Is the Curve C Regular for Different Values of d and r?

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Hello !

Homework Statement


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 ?

Homework Equations

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
 
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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.
 
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)])
 
Dassinia said:
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)])
So what are the maximum and minimum value of that expression?
 
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
 
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:
Dassinia said:
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
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?
 
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