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Curve of a circle

  • Thread starter Lee7
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  • #1
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Homework Statement



Find the curve whose curvature is 2, passes through the point (1,0) and whose tangent vector at (1,0) is [1/2 , (√3)/2 ].


The Attempt at a Solution



I know I must use the Fundamental theorem of plane curves but I don't know how to apply it correctly here. Another way to do it is the following but I don't know how to finish it:

Let us call our curve f(t). The curvature for f(t) is given by k=|f''(t)|=2. Since our curvature is a constant 2, we can think of this curve f(t) as a circle with radius r = 1/2. The equation for a circle is:

(x-a)^2 + (y-b)^2 = r^2, since the circle passes through (1,0) I can plug it in for (x,y) I get
(1-a)^2 + (0-b)^2 = (1/2)^2
(1-a)^2 + (b)^2 = 1/4

So we are given the points on the circle but we don't know what the center is. We are given the tangent vector.

So at (1,0) the tangent line is (1/2, (√3)/2). I can get the slope of this tangent line by rise over run (simple enough). Thus slope equals m = √3.

But how can I continue?
 

Answers and Replies

  • #2
arildno
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Well, let us represent your circle in parametric equations, shall we?
We have:
[tex]x(t)=x_{c}+frac{1}{2}\cos(t), y(t)=y_{c}+\frac{1}{2}\sin(t)[/tex]
Thus, we have, at t*:
[tex]1=x_{c}+frac{1}{2}\cos(t*) (1), 0=y_{c}+\frac{1}{2}\sin(t*) (2)[/tex]
Furthermore, at t*, we have that rise over run satisfies, for example:
[tex]\cot(t*)=\sqrt{3} (3)[/tex]
(1), (2) and (3) gives you three equations for your three unknowns.
 
  • #3
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Finding the unknowns is what I am having the most trouble in. Can you elaborate more please?
 
  • #4
arildno
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Well, solve (3) first, for what cos(t*) and sin(t*) must be.
Then insert those relations into (1) and (2), and solve for the coordinates of the center.
 
  • #5
pasmith
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Homework Statement



Find the curve whose curvature is 2, passes through the point (1,0) and whose tangent vector at (1,0) is [1/2 , (√3)/2 ].


The Attempt at a Solution



I know I must use the Fundamental theorem of plane curves but I don't know how to apply it correctly here. Another way to do it is the following but I don't know how to finish it:

Let us call our curve f(t). The curvature for f(t) is given by k=|f''(t)|=2. Since our curvature is a constant 2, we can think of this curve f(t) as a circle with radius r = 1/2.
Correct.

So we are given the points on the circle but we don't know what the center is. We are given the tangent vector.
The tangent is perpendicular to the radius of the circle through (1,0), so the centre of the circle must lie on the line through (1,0) perpendicular to the tangent. That, together with the knowledge that the radius of the circle is 1/2, gives you two possible locations for the centre of the circle.

But I think you are intended to start from
[tex]
\frac{d\psi}{ds} = 2
[/tex]
and use the fact that
[tex]
x'(s) = \cos\psi(s) \\
y'(s) = \sin \psi(s)
[/tex]
and you can solve for [itex]x(s)[/itex] and [itex]y(s)[/itex] by imposing the initial conditions
[tex]
x(0) = 1, \quad y(0) = 0, \quad x'(0) = \frac12, \quad y'(0) = \frac{\sqrt{3}}2
[/tex]
 
  • #6
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Pasmith - You got d/ds =2 from the radius correct?
 
  • #7
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arildno - cos is just ((√3)/2). and sin is 1/2
 
  • #8
arildno
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That's right!
Meaning, when you insert those in (1) and (2)?
 
  • #9
pasmith
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