Q. 50 - 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.

Step-by-Step Solution

Answer

The directional derivative of the given function is $-\frac{180}{13}$.

Step 1. Given information.

The given function is

$z=3x^4-5y^2$

Step 2. Calculation.

The given vector is  .

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

$\sqrt{{\left(\frac{5}{13}\right)}^2+{\left(\frac{12}{13}\right)}^2}=1$

so, the unit vector is $\hat{\mathrm{n}}=\left(\frac{5}{13}\mathbf{i}-\frac{12}{13}\mathbf{j}\right)$

Now we have to find the gradient of the function. 

$$\begin{gathered} \nabla z=\frac{\partial \left(3x^4-5y^2\right)}{\partial x}\mathbf{i}+\frac{\partial \left(3x^4-5y^2\right)}{\partial y}\mathbf{j} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }=12{\mathrm{x}}^3\mathbf{i}-10\mathrm{y}\mathbf{j} \end{gathered}$$

Therefore the required directional derivative is equal to:

$$\begin{gathered} \hat{\mathrm{n}}·\left(\nabla \mathrm{z}\right)=\left(\frac{5}{13}\mathbf{i}-\frac{12}{13}\mathbf{j}\right)·\left(12x^3\mathbf{i}-10y\mathbf{j}\right) \\ =\frac{60}{13}{x}^3+\frac{120}{13}y \end{gathered}$$

Step 3. Calculation.

Now we find directional derivative of the given function at the point  $\left(1,-2\right)$.

$$\begin{gathered} \hat{\mathrm{n}}·{\overline{)\nabla z}}_{\left(1,-2\right)}=\frac{60}{13}{\left(1\right)}^3+\frac{120}{13}\left(-2\right) \\ =\frac{60}{13}-\frac{240}{13} \\ =-\frac{180}{13} \end{gathered}$$

Step4. Conclusion.

The directional derivative of the given function is $-\frac{180}{13}$.