# Show that a curve is regular

1. Sep 26, 2015

### Dassinia

Hello !

1. The problem statement, all variables and given/known data
Consider a parametrized curve
C(θ)=( (R+r)*cos(θ) - d*cos(θ(R+r)/r) ; (R+r)*sin(θ) - d*sin(θ(R+r)/r) )
Show that C is regular for d<r. Is it regular if d=r ?

2. Relevant equations

3. The attempt at a solution
C'(θ)=( -(R+r)*sin(θ) +d*(R+r)/r*sin(θ(R+r)/r) ; (R+r)*cos(θ) - d*(R+r)/r*cos(θ(R+r)/r) )
I don't know how to show that it is regular for d<r, i am supposed to show that C'(θ)≠0 but I don't know how to introduce the condition to prove that

Thanks

2. Sep 26, 2015

### Svein

You have found $\frac{dx}{d\theta}, \frac{dy}{d\theta}$, which are the components of the tangent. Now find the length of that vector.

3. Sep 26, 2015

### Dassinia

By using sin²x+cos²x=1 and using formulas for cos(a)*cos(b) and sin(a)*sin(b) I ended up with
d²(R+r)²/r²+(R+r)²-2d(R+r)²/r * cos(θ[(R+r)/r -(R+r)])

4. Sep 26, 2015

### Svein

So what are the maximum and minimum value of that expression?

5. Sep 26, 2015

### Dassinia

Max when cos(θ[(R+r)/r -(R+r)])=0 we have d²(R+r)²/r²+(R+r)²
and Min when cos(θ[(R+r)/r -(R+r)])=1 we have d²(R+r)²/r²+(R+r)²-2d(R+r)²/r

6. Sep 26, 2015

### Dassinia

There's a mistake in my equations it is
d²(R+r)²/r²+(R+r)²-2d(R+r)²/r * cos(θ[(R+r)/r -1])

Max when cos(θ[R/r])=0 we have d²(R+r)²/r²+(R+r)²
and Min when cos(θ[R/r])=1 we have d²(R+r)²/r²+(R+r)²-2d(R+r)²/r

Last edited: Sep 26, 2015
7. Sep 26, 2015

### Svein

So, the maximum is >0. The minimum is (R+r)²(d²/r² -2d/r + 1) = (R+r)²(d/r -1)2 which is ≥0. When is the minimum = 0?