Question

$\rm Given :f(x ,y) = {e}^{x} \: cos \: y \: (0,0), \: \theta = \frac{\pi}{4} \\ \rm find \: the \: directional \: derivative \: of \: f \: at \\ \rm the \: given \: point \: in \: the \: direction \\ \rm indicated \: by \: the \: angle \: \theta$​

Answer 1

Answer:

$\huge{Answer}$

1/2

The directional derivative of f at the given point in the direction indicated by the angle 0 is expressed as Vf(x, y) * u where u is the unit vector in the direction 0. Lets first calculate ▼ f(x, y) at (0, 1)

V=7i+jVƒ(z,y) = (ycos(ry))zi+7(cos(zy))yjVf(x,y)=-y2sinayi+(coszy zysinry)j

Vf(x, y) at (0, 1) = -12sin0 i + (coso - Osin0)\j = 0i+j

The unit vector u in the direction of is expressed as costi + sine j unit vector u at е = π/6 is cos π/6i + sin π/6 j u= √3/2 i +1/2 jVf(x, y) * u

Taking the dot product i.e

= (0i+j)*(√3/2 i +1/2 j) =

= 1/2

The directional derivative of f

is 1/2

Answer 2

The directional derivative of f(x,y) at the point (0,0) in the direction of the angle $\rm \theta = \frac{\pi}{4}$ is given by:

$\rm D_{\theta} f(0,0) = \nabla f(0,0) \cdot \mathbf{u}_{\theta}$

where $\rm\nabla f(x,y)$is the gradient of f at the point (x,y), and $\mathbf{u}_{\theta}$ is the unit vector in the direction of the angle $\theta$. In this case, we have:

$\rm\nabla f(x,y) = \begin{pmatrix} \frac{\partial f}{\partial x}(x,y) \\ \frac{\partial f}{\partial y}(x,y)\end{pmatrix} = \begin{pmatrix} e^x \cos y \\ -e^x \sin y \end{pmatrix}$

and

$\mathbf{u}_{\theta} =\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}$

Therefore, we have:

$\begin{align}D_{\theta} f(0,0) &=\nabla f(0,0) \cdot \mathbf{u}_{\theta} \\&= \begin{pmatrix} e^0 \cos 0 \\ -e^0 \sin 0 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \\&= \begin{pmatrix} 1 \\ 0 \end{pmatrix}\cdot\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \\&= \frac{1}{\sqrt{2}}\end{align}$

Therefore, the directional derivative of f(x,y) at the point (0,0) in the direction of the angle $\theta = \frac{\pi}{4}$is $\frac{1}{\sqrt{2}}$.