Q. 51 - 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.

$z=\ln \left(\frac{x}{y^2}\right), P=\left(7,1\right), \mathbf{v}=-\mathbf{i}-4\mathbf{j}$

Step-by-Step Solution

Answer

The directional derivative of the given function is $\frac{55\sqrt{17}}{119}$.

Step 1. Given information.

The given function is

$z=\ln \left(\frac{x}{y^2}\right)$

Step 2. Calculation.

The given vector is $\mathbf{v}=-\mathbf{i}-4\mathbf{j}$.

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

$\sqrt{{\left(-1\right)}^2+{\left(-4\right)}^2}=\sqrt{17}$

so, the unit vector is $\hat{\mathrm{n}}=\frac{1}{\sqrt{17}}\left(-\mathbf{i}-4\mathbf{j}\right)$
Now we have to find the gradient of the function.

$$\begin{gathered} \nabla z=\frac{\partial \left(\ln \left({\displaystyle \frac{x}{y^2}}\right)\right)}{\partial x}\mathbf{i}+\frac{\partial \left(\ln \left(\frac{x}{y^2}\right)\right)}{\partial y}\mathbf{j} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }=\frac{1·\left({\displaystyle \frac{1}{{\mathrm{y}}^2}}\right)}{\left({\displaystyle \frac{\mathrm{x}}{{\mathrm{y}}^2}}\right)}\mathbf{i}\mathbf{+}\frac{1·{\displaystyle \left(\frac{-2\mathrm{x}}{{\mathrm{y}}^3}\right)}}{\left({\displaystyle \frac{\mathrm{x}}{{\mathrm{y}}^2}}\right)}\mathbf{j} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }=\frac{1}{\mathrm{x}}\mathbf{i}-\frac{2}{\mathrm{y}}\mathbf{j} \end{gathered}$$

Therefore the required directional derivative is equal to:

$$\begin{gathered} \hat{\mathrm{n}}·\left(\nabla \mathrm{z}\right)=\frac{1}{\sqrt{17}}\left(-\mathbf{i}-4\mathbf{j}\right)·\left(\frac{1}{\mathrm{x}}\mathbf{i}-\frac{2}{\mathrm{y}}\mathbf{j}\right) \\ =-\frac{1}{x\sqrt{17}}+\frac{8}{y\sqrt{17}} \end{gathered}$$

Step 3. Calculation.

Now we find directional derivative of the given function at the point $\left(7,1\right)$.

$$\begin{gathered} \hat{\mathrm{n}}·{\overline{)\nabla z}}_{\left(7,1\right)}=-\frac{1}{x\sqrt{17}}+\frac{8}{y\sqrt{17}} \\ =-\frac{1}{7\sqrt{17}}+\frac{8}{\sqrt{17}} \\ =\frac{55}{7\sqrt{17}} \\ =\frac{55\sqrt{17}}{119} \end{gathered}$$

Step4. Conclusion.

The directional derivative of the given function is $\frac{55\sqrt{17}}{119}$.