Exploring the Geometric Property of a Planar Curve [gamma]

AI Thread Summary
The discussion focuses on deriving a first-order differential equation for a planar curve where the radius vector and tangent intersect at a fixed angle. Participants explore the relationship between angles and tangents, leading to the formulation of a differential equation that incorporates the angle parameter. For the specific case where the angle is π/2, it is concluded that the solution represents a circle. The conversation also touches on applying Euler's method for numerical solutions and suggests using polar coordinates to simplify the problem. Overall, the discussion emphasizes the geometric properties of curves and the mathematical techniques for solving related differential equations.
yoyo
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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)

dont know how to derive the differential from here...and not sure about the rest of the problem...please help...!

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?
 
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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.
 
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&#039;+ yy&#039;}{\sqrt{x^2+ y^2}\sqrt{x&#039;^2+ y&#039;^2}}= cos(\alpha)
or
xy&#039;+ yy&#039;= cos(\alpha)\sqrt{x^2+ y^2}\sqrt{x&#039;^2+ y&#039;^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}

That's your differential equation.

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.)
 
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yoyo said:
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)
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 definitely switch to polar coordinates.
 
Galileo said:
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 definitely switch to polar coordinates.


I understand how to plug in and get to this point, but how would i separate 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 I am not sure

help??
 
Does anyone know how to convert the above equation into polar coordinates ^^^^
 
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