in Exercises

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

$\omega =x\sin y\cos z, P=\left(3,\frac{\pi }{4},-\frac{\pi }{2}\right), v=i-2 j+3 k$

Step-by-Step Solution

Answer

The required directional derivative of the function is $\nabla z·u=1.7$

Step 1: Given information

Think about the following function.

$\omega =x\sin y\cos z$

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

$$\begin{gathered} \nabla \omega =i\frac{\partial \omega }{\partial x}+j\frac{\partial \omega }{\partial y}+k\frac{\partial \omega }{\partial z} \\ \omega =x\sin y\cos z \\ \frac{\partial \omega }{\partial x}=\sin y\cos z \\ \Rightarrow \frac{\partial \omega }{\partial y}=x\cos y\cos z \\ \Rightarrow \frac{\partial \omega }{\partial z}=-x\sin y\sin z \end{gathered}$$

When it comes to the next step,

$$\begin{gathered} \nabla \omega =i\frac{\partial \omega }{\partial x}+j\frac{\partial \omega }{\partial y}+k\frac{\partial \omega }{\partial z} \\ \Rightarrow \nabla \omega =i\left(\sin y\cos z\right)+j\left(x\cos y\cos z\right)+k(-x\sin y\sin z) \end{gathered}$$

Consider the vector below.

$v=i-2 j+3 k$

Along the vector$v=i-2 j+3 k$, there is a unit vector.

$$\begin{gathered} u=\frac{i-2j+3k}{\sqrt{(1{)}^2+(-2{)}^2+3^2}} \\ \Rightarrow u=\frac{i-2j+3k}{\sqrt{14}} \\ \Rightarrow u=\frac{i-2j+3k}{\sqrt{14}} \end{gathered}$$

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

$$\begin{gathered} \nabla \omega ·u=\left(i\right(\sin y\cos z)+j(x\cos y\cos z)+k(-x\sin y\sin z\left)\right)·\left(\frac{i-2j+3k}{\sqrt{14}}\right) \\ \Rightarrow \nabla \omega ·u=\frac{\left(1\right(\sin y\cos z)-2(x\cos y\cos z)+3(-x\sin y\sin z\left)\right)}{\sqrt{14}} \end{gathered}$$

At point $P=\left(3,\frac{\pi }{4},-\frac{\pi }{2}\right)$

$$\begin{gathered} \nabla \omega ·u=\frac{\left(1\left(\sin \left(\frac{\pi }{4}\right)\cos \left(-\frac{\pi }{2}\right)\right)-2\left(3\cos \left(\frac{\pi }{4}\right)\cos \left(-\frac{\pi }{2}\right)\right)+3\left(-3\sin \left(\frac{\pi }{4}\right)\sin \left(-\frac{\pi }{2}\right)\right)\right)}{\sqrt{14}} \\ \Rightarrow \nabla z·u=\frac{6.36}{\sqrt{14}} \end{gathered}$$

Hence, the directional derivative of the function is

$\nabla z·u=1.7$