Chapter 3.2, Problem 46E

(a)

To determine

To find : equation of normal and tangent line to the cissoid of Diocles at indicated point.

Answer to Problem 46E

The normal and tangent line at $\left(1,1\right)$ is $y=-\frac{x}{2}+\frac{3}{2}$ and $y=2x-1$ respectively.

Explanation of Solution

Given information :

The equation of cissoid of Diocles is $y^2\left(2-x\right)=x^3$ and point $\left(1,1\right)$ .

Formula needed :

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Tangent and normal line at point $\left(a,b\right)$ is $y-b=m\left(x-a\right)$ and $y-b=-\frac{1}{m}\left(x-a\right)$ respectively, where $m={\frac{dy}{dx}}_{\left(a,b\right)}$ .

Differentiate $y^2\left(2-x\right)=x^3$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}{y}^2\left(2-x\right)=\frac{d}{dx}{x}^3\\ y^2\cdot \frac{d}{dx}\left(2-x\right)+\left(2-x\right)\cdot \frac{d}{dx}{y}^2=\frac{d}{dx}{x}^3\\ y^2\cdot \left(-1\right)+\left(2-x\right)\cdot 2y\frac{dy}{dx}=3x^2\\ -y^2+2\left(2-x\right)y\frac{dy}{dx}=3x^2\end{array}$

Solve further,

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

Substitute $\left(1,1\right)$ in the $\frac{dy}{dx}=\frac{3x^2+y^2}{2\left(2-x\right)y}$ in order to find slope at $\left(\frac{\sqrt{3}}{4},\frac{\sqrt{3}}{2}\right)$ .

  $\begin{array}{c}{\frac{dy}{dx}}_{\left(1,1\right)}=\frac{3\cdot 1^2+1^2}{2\left(2-1\right)1}\\ m=\frac{3\cdot 1+1}{2\left(1\right)1}\\ =\frac{3+1}{2}\\ =2\end{array}$

The tangent line at point $\left(1,1\right)$ is given by $y-1=2\left(x-1\right)$ .

  $\begin{array}{c}y-1=2x-2\\ y=2x-2+1\\ y=2x-1\end{array}$

The normal line at point $\left(1,1\right)$ is given by $y-1=-\frac{1}{2}\left(x-1\right)$ .

  $\begin{array}{c}y-1=-\frac{x}{2}+\frac{1}{2}\\ y=-\frac{x}{2}+\frac{1}{2}+1\\ y=-\frac{x}{2}+\frac{3}{2}\end{array}$

(b)

To determine

To explain : procedure to draw a graph on grapher.

Explanation of Solution

Given information :

The equation of cissoid of Diocles is $y^2\left(2-x\right)=x^3$ and point $\left(1,1\right)$ .

Break the curve equation into $y=f\left(x\right)$ .

  $y^2=\frac{x^3}{\left(2-x\right)}$

Take square root on both sides,

  $\begin{array}{c}\sqrt{y^2}=\pm \sqrt{\frac{x^3}{\left(2-x\right)}}\\ y=-\sqrt{\frac{x^3}{\left(2-x\right)}},\sqrt{\frac{x^3}{\left(2-x\right)}}\end{array}$

The equation of curve breaks into two functions $f\left(x\right)=\sqrt{\frac{x^3}{\left(2-x\right)}}$ and $g\left(x\right)=-\sqrt{\frac{x^3}{\left(2-x\right)}}$ .

Insert function one by one and equation of tangent and normal.