Question

The graph in Figure 1.1.7 is the member of the family of folia in Problem 51 corresponding to c=1. Discuss: How can the DE in Problem 51 help in finding points on the graph of $x^3+y^3=3xy$where the tangent line is vertical? How does knowing where a tangent line is vertical help in determining an interval I of definition of a solution$\varphi$of the DE? Carry out your ideas and compare with your estimates of the intervals in Problem 50.

Short answer

The points are (0,0) and $(2^{1/3},4^{1/3})$.

Step-by-step solution

Definition of differential equation.

An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE).

Finding y-coordinate.

Given that,$x^3+y^3=3xy$....(1)

The points in which the curve has vertical tangent lines are the ones that satisfy:

$x(2y^3-x^3)=0$

The equation has two cases, first when x=0, from (1) we obtain the y coordinate,

$$\begin{gathered} 0^3+y^3=3\left(0\right)y \\ {y}^3=0 \end{gathered}$$

So, y = 0

Therefore, one vertical tangent line is obtained in (0,0).

Further, when $(2y^3-x^3)=0$from (1) we obtain the y coordinate,:

$$\begin{gathered} 2y^3+y^3=3y^{1/3}\left(2^{1/3}\right)y \\ 3y^3=3\left(2^{1/3}\right)y^{4/3} \\ {y}^3=2^{1/3}{y}^{4/3} \end{gathered}$$

Simplify,

y = 2^1/3

Finding x-coordinate.

By applying back substitution, we obtain the x-coordinate : x = 4^1/3 .

Therefore, the other vertical tangent line is obtained in $(2^{1/3},4^{1/3})$, which is approximately $(1.59,1.25)$

The points in which the curve has vertical tangent lines are the ones that satisfy:

$x(2y^3-x^3)=0$

Therefore, This are the points (0,0) and $(2^{1/3},4^{1/3})$.