Solve Differential Equation of Family of Curves and Orthogonal Trajectories

[brit123]

Doing some extra credit and got stuck on this one.


Find the differential equation of the family of curves and of the orthogonal trajectories.

y = c - 2x

Needing a little help on this one...

Thanks

 

[HallsofIvy]

"The differential equation of a family of curves" means a differential equation such that that the general solution to the differential equation is that family of curves. Can you think of a differential equation that has
y= c- 2x as its general solution?

One curve is orthogonal to another if there tangent lines are perpendicular where they intersect. What does this tell you about there derivatives?

(Actually, all "curves" satisfying y= c-2x are straight lines.)

 

[tacman]

for reaching out for help on this problem! To solve this, we first need to understand what a family of curves and orthogonal trajectories are.

A family of curves is a group of curves that have a common form or structure, but with different values for the constants. In this case, the equation y = c - 2x represents a family of straight lines where c is the constant that determines the specific line in the family.

On the other hand, orthogonal trajectories are curves that intersect the family of curves at right angles. This means that the slope of the tangent line of the orthogonal trajectory at the point of intersection is the negative reciprocal of the slope of the tangent line of the original curve at the same point.

To find the differential equation of the family of curves, we can use the standard form of a straight line y = mx + b, where m is the slope and b is the y-intercept. In this case, we can rewrite the equation as y = -2x + c, where c is the y-intercept.

To find the differential equation of the orthogonal trajectories, we can use the fact that the slopes of the tangent lines of the two curves at the point of intersection are negative reciprocals. So, if the slope of the original curve is m, the slope of the orthogonal trajectory would be -1/m.

Now, we can use the derivative to find the slope of the original curve at any point. The derivative of y = c - 2x is -2, which represents the slope of the original curve. Therefore, the slope of the orthogonal trajectory would be -1/(-2) = 1/2.

Using this slope and the point of intersection (x,y), we can write the equation of the orthogonal trajectory as y - y = 1/2(x - x). Simplifying this, we get y = 1/2x + d, where d is the y-intercept.

Now, to find the differential equation of the orthogonal trajectories, we can differentiate this equation to get the slope of the orthogonal trajectory at any point. The derivative of y = 1/2x + d is 1/2, which represents the slope of the orthogonal trajectory.

Therefore, the differential equation of the orthogonal trajectories is dy/dx = 1/2.

I hope this helps you understand the process of finding the differential equation of a family of curves and orthogonal trajectories. Good luck