Chapter 14, Problem 1P

Find the directional derivative of

$f\left(x,y\right)=x^2+2y^2$

at $x=2\text{and }y=2$ in the direction of $h=2i+3j$.

To determine

To calculate: The directional derivative of the function $f\left(x,y\right)=x^2+2y^2$ at $x=2\text{ and }y=2$ in the direction of $h=2i+3j$.

Answer to Problem 1P

Solution:

The directional derivative of the function $f\left(x,y\right)=x^2+2y^2$ at $x=2\text{ and }y=2$ in the direction of $h=2i+3j$ is 8.8752.

Explanation of Solution

Given:

The function $f\left(x,y\right)=x^2+2y^2$ at $x=2\text{ and }y=2$ in the direction of $h=2i+3j$.

Formula used:

The directional derivative of a function f is given by,

$g^{\prime }\left(0\right)=\frac{\partial f}{\partial x}\cos \theta +\frac{\partial f}{\partial y}\sin \theta$

Where, $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$ for the direction $\left(xi+yj\right)$.

Calculation:

Consider the function,

$f\left(x,y\right)=x^2+2y^2$

Partially differentiate the above function with respect to x,

$\begin{array}{c}\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left(x^2+2y^2\right)\\ =\frac{\partial }{\partial x}\left(x^2\right)+\frac{\partial }{\partial x}\left(2y^2\right)\\ =2x+0\\ =2x\end{array}$

Now, partially differentiate the function with respect to y,

$\begin{array}{c}\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left(x^2+2y^2\right)\\ =\frac{\partial }{\partial y}\left(x^2\right)+\frac{\partial }{\partial y}\left(2y^2\right)\\ =0+2\times 2y\\ =4y\end{array}$

The direction is $h=2i+3j$. Therefore, the angle in the direction is,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{3}{2}\right)\\ ={\tan }^{-1}\left(1.5\right)\\ =56.31°\end{array}$

Thus, the directional derivative of a function f is given by,

$g^{\prime }\left(0\right)=\left(2x\right)\cos \left(56.31\right)+\left(4y\right)\sin \left(56.31\right)$

Substitute the values $x=2\text{ and }y=2$ in the above function and simplify as below,

$\begin{array}{c}{g}^{\prime }\left(0\right)=\left(2\times 2\right)\cos \left(56.31\right)+\left(4\times 2\right)\sin \left(56.31\right)\\ =4\times 0.5547+8\times 0.8321\\ =2.2188+6.6564\\ =8.8752\end{array}$

Hence, the directional derivative of the function $f\left(x,y\right)=x^2+2y^2$ at $x=2\text{ and }y=2$ in the direction of $h=2i+3j$ is 8.8752.