Chapter 4.2, Problem 46E

a.

To determine

To find: the equations for the tangent and normal to the cissoids of Diocles,

Answer to Problem 46E

Tangent line: $y=2x-1.$

Normal line: $y=\frac{-1}{2}x+\frac{3}{2}.$

Explanation of Solution

Given information: The given cissoids of Diocles $y^2(2-x)=x^3$ at the point (1, 1) as

Pictured below.

  

Calculation:

The given cissoids of Diocles $y^2(2-x)=x^3$ .

Differentiate implicitly using the product rule.

  $\begin{array}{c}{y}^2(2-x)=x^3.\\ 2y\frac{dy}{dt}(2-x)+y^2(-1)=3x^2.\\ 2y(2-x)\frac{dy}{dt}-y^2=3x^2.\\ 2y(2-x)\frac{dy}{dt}=3x^2+y^2\\ \frac{dy}{dt}=\frac{3x^2+y^2}{2y(2-x)}\mathrm{...........}(1)\\ \text{Putting}x=1\text{and}y=1.\\ $ {(\frac{dy}{dt})}_{(1,1)}=\frac{3{(1)}^2+1^2}{2(1)(2-1)}=\frac{4}{2}=2$\text{Find}\text{the}\text{equation}\text{of}\text{the}\text{tangent}\text{line}\text{using}\text{point-slope}\text{form,}\\ y-y_1=m(x-x_1)\text{where}m=2\text{and}(x_1,y_1)=(1,1).\text{Then}\text{solve}\text{for}y.\\ y-1=2(x-1).\\ y-1=2x-2.\\ y=2x-1.\end{array}$

The slope of the normal line is the negative reciprocal of the tangent line slope. Therefore the slope of the normal line is -1/2.

  $\begin{array}{c}\text{Find}\text{the}\text{equation}\text{of}\text{the}\text{normal}\text{line}\text{using}\text{point-slope}\text{form,}\\ y-y_1=m(x-x_1)\text{where}m=\frac{-1}{2}\text{and}(x_1,y_1)=(1,1).\text{Then}\text{solve}\text{for}y.\\ y-1=\frac{-1}{2}(x-1).\\ y-1=\frac{-1}{2}x+\frac{1}{2}.\\ y=\frac{-1}{2}x+\frac{3}{2}.\end{array}$

b.

To determine

To explain : how to reproduce the graph on the grapher.

Answer to Problem 46E

Explanation of Solution

Given information: The given cissoids of Diocles $y^2(2-x)=x^3$ at the point (1, 1) as

Pictured below.

  

Calculation:

  $y^2(2-x)=x^3$ : $y=\pm \frac{\sqrt{x^3}}{\sqrt{(2-x)}}.$

Solve for y . he graph does not pass the vertical −line test , judging whether it is in −fact a

function .When solve for y, will see that it is two functions and graphed together, they will give the graph shown