Recent content by Lee7

So it must cross itself? Am I overthinking this?

Wow, this is harder than I thought.

Yup, I am meant circle. Will an number 8 satisfy it where the top is a bit more curved?

arildno - cos is just ((√3)/2). and sin is 1/2

If it were convex then it is just a sphere? But a non-convex with positive curvature is confusing me.

Can you give a hint please?

Pasmith - You got d/ds =2 from the radius correct?

Ray - You are right! So how can I draw one that is positive curvature?

Finding the unknowns is what I am having the most trouble in. Can you elaborate more please?

since it is not convex I think it can cross itself.

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...

Homework Statement How can I draw a closed plane curve with positive curvature that is not convex The Attempt at a Solution I was thinking drawing it like a banana but more curved, will that do?

I just never had a concrete example of how to do this. We do a lot of proofs instead of examples like these, but I am going to start over again. 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...

I am not getting it. If you show me how to construct it then I will have a concrete example to follow on which then I can use for future use for examples like these.

"So if you can just figure out where this circle is centered, you are done. " - So if I apply my equation (x-a)^2 + (y-b)^2 = r^2, I get (1-a)^2 + (0-b)^2 = (1/2)^2 (1-a)^2 + (b)^2 = 1/4 From this how can I continue?