The curvature of the helix r(t)equals(a cosine t )iplus(a sine t )jplusbt k (a, bgreater than or equals0) is kappaequalsstartfraction a over a squared plus b squared endfraction . what is the largest value kappa can have for a given value of b?

Explanation:

The equation for kappa ( κ) is

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

Now, the conditions to find a maximum at are:

Taking the first derivative:

This clearly will be zero when

as both are greater (or equal) than zero, this implies

The second derivative is

We dcan skip solving the equation noting that, if a=b, then

at this point, this give us only the first term

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is