Chapter 4, Problem 79RE

a.

To determine

To find: The $\frac{dy}{dx}$ at the point $(1,1)$ .

Answer to Problem 79RE

The value of $\frac{dy}{dx}$ at the point $(1,1)$ is $-\frac{2}{3}$ .

Explanation of Solution

Given:

Equation is $x^2+2xy+2y^2=5$ .

Calculation: Equation is $x^2+2xy+2y^2=5$ .

First derivative is:

  $\begin{array}{l}2x+2x\frac{dy}{dx}+y(2)+4y\frac{dy}{dx}=0\\ 2x\frac{dy}{dx}+4y\frac{dy}{dx}=-2x-2y\\ \frac{dy}{dx}(2x+4y)=-(2x+2y)\\ \frac{dy}{dx}=\frac{-(2x+2y)}{2x+4y}\end{array}$

Put $x=1\mathrm{and}y=1\mathrm{in}\frac{dy}{dx}$ we get:

  $\begin{array}{l}\frac{dy}{dx}=\frac{-(2x+2y)}{2x+4y}\\ \frac{dy}{dx}=\frac{-(2\times 1+2\times 1)}{2\times 1+4\times 1}\\ \frac{dy}{dx}=\frac{-(2+2)}{2+4}\\ \frac{dy}{dx}=-\frac{-4}{6}=-\frac{2}{3}\end{array}$

Thus, the value of $\frac{dy}{dx}$ at the point $(1,1)$ is $-\frac{2}{3}$ .

b.

To determine

To find: The $\frac{d^{2}y}{d{x}^2}$ at the point $(1,1)$ .

Answer to Problem 79RE

The value of $\frac{d^{2}y}{d{x}^2}$ at the point $(1,1)$ is $-\frac{2}{3}$ .

Explanation of Solution

Given:

Equation is $x^2+2xy+2y^2=5$ .

In part a we find the value of $\frac{dy}{dx}=\frac{-(2x+2y)}{2x+4y}$ .

Calculation: Equation is $x^2+2xy+2y^2=5$ .

In part a we find the value of $\frac{dy}{dx}=\frac{-(2x+2y)}{2x+4y}$ .

second derivative is:

  $\frac{dy}{dx}=\frac{-(2x+2y)}{2x+4y}$

  $\frac{d^{2}y}{d{x}^2}=-\frac{(2x+4y)\frac{d}{dx}(2x+2y)-(2x+2y)\frac{d}{dx}(2x+4y)}{{(2x+4y)}^2}$

  $\frac{d^{2}y}{d{x}^2}=-\frac{(2x+4y)\left(2+2\frac{dy}{dx}\right)-(2x+2y)\left(2+4\frac{dy}{dx}\right)}{{(2x+4y)}^2}$

  $\frac{d^{2}y}{d{x}^2}=-\frac{4x+4x\frac{dy}{dx}+8y+8y\frac{dy}{dx}-4x-8x\frac{dy}{dx}-4y-8y\frac{dy}{dx}}{{(2x+4y)}^2}$

  $\frac{d^{2}y}{d{x}^2}=-\frac{-4x\frac{dy}{dx}+4y}{{(2x+4y)}^2}$put $x=1,y=1$ and $\frac{dy}{dx}=-\frac{2}{3}$ in $\frac{d^{2}y}{d{x}^2}$

  $\frac{d^{2}y}{d{x}^2}=-\frac{-4(1)\left(\frac{-2}{3}\right)+4(1)}{{(2\times 1+4\times 1)}^2}$

  $\frac{d^{2}y}{d{x}^2}=-\frac{\frac{8}{3}+4}{{(2+4)}^2}=-\frac{\frac{8+12}{3}}{6^2}$

  $\frac{d^{2}y}{d{x}^2}=-\frac{\frac{20}{3}}{36}=-\frac{20}{3\times 36}=-\frac{5}{27}$

Thus, the value of $\frac{d^{2}y}{d{x}^2}$ at the point $(1,1)$ is $-\frac{5}{27}$ .