curvature

[corky0777]

I am so lost:
Let r(t)= <a cos (f(t)), a sin (f(t))>, where a is a constant and f (t) is some function with f'(t)>0 for all t. Since the path of r is a circle of radius a, the curvature should be constant with K=1/a. show this by calculating K

 

[Deleted member 4993]

I am so lost:
Let r(t)= <a cos (f(t)), a sin (f(t))>, where a is a constant and f (t) is some function with f'(t)>0 for all t. Since the path of r is a circle of radius a, the curvature should be constant with K=1/a. show this by calculating K

What is the equation for K?

 

[corky0777]

K is the curvature. There are many different formulas to use to find the curvature

 

[Deleted member 4993]

K is the curvature. There are many different formulas to use to find the curvature

Show us some that you think would be relevant - then we can discuss.

 

[corky0777]

K= abs(x' y''-y' x'')/[(x')[sup:3lqz42xd]2[/sup:3lqz42xd] + (y')[sup:3lqz42xd]2[/sup:3lqz42xd]][sup:3lqz42xd]3/2[/sup:3lqz42xd]

 

[galactus]

You have the right formula.

\(\displaystyle {\kappa}=\frac{|x'y''-y'x''|}{((x')^{2}+(y')^{2})^{\frac{3}{2}}}\)

where the prime is the differentiation w.r.t t.

It will simplify down and you should see it.

\(\displaystyle x(t)=a\cdot cos(t), \;\ y(t)=a\cdot sin(t)\)

Find the respective first and second derivatives and plug them into the formula.

 

[Deleted member 4993]

K= abs(x' y''-y' x'')/[(x')[sup:12mwsvuh]2[/sup:12mwsvuh] + (y')[sup:12mwsvuh]2[/sup:12mwsvuh]][sup:12mwsvuh]3/2[/sup:12mwsvuh]

In your case:

x = a * sin[f(t)]

x'(t) = a * f'(t) * cos[f(t)]

x"(t) = a * {f"(t) * cos[f(t)] - [f'(t)][sup:12mwsvuh]2[/sup:12mwsvuh] * sin[f(t)]}

and so on ....

just a bunch of messy algebra left

 

[galactus]

Yes, it is icky algebra. But if done correctly, it will whittle down to a nice, little, itty-bitty term.