# Thread: How to find the maximum value of the curvature?

1. ## How to find the maximum value of the curvature?

r(t) =<cost, asint, bt>

If a = 2 and b = 1, what is the maximum value of the curvature?

*I plugged everything into the curvature formula. I was left with a very complex equation. I wanted to take the derivative of this equation, but it was way too messy. There must be an alternate way to do this without taking the derivative of . Thanks in advance!

2. Originally Posted by bambamm
r(t) =<cost, asint, bt>

If a = 2 and b = 1, what is the maximum value of the curvature?

*I plugged everything into the curvature formula. I was left with a very complex equation. I wanted to take the derivative of this equation, but it was way too messy. There must be an alternate way to do this without taking the derivative of . Thanks in advance!

By definition , $\kappa(t):=\frac{|\dot{r}\times \ddot{r}|}{|\dot{r}|^3}$ , and if I didn't make a mistake, $\kappa(t)=6\sin t\cos t\,\frac{17+3(1+2\sin^2t)}{(3\cos^2t+2)^4}$ , and then $\kappa(t)=0\iff \cos t=0\,\,\,or\,\,\,\sin t=0$ , which is already pretty easy to check.
Of course, the calculations above are pretty nasty, so you may want to parametrize your curve by means of the arc length, which then yields a much easier expression for the curvature. You can read more about this here: http://homepage.smc.edu/kennedy_john...etrization.PDF

Tonio

3. I got

$\kappa(t)=\frac {(5+15sin^2t)^{\frac 12}}{(2+3cos^2t)^{\frac 32}}$

Now instead of differentiating this I simply note that the numerator is maximum for the same t as makes the denominator minimum, hence maximising the curvature.

ie for $t=\frac{\pi}{2}$

The curve that you are considering is a deformed helix (it would be a helix for a=1). So it makes sense that there can be no value of t for which the curvature can be zero.