Chapter 4.2, Problem 54E

To determine

To find: The slope of the Devil’s curve at the four indicated points.

Answer to Problem 54E

The slope at the point $\left(-3,2\right)$ is $-\frac{27}{8}$ .

The slope at the point $\left(-3,-2\right)$ is $\frac{27}{8}$ .

The slope at the point $\left(3,2\right)$ is $\frac{27}{8}$ .

The slope at the point $\left(3,-2\right)$ is $-\frac{27}{8}$ .

Explanation of Solution

Given Information: The equation of the Devil’s curve is $y^4-4y^2=x^4-9x^2$ at the points $\left(-3,2\right),\left(3,2\right),\left(-3,-2\right),\left(3,-2\right)$ .

Calculation:

The slope of the Devil’s curve $\frac{dy}{dx}$ is,

  $\begin{array}{c}{y}^4-4y^2=x^4-9x^2\\ 4y^3\frac{dy}{dx}-8y\frac{dy}{dx}=4x^3-18x\\ 2y\left(2y^2-4\right)\frac{dy}{dx}=2x\left(2x^2-9\right)\\ \frac{dy}{dx}=\frac{2x\left(2x^2-9\right)}{2y\left(2y^2-4\right)}\\ =\frac{x\left(2x^2-9\right)}{y\left(2y^2-4\right)}\end{array}$

The slope at the point $\left(-3,2\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=\frac{x\left(2x^2-9\right)}{y\left(2y^2-4\right)}\\ =\frac{-3\left(2{\left(-3\right)}^2-9\right)}{2\left(2{\left(2\right)}^2-4\right)}\\ =\frac{-3\left(18-9\right)}{2\left(8-4\right)}\\ =\frac{-3\times 9}{2\times 4}\\ =-\frac{27}{8}\end{array}$

The slope at the point $\left(3,2\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=\frac{x\left(2x^2-9\right)}{y\left(2y^2-4\right)}\\ =\frac{3\left(2{\left(3\right)}^2-9\right)}{2\left(2{\left(2\right)}^2-4\right)}\\ =\frac{3\left(18-9\right)}{2\left(8-4\right)}\\ =\frac{3\times 9}{2\times 4}\\ =\frac{27}{8}\end{array}$

The slope at the point $\left(3,-2\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=\frac{x\left(2x^2-9\right)}{y\left(2y^2-4\right)}\\ =\frac{3\left(2{\left(3\right)}^2-9\right)}{-2\left(2{\left(-2\right)}^2-4\right)}\\ =\frac{3\left(18-9\right)}{-2\left(8-4\right)}\\ =\frac{3\times 9}{-2\times 4}\\ =-\frac{27}{8}\end{array}$

The slope at the point $\left(-3,-2\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=\frac{x\left(2x^2-9\right)}{y\left(2y^2-4\right)}\\ =\frac{-3\left(2{\left(-3\right)}^2-9\right)}{-2\left(2{\left(-2\right)}^2-4\right)}\\ =\frac{3\left(18-9\right)}{2\left(8-4\right)}\\ =\frac{3\times 9}{2\times 4}\\ =\frac{27}{8}\end{array}$

Conclusion:

The slope at the point $\left(-3,2\right)$ is $-\frac{27}{8}$ .

The slope at the point $\left(-3,-2\right)$ is $\frac{27}{8}$ .

The slope at the point $\left(3,2\right)$ is $\frac{27}{8}$ .

The slope at the point $\left(3,-2\right)$ is $-\frac{27}{8}$ .