Question

Expand e^(x)cos y in powers of x and y as far as the terms of second degree.​

Answer 1

Answer:

hope this helps you....

Answer 2

Given:

An equation $e^{x} \cos y$.

To find:

Expand $e^{x} \cos y$ in powers of x and y up to a second degree.

Solution:

First, the partial derivatives of $e^{x} \cos y$ at $\left(1, \frac{\pi}{4}\right)$ are:

$\begin{array}{c}\\f(x, y)=e^{x} \cos y \\\\f_{x}=e^{x} \cos y \quad \rightarrow \quad \frac{e}{\sqrt{2}} \\\\f_{y}=-e^{x} \sin y \rightarrow \frac{-e}{\sqrt{2}} \\\\f_{x x}=e^{x} \cos y \rightarrow \frac{e}{\sqrt{2}} \\\\f_{x y}=-e^{x} \sin y \rightarrow \frac{-e}{\sqrt{2}} \\\\f_{y y}=-e^{x} \cos y \rightarrow \frac{-e}{\sqrt{2}}\end{array}$

We know that,

$\begin{aligned}f(x, y)=& f\left(1, \frac{\pi}{4}\right)+\left[(x-1) f_{x}\left(1, \frac{\pi}{4}\right)+\left(y-\frac{\pi}{4}\right) f_{y}\left(1, \frac{\pi}{4}\right)\right] \\&+\frac{1}{2 !}\left[(x-1)^{2} f_{x x}\left(1, \frac{\pi}{4}\right)+2(x-1)\left(y-\frac{\pi}{4}\right) f_{x y}\left(1, \frac{\pi}{4}\right)\right.\\&+\left(y-\frac{\pi}{4}\right)^{2} f_{x}\left(1, \frac{\pi}{4}\right)+\cdots\end{aligned}$

at point $\left(1, \frac{\pi}{4}\right)$.

On substituting these values, we get:

$\begin{array}{l}e^{x} \cos y=\frac{e}{\sqrt{2}}\left\{1+\left[(x-1)-\left(y-\frac{\pi}{4}\right)\right]\right\left.+\frac{1}{2}\left[(x-1)^{2}-2(x-1)\left(y-\frac{\pi}{4}\right)-\left(y-\frac{\pi}{4}\right)^{2}\right]\right\}\end{array}$

Therefore, $\begin{array}{l}e^{x} \cos y=\frac{e}{\sqrt{2}}\left\{1+\left[(x-1)-\left(y-\frac{\pi}{4}\right)\right]\right.\left.+\frac{1}{2}\left[(x-1)^{2}-2(x-1)\left(y-\frac{\pi}{4}\right)-\left(y-\frac{\pi}{4}\right)^{2}\right]\right\}\end{array}$is the expansion of $e^{x} \cos y$ at the point $\left(1, \frac{\pi}{4}\right)$.