# Homework Help: Planar curves

1. Mar 3, 2005

### yoyo

Recall that every point (x, y) in the plane is described by its radius-
vector r = xi + yj. A planar curve [gamma] has the following geometric property: at every point on the curve the radius vector and the tangent intersect at a fixed angle [alpha].

1. Derive a first order differential equation dy/dx = f(x, y, [alpha]) describing [gamma]
.
Note that the right-hand side should depend on the parameter [alpha].

For part 1, I drew a circle(not sure if it's suppose to be a circle) on the (x, y) plane. Then i drew a radial vector from the origin to any point. i let the radial vector make an angle [phi] with x-axis. extended the tangent all the way to the x axis and let the tangent make an angle theta with the x axis.

then i got

tan [phi]= y/x......(1)

tan [theta]= dy/dx...(2)

so [alpha]= [theta] - [phi]

tan [alpha]= tan ([theta] - [phi]) = (tan [theta] - tan [phi])/(1+tan[theta]*tan[phi])......(3)

2. Solve the equation from Part 1 for the case when [alpha] = pi/2 assuming that [gamma] passes through (1, 0). For which curve is the radius vector always perpendicular to the tangent?

I think the answer is a circle but not sure how to solve it???/

3. Assume that [alpha] = 3pi/4 and the curve passes through (1, 0). Apply Euler's method with step h = 0.5 to trace out the curve for this case.

is this step function? if so how do i go about solving this???

4. Solve the equation from Part 1 for [alpha]= 3pi/4 with (1, 0) as the initial condition. Superimpose the exact solution from this part with the numerical
solution from Part 3. Do you know the name of this curve?

I think in order to plot the curve, i have to switch to polar coordinates
x = r cos [theta]; y = r sin [theta]:

help???

2. Mar 4, 2005

### einstone

Answer to Question1: Put theta=arctan(dy/dx) & phi=arctan(y/x) in the formula you've got.
But why not use polar coordinates? That'd solve all the problems.
I'm,with great respect,
Einstone.

3. Mar 4, 2005

### HallsofIvy

Sounds to me like you are making this a lot more complicated than necessary.

If the radius vector is r(t) then the tangent vector is r'(t) Since than angle between any two vectors, u, v, is given by cos(θ)= u.v/(|u||v|), the condition that r and r' have constant angle, α, between them is that r'(t).r(t)/(|r||r'|)= cos(α). Since r= <x, y>, r'= <x', y'>, that becomes
$$\frac{xx'+ yy'}{\sqrt{x^2+ y^2}\sqrt{x'^2+ y'^2}}= cos(\alpha)$$
or
$$xy'+ yy'= cos(\alpha)\sqrt{x^2+ y^2}\sqrt{x'^2+ y'^2}$$

"Dividing" both sides by x' (It's really using the chain rule.)
$$x+ y\frac{dy}{dx}= cos(\alpha)\sqrt{x^2+ y^2}\sqrt{1+(\frac{dy}{dx})^2}$$

Of course, if $\alpha= \frac{\pi}{2}$ the cos(&alpha;)= 0 and your differential equation is just
$$x+ y\frac{dy}{dx}= 0$$.
That's a simple separable differential equation (and, yes, the solution is a circle.)

Last edited by a moderator: Mar 4, 2005
4. Mar 4, 2005

### Galileo

This is okay. If you write it out in x and y:

$$\frac{dy}{dx}=\tan(\alpha + \phi)=\frac{\tan(\alpha)+\tan(\phi)}{1-\tan(\alpha)\tan(\phi)}=\frac{y/x+\tan(\alpha)}{1-\tan(\alpha)y/x}$$

There's your differential equation. But definately switch to polar coordinates.

5. Mar 5, 2005

### yoyo

I understand how to plug in and get to this point, but how would i seperate the variables in this case? I tried plugging in a lot of different things but keep coming up with an unsolvable differential.

i get tan(a) when i do it and when i plug in pi/2 i get undefined. i'm sure i'm doing somethign wrong but im not sure

help??

6. Mar 6, 2005

### yoyo

Does anyone know how to convert the above equation in to polar coordinates ^^^^