Q. 49 - Calculus

Question

In Exercises 49–54, find the directional derivative of the given function at the specified point P and in the specified direction v. Note that some of the direction vectors are not unit vectors. 

$z=e^x\tan y, P=\left(0,\frac{\pi }{4}\right), \mathbf{v}=3\mathbf{i}-\mathbf{j}$ 

Step-by-Step Solution

Answer

The directional derivative of the given function is $\frac{\sqrt{10}}{10}$.

Step 1. Given information.

The given function is

 $z=e^x\tan y$.

Step 2. Calculation.

The given vector is $\mathbf{v}=3\mathbf{i}-\mathbf{j}$.

First we find the magnitude of the given vector. The magnitude of the given vector is:

$\sqrt{3^2+{(-1)}^2}=\sqrt{10}$

so, the unit vector is $\hat{\mathrm{n}}=\frac{1}{\sqrt{10}}\left(3\mathrm{i}-\mathrm{j}\right)$

Now we have to find the gradient of the function.

$$\begin{gathered} \nabla z=\frac{\partial \left(e^x\tan y\right)}{\partial x}\mathbf{i}+\frac{\partial \left(e^x\tan y\right)}{\partial y}\mathbf{j} \\ ={\mathrm{e}}^{\mathrm{x}}\mathrm{tany} \mathbf{i}+{\mathrm{e}}^{\mathrm{x}}{\sec }^2\mathrm{y}\mathbf{j} \end{gathered}$$

Therefore the required directional derivative is equal to:

$$\begin{gathered} \hat{\mathrm{n}}·\left(\nabla \mathrm{z}\right)=\frac{1}{\sqrt{10}}\left(3\mathbf{i}-\mathbf{j}\right)·{\mathrm{e}}^{\mathrm{x}}\mathrm{tany} \mathbf{i}+{\mathrm{e}}^{\mathrm{x}}{\sec }^2\mathrm{y}\mathbf{j} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }=\frac{3}{\sqrt{10}}{\mathrm{e}}^{\mathrm{x}}\mathrm{tany}-\frac{1}{\sqrt{10}}{\mathrm{e}}^{\mathrm{x}}{\sec }^2\mathrm{y} \end{gathered}$$

Step 3. Calculation.

Now we find directional derivative of the given function at the point $\left(0,\frac{\mathrm{\pi }}{4}\right)$.

$$\begin{gathered} \hat{\mathrm{n}}·{\overline{)\nabla z}}_{\left(0,\frac{\mathrm{\pi }}{4}\right)}=\frac{3}{\sqrt{10}}{e}^0\tan \left(\frac{\mathrm{\pi }}{4}\right)-\frac{1}{\sqrt{10}}{e}^{0}se{c}^2\frac{\mathrm{\pi }}{4} \\ =\frac{1}{\sqrt{10}} \\ =\frac{\sqrt{10}}{10} \end{gathered}$$

Step4. Conclusion.

The directional derivative of the given function is $\frac{\sqrt{10}}{10}$.