Orthogonal Trojectories

[snowJT]

Homework Statement

Write the equation to find the orthogonal trojectory for $$x^2 + y^2 = r^2$$

2. The attempt at a solution

Well.. you solve this by just integrating.. but, I don't know what to do with the r^2 ?

$$x^2 + y^2 = r^2$$
$$x^2 + y^2 \frac{dy}{dx} = r^2 \frac{dy}{dx}$$

is that even right? then... integrate r^2 to become $$\frac{r^3}{3}$$ ?

 

[Dick]

I believe you are supposed to consider x^2+y^2=r^2 as a family of curves (one curve for each value of r). Now you are supposed to find another family of curves whose intersections with these are always orthogonal. Hint: derive a differential equation satisfied by the first set. How would you need to change the differential equation to force orthogonality? Hint 2): check your differentiation, r is a CONSTANT.

 

[HallsofIvy]

You might also note that the family of curves is a family of cocentric circles! What curves are perpendicular to all of them?

And your differentiation is awful! The derivative of x² is NOT x², the derivative of y² is NOT y² dy/dx and the derivative of r² is NOT r² dy/dx!

 

[snowJT]

ohh??

so let me give this a shot...
$$x^2 + y^2 = r^2$$

$$2x + 2y \frac{dy}{dx} = 2r$$

$$r = x + y \frac{dy}{dx}$$

$$\sqrt{x^2 + y^2} = x + y \frac{dy}{dx}$$

is this right so far...?

 

[Dick]

This is the last time I'm saying this. r is a CONSTANT. What does this say about it's derivative?

 

[chaoseverlasting]

dy/dx is the tangent of the angle of the curve right? So you could find the tangent of the angle perpendicular to that and derive a new DE, which you could solve for your answer... I guess there's a geometric way to do this as well (without using calculus), like HallsofIvy said, what family of curves is perpendicular to that family of circles?

 

[Dick]

Yes, as Halls said - it's pretty easy to just see the answer in this case. But I'm guessing this a HW for a calculus course so you might want to do it that way. Actually, doing it that way is not that much harder than guessing...

 

[HallsofIvy]

ohh??

so let me give this a shot...
$$x^2 + y^2 = r^2$$

$$2x + 2y \frac{dy}{dx} = 2r$$

$$r = x + y \frac{dy}{dx}$$

$$\sqrt{x^2 + y^2} = x + y \frac{dy}{dx}$$

is this right so far...?

No, it's not right. If you want to be really technical, the derivative of r² with respect to x is $2r\frac{dr}{dx}$, just as you did with y- but since r is a constant, what is $\frac{dr}{dx}$?
Once you have found $\frac{dy}{dx}$ remember that a line is orthogonal to a line with slope m if and only if it has slope -1/m.